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Alleged trick for Fourier transform of $fracx^41+x^4$
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I was supposed to evaluate the Fourier transform of the function $$f(x) = fracx^41+x^4$$ and apart from the standard way of integrating with complex contours, which I find pretty tedious, I thought that it could be a better way.
My first idea was to use the convolution theorem treating $f$ as a product of two functions $$g(x) = x^4;;;; h(x) = frac11+x^4$$ using the fact that $$mathcalFx^n f(x) = i^n(mathcalFf)^(n)$$This seems to me the fastest way. Or maybe one could simply decompose the fraction in $$f(x) = 1+frac11+x^4$$ In either cases the problem now is on the evaluation of the Fourier transform of $h$ which, even in this case, I would prefer, if possible, to not use contour integration. My mind jumped on the idea that maybe there would be some connection with the Fourier transform of $$frac11+x^2$$ maybe by substitution of $y=x^2$. I get then $$h(y) = frac11+y^2$$ but there's not a property for the Fourier transform on such "scaling". Is there a property similar to the translation and scaling for Fourier transform that holds even for substitution like the latter?
If this is not the right way, is there some other properties or tricks that I could use to evaluate that transform?
Edit
What if I decompose the fraction in this manner
$$frac11+x^4=frac1(1+ix^2)(1-ix^2) = frac1i^2(frac1i+x^2)(frac1i-x^2) = fraci2(i+x^2)-fraci2(-i+x^2)$$
would this be a suitable way?
Using this "trick" I would get $$iover 2mathcalFleftfrac1i+x^2right = iover2sqrtpiover 2ie^sqrti$$ and a similar transform for the other factor
Just to be on the same page, I define the Fourier transform in the following way $$hatf(k) = frac1sqrt2piint_mathbbRe^-ikxf(x)dx$$
fourier-analysis fourier-transform
asked Aug 2 at 20:27
Davide Morgante
1,632220
 |Â
show 7 more comments
up vote
4
down vote
favorite
I was supposed to evaluate the Fourier transform of the function $$f(x) = fracx^41+x^4$$ and apart from the standard way of integrating with complex contours, which I find pretty tedious, I thought that it could be a better way.
My first idea was to use the convolution theorem treating $f$ as a product of two functions $$g(x) = x^4;;;; h(x) = frac11+x^4$$ using the fact that $$mathcalFx^n f(x) = i^n(mathcalFf)^(n)$$This seems to me the fastest way. Or maybe one could simply decompose the fraction in $$f(x) = 1+frac11+x^4$$ In either cases the problem now is on the evaluation of the Fourier transform of $h$ which, even in this case, I would prefer, if possible, to not use contour integration. My mind jumped on the idea that maybe there would be some connection with the Fourier transform of $$frac11+x^2$$ maybe by substitution of $y=x^2$. I get then $$h(y) = frac11+y^2$$ but there's not a property for the Fourier transform on such "scaling". Is there a property similar to the translation and scaling for Fourier transform that holds even for substitution like the latter?
If this is not the right way, is there some other properties or tricks that I could use to evaluate that transform?
Edit
What if I decompose the fraction in this manner
$$frac11+x^4=frac1(1+ix^2)(1-ix^2) = frac1i^2(frac1i+x^2)(frac1i-x^2) = fraci2(i+x^2)-fraci2(-i+x^2)$$
would this be a suitable way?
Using this "trick" I would get $$iover 2mathcalFleftfrac1i+x^2right = iover2sqrtpiover 2ie^sqrti$$ and a similar transform for the other factor
Just to be on the same page, I define the Fourier transform in the following way $$hatf(k) = frac1sqrt2piint_mathbbRe^-ikxf(x)dx$$
fourier-analysis fourier-transform
asked Aug 2 at 20:27
Davide Morgante
1,632220
The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
1
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
1
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48
 |Â
show 7 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was supposed to evaluate the Fourier transform of the function $$f(x) = fracx^41+x^4$$ and apart from the standard way of integrating with complex contours, which I find pretty tedious, I thought that it could be a better way.
My first idea was to use the convolution theorem treating $f$ as a product of two functions $$g(x) = x^4;;;; h(x) = frac11+x^4$$ using the fact that $$mathcalFx^n f(x) = i^n(mathcalFf)^(n)$$This seems to me the fastest way. Or maybe one could simply decompose the fraction in $$f(x) = 1+frac11+x^4$$ In either cases the problem now is on the evaluation of the Fourier transform of $h$ which, even in this case, I would prefer, if possible, to not use contour integration. My mind jumped on the idea that maybe there would be some connection with the Fourier transform of $$frac11+x^2$$ maybe by substitution of $y=x^2$. I get then $$h(y) = frac11+y^2$$ but there's not a property for the Fourier transform on such "scaling". Is there a property similar to the translation and scaling for Fourier transform that holds even for substitution like the latter?
If this is not the right way, is there some other properties or tricks that I could use to evaluate that transform?
Edit
What if I decompose the fraction in this manner
$$frac11+x^4=frac1(1+ix^2)(1-ix^2) = frac1i^2(frac1i+x^2)(frac1i-x^2) = fraci2(i+x^2)-fraci2(-i+x^2)$$
would this be a suitable way?
Using this "trick" I would get $$iover 2mathcalFleftfrac1i+x^2right = iover2sqrtpiover 2ie^sqrti$$ and a similar transform for the other factor
Just to be on the same page, I define the Fourier transform in the following way $$hatf(k) = frac1sqrt2piint_mathbbRe^-ikxf(x)dx$$
fourier-analysis fourier-transform
asked Aug 2 at 20:27
Davide Morgante
1,632220
I was supposed to evaluate the Fourier transform of the function $$f(x) = fracx^41+x^4$$ and apart from the standard way of integrating with complex contours, which I find pretty tedious, I thought that it could be a better way.
My first idea was to use the convolution theorem treating $f$ as a product of two functions $$g(x) = x^4;;;; h(x) = frac11+x^4$$ using the fact that $$mathcalFx^n f(x) = i^n(mathcalFf)^(n)$$This seems to me the fastest way. Or maybe one could simply decompose the fraction in $$f(x) = 1+frac11+x^4$$ In either cases the problem now is on the evaluation of the Fourier transform of $h$ which, even in this case, I would prefer, if possible, to not use contour integration. My mind jumped on the idea that maybe there would be some connection with the Fourier transform of $$frac11+x^2$$ maybe by substitution of $y=x^2$. I get then $$h(y) = frac11+y^2$$ but there's not a property for the Fourier transform on such "scaling". Is there a property similar to the translation and scaling for Fourier transform that holds even for substitution like the latter?
If this is not the right way, is there some other properties or tricks that I could use to evaluate that transform?
Edit
What if I decompose the fraction in this manner
$$frac11+x^4=frac1(1+ix^2)(1-ix^2) = frac1i^2(frac1i+x^2)(frac1i-x^2) = fraci2(i+x^2)-fraci2(-i+x^2)$$
would this be a suitable way?
Using this "trick" I would get $$iover 2mathcalFleftfrac1i+x^2right = iover2sqrtpiover 2ie^sqrti$$ and a similar transform for the other factor
Just to be on the same page, I define the Fourier transform in the following way $$hatf(k) = frac1sqrt2piint_mathbbRe^-ikxf(x)dx$$
fourier-analysis fourier-transform
asked Aug 2 at 20:27
Davide Morgante
1,632220
edited Aug 3 at 9:20
asked Aug 2 at 20:27
Davide Morgante
1,632220
asked Aug 2 at 20:27
Davide Morgante
1,632220
asked Aug 2 at 20:27
Davide Morgante
1,632220
1,632220
The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
1
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
1
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48
 |Â
show 7 more comments
The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
1
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
1
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48
The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
1
1
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
1
1
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48
 |Â
show 7 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For the sake of simplicity, let $tinmathbbCsetminusmathbbR_geq 0$ so that the polynomial $X^2-tinmathbbC[X]$ does not have a real root, but you can use the Cauchy principal values when $tinmathbbR_geq 0$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=dfrac1z^2-ttext for all zinmathbbCsetminusbig-sqrtt,+sqrttbig,,$$
where $sqrtt$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=frac1sqrt2pi,int_-infty^+infty,exp(textikx),f_t(x),textdxtext for kinmathbbR,.$$
For simplicity, I shall write $c:=dfrac1sqrt2pi$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $kgeq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]cupBigR,exp(textitheta),Big,.$$
Observe that
$$2pitexti,textRes_z=+sqrttbig(f_t(z)big)=lim_Rtoinfty,oint_C_R,fracexp(textikz)z^2-t,textdz=frac1c,F_t(k),.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=fraccpitextisqrtt,expBig(texti|k|sqrttBig)text for each kinmathbbR,.$$
For example, $sqrt+texti=frac+1+textisqrt2$ and $sqrt-texti=frac-1+textisqrt2$. This means
$$F_stexti(k)=fraccpitextileft(fracs+textisqrt2right),expBiggl(texti,left(fracs+textisqrt2right),|k|Biggr)=cpileft(frac1+stextisqrt2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),,$$
where $sin-1,+1$. Now, let $varphi(z):=dfrac1z^4+1$. Then, as you found out,
$$varphi(z)=-fractexti2,sum_sin-1,+1,fracsz^2-stexti,.$$
Hence, the Fourier transform $hatvarphi$ of $varphi$ is given by
$$hatvarphi(k)=-fractexti2,sum_sin-1,+1,s,F_stexti(k),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,sum_sin-1,+1,left(frac1-stexti2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Biggr),.$$
Now, if you want to determine the Fourier transform $hatPhi$ of $Phi$, where $$Phi(z):=dfracz^4z^4+1=1-dfrac1z^4+1,,$$ you will get a distribution. Note that the Fourier transformation $hatkappa_epsilon$ of the constant function $kappa_epsilonequiv epsilon$ is $$hatkappa_epsilon(k)=2cpi,epsilon,delta(k),,$$
where $delta$ is the Dirac delta distribution.
Thus, using $Phi=kappa_1-phi$, we conclude that
$$hatPhi(k)=2cpi,left(delta(k)-frac12sqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Bigg)right),.$$
Nonetheless, be warned that $hatPhi$ is not a function.
answered Aug 2 at 22:41
Batominovski
22.7k22776
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
add a comment |Â
up vote
1
down vote
Note that
$$
fracx^4x^4+1
=
1-underbracefrac1x^4+1_f(x) ,
$$
so there is a simpler way to reduce to the function $f$
isolated above. Now it is a matter of taste to go.
Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $tge 0$.
(Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.)
$$
beginaligned
hat f(t)
&=
frac1sqrt 2piint_Bbb R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2piint_-R^R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2pi
int_C
frace^itz1+z^4; dz
-
lim_Rto infty
frac1sqrt 2pi
underbrace
int_S
frace^itz1+z^4; dz_to 0
\
&qquadqquadtextwhere $C=Jsqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,
\
&=
lim_Rto infty
frac1sqrt 2pi
cdot 2pi i
sum_atext pole inside C
operatornameRes_z=a
frace^itz1+z^4
\
&=
frac1sqrt 2pi
cdot 2pi i
sum_ainz_1,z_2=(pm1+i)/sqrt 2
operatornameRes_z=a
frace^itzz^4+1
\
&qquadqquadtextand we can isolate the residues in $z_1,z_2=frac 1sqrt 2(pm1+i)$,
\
&qquadqquadtextby factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
\
&=
frac1sqrt 2pi
cdot 2pi i
left(
exp(itz_1)cdotfrac 1(z_1-z_2)(z_1-z_3)(z_1-z_4)
+
exp(itz_2)cdotfrac 1(z_2-z_1)(z_2-z_3)(z_2-z_4)
right)
\
&qquadqquadtextnow we draw the unit circle and the four roots $z_1,2,3,4$ on it to simply see the relations
\
&qquadqquadtext$z_1-z_2=sqrt2$, $z_1-z_3=sqrt2(1+i)$, $z_1-z_4=isqrt 2$, and
\
&qquadqquadtext$z_2-z_1=-sqrt2$, $z_2-z_4=-sqrt2(1-i)$, $z_2-z_3=isqrt 2$,
\
&=
frac1sqrt 2pi
cdot 2pi i
cdot
sum_pm
frac 1sqrt 2cdot isqrt 2cdot (1pm i)sqrt 2
expleft(frac tsqrt 2(ipm 1)right) .
endaligned
$$
A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...
In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).
answered Aug 3 at 0:22
dan_fulea
3,9571211
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the sake of simplicity, let $tinmathbbCsetminusmathbbR_geq 0$ so that the polynomial $X^2-tinmathbbC[X]$ does not have a real root, but you can use the Cauchy principal values when $tinmathbbR_geq 0$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=dfrac1z^2-ttext for all zinmathbbCsetminusbig-sqrtt,+sqrttbig,,$$
where $sqrtt$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=frac1sqrt2pi,int_-infty^+infty,exp(textikx),f_t(x),textdxtext for kinmathbbR,.$$
For simplicity, I shall write $c:=dfrac1sqrt2pi$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $kgeq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]cupBigR,exp(textitheta),Big,.$$
Observe that
$$2pitexti,textRes_z=+sqrttbig(f_t(z)big)=lim_Rtoinfty,oint_C_R,fracexp(textikz)z^2-t,textdz=frac1c,F_t(k),.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=fraccpitextisqrtt,expBig(texti|k|sqrttBig)text for each kinmathbbR,.$$
For example, $sqrt+texti=frac+1+textisqrt2$ and $sqrt-texti=frac-1+textisqrt2$. This means
$$F_stexti(k)=fraccpitextileft(fracs+textisqrt2right),expBiggl(texti,left(fracs+textisqrt2right),|k|Biggr)=cpileft(frac1+stextisqrt2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),,$$
where $sin-1,+1$. Now, let $varphi(z):=dfrac1z^4+1$. Then, as you found out,
$$varphi(z)=-fractexti2,sum_sin-1,+1,fracsz^2-stexti,.$$
Hence, the Fourier transform $hatvarphi$ of $varphi$ is given by
$$hatvarphi(k)=-fractexti2,sum_sin-1,+1,s,F_stexti(k),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,sum_sin-1,+1,left(frac1-stexti2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Biggr),.$$
Now, if you want to determine the Fourier transform $hatPhi$ of $Phi$, where $$Phi(z):=dfracz^4z^4+1=1-dfrac1z^4+1,,$$ you will get a distribution. Note that the Fourier transformation $hatkappa_epsilon$ of the constant function $kappa_epsilonequiv epsilon$ is $$hatkappa_epsilon(k)=2cpi,epsilon,delta(k),,$$
where $delta$ is the Dirac delta distribution.
Thus, using $Phi=kappa_1-phi$, we conclude that
$$hatPhi(k)=2cpi,left(delta(k)-frac12sqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Bigg)right),.$$
Nonetheless, be warned that $hatPhi$ is not a function.
answered Aug 2 at 22:41
Batominovski
22.7k22776
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
add a comment |Â
up vote
2
down vote
accepted
For the sake of simplicity, let $tinmathbbCsetminusmathbbR_geq 0$ so that the polynomial $X^2-tinmathbbC[X]$ does not have a real root, but you can use the Cauchy principal values when $tinmathbbR_geq 0$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=dfrac1z^2-ttext for all zinmathbbCsetminusbig-sqrtt,+sqrttbig,,$$
where $sqrtt$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=frac1sqrt2pi,int_-infty^+infty,exp(textikx),f_t(x),textdxtext for kinmathbbR,.$$
For simplicity, I shall write $c:=dfrac1sqrt2pi$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $kgeq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]cupBigR,exp(textitheta),Big,.$$
Observe that
$$2pitexti,textRes_z=+sqrttbig(f_t(z)big)=lim_Rtoinfty,oint_C_R,fracexp(textikz)z^2-t,textdz=frac1c,F_t(k),.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=fraccpitextisqrtt,expBig(texti|k|sqrttBig)text for each kinmathbbR,.$$
For example, $sqrt+texti=frac+1+textisqrt2$ and $sqrt-texti=frac-1+textisqrt2$. This means
$$F_stexti(k)=fraccpitextileft(fracs+textisqrt2right),expBiggl(texti,left(fracs+textisqrt2right),|k|Biggr)=cpileft(frac1+stextisqrt2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),,$$
where $sin-1,+1$. Now, let $varphi(z):=dfrac1z^4+1$. Then, as you found out,
$$varphi(z)=-fractexti2,sum_sin-1,+1,fracsz^2-stexti,.$$
Hence, the Fourier transform $hatvarphi$ of $varphi$ is given by
$$hatvarphi(k)=-fractexti2,sum_sin-1,+1,s,F_stexti(k),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,sum_sin-1,+1,left(frac1-stexti2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Biggr),.$$
Now, if you want to determine the Fourier transform $hatPhi$ of $Phi$, where $$Phi(z):=dfracz^4z^4+1=1-dfrac1z^4+1,,$$ you will get a distribution. Note that the Fourier transformation $hatkappa_epsilon$ of the constant function $kappa_epsilonequiv epsilon$ is $$hatkappa_epsilon(k)=2cpi,epsilon,delta(k),,$$
where $delta$ is the Dirac delta distribution.
Thus, using $Phi=kappa_1-phi$, we conclude that
$$hatPhi(k)=2cpi,left(delta(k)-frac12sqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Bigg)right),.$$
Nonetheless, be warned that $hatPhi$ is not a function.
answered Aug 2 at 22:41
Batominovski
22.7k22776
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the sake of simplicity, let $tinmathbbCsetminusmathbbR_geq 0$ so that the polynomial $X^2-tinmathbbC[X]$ does not have a real root, but you can use the Cauchy principal values when $tinmathbbR_geq 0$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=dfrac1z^2-ttext for all zinmathbbCsetminusbig-sqrtt,+sqrttbig,,$$
where $sqrtt$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=frac1sqrt2pi,int_-infty^+infty,exp(textikx),f_t(x),textdxtext for kinmathbbR,.$$
For simplicity, I shall write $c:=dfrac1sqrt2pi$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $kgeq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]cupBigR,exp(textitheta),Big,.$$
Observe that
$$2pitexti,textRes_z=+sqrttbig(f_t(z)big)=lim_Rtoinfty,oint_C_R,fracexp(textikz)z^2-t,textdz=frac1c,F_t(k),.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=fraccpitextisqrtt,expBig(texti|k|sqrttBig)text for each kinmathbbR,.$$
For example, $sqrt+texti=frac+1+textisqrt2$ and $sqrt-texti=frac-1+textisqrt2$. This means
$$F_stexti(k)=fraccpitextileft(fracs+textisqrt2right),expBiggl(texti,left(fracs+textisqrt2right),|k|Biggr)=cpileft(frac1+stextisqrt2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),,$$
where $sin-1,+1$. Now, let $varphi(z):=dfrac1z^4+1$. Then, as you found out,
$$varphi(z)=-fractexti2,sum_sin-1,+1,fracsz^2-stexti,.$$
Hence, the Fourier transform $hatvarphi$ of $varphi$ is given by
$$hatvarphi(k)=-fractexti2,sum_sin-1,+1,s,F_stexti(k),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,sum_sin-1,+1,left(frac1-stexti2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Biggr),.$$
Now, if you want to determine the Fourier transform $hatPhi$ of $Phi$, where $$Phi(z):=dfracz^4z^4+1=1-dfrac1z^4+1,,$$ you will get a distribution. Note that the Fourier transformation $hatkappa_epsilon$ of the constant function $kappa_epsilonequiv epsilon$ is $$hatkappa_epsilon(k)=2cpi,epsilon,delta(k),,$$
where $delta$ is the Dirac delta distribution.
Thus, using $Phi=kappa_1-phi$, we conclude that
$$hatPhi(k)=2cpi,left(delta(k)-frac12sqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Bigg)right),.$$
Nonetheless, be warned that $hatPhi$ is not a function.
answered Aug 2 at 22:41
Batominovski
22.7k22776
For the sake of simplicity, let $tinmathbbCsetminusmathbbR_geq 0$ so that the polynomial $X^2-tinmathbbC[X]$ does not have a real root, but you can use the Cauchy principal values when $tinmathbbR_geq 0$ to deal with the job. This job is to compute the Fourier transform $F_t$ of the function $f_t$, where $$f_t(z):=dfrac1z^2-ttext for all zinmathbbCsetminusbig-sqrtt,+sqrttbig,,$$
where $sqrtt$ is so chosen that the imaginary part is positive (this is possible as $X^2-t$ has no real roots).
That is,
$$F_t(k)=frac1sqrt2pi,int_-infty^+infty,exp(textikx),f_t(x),textdxtext for kinmathbbR,.$$
For simplicity, I shall write $c:=dfrac1sqrt2pi$ (since $c$ may be defined to be a different constant for a different user, I shall leave this constant as just $c$).
Without loss of generality, suppose that $kgeq 0$. For $R>0$, let $C_R$ to be the positively oriented contour given by
$$[-R,+R]cupBigR,exp(textitheta),Big,.$$
Observe that
$$2pitexti,textRes_z=+sqrttbig(f_t(z)big)=lim_Rtoinfty,oint_C_R,fracexp(textikz)z^2-t,textdz=frac1c,F_t(k),.$$
Hence, taking the sign of $k$ into account, we get
$$F_t(k)=fraccpitextisqrtt,expBig(texti|k|sqrttBig)text for each kinmathbbR,.$$
For example, $sqrt+texti=frac+1+textisqrt2$ and $sqrt-texti=frac-1+textisqrt2$. This means
$$F_stexti(k)=fraccpitextileft(fracs+textisqrt2right),expBiggl(texti,left(fracs+textisqrt2right),|k|Biggr)=cpileft(frac1+stextisqrt2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),,$$
where $sin-1,+1$. Now, let $varphi(z):=dfrac1z^4+1$. Then, as you found out,
$$varphi(z)=-fractexti2,sum_sin-1,+1,fracsz^2-stexti,.$$
Hence, the Fourier transform $hatvarphi$ of $varphi$ is given by
$$hatvarphi(k)=-fractexti2,sum_sin-1,+1,s,F_stexti(k),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,sum_sin-1,+1,left(frac1-stexti2right),expleft(-fracksqrt2right),expleft(fracksqrt2right),.$$
That is,
$$hatvarphi(k)=fraccpisqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Biggr),.$$
Now, if you want to determine the Fourier transform $hatPhi$ of $Phi$, where $$Phi(z):=dfracz^4z^4+1=1-dfrac1z^4+1,,$$ you will get a distribution. Note that the Fourier transformation $hatkappa_epsilon$ of the constant function $kappa_epsilonequiv epsilon$ is $$hatkappa_epsilon(k)=2cpi,epsilon,delta(k),,$$
where $delta$ is the Dirac delta distribution.
Thus, using $Phi=kappa_1-phi$, we conclude that
$$hatPhi(k)=2cpi,left(delta(k)-frac12sqrt2,expleft(-fracksqrt2right),Biggl(cosleft(fracksqrt2right)+sinleft(fracksqrt2right)Bigg)right),.$$
Nonetheless, be warned that $hatPhi$ is not a function.
answered Aug 2 at 22:41
Batominovski
22.7k22776
edited Aug 2 at 22:47
answered Aug 2 at 22:41
Batominovski
22.7k22776
answered Aug 2 at 22:41
Batominovski
22.7k22776
answered Aug 2 at 22:41
Batominovski
22.7k22776
22.7k22776
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
add a comment |Â
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Yeah! I very much like your approach, it's ingenious. Tomorrow i'll follow with some pencil an paper, thank you so much
– Davide Morgante
Aug 2 at 22:46
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
Doesn't that $sin$ in the last expression lack a factor $i$?
– Davide Morgante
Aug 3 at 8:43
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
No, it is canceled out as $$sin(theta)=fracexp(+textitheta)-exp(-textitheta)2colorredtexti,.$$
– Batominovski
Aug 3 at 8:44
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
Oh yes, that's true! Thank you very much again
– Davide Morgante
Aug 3 at 8:45
add a comment |Â
up vote
1
down vote
Note that
$$
fracx^4x^4+1
=
1-underbracefrac1x^4+1_f(x) ,
$$
so there is a simpler way to reduce to the function $f$
isolated above. Now it is a matter of taste to go.
Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $tge 0$.
(Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.)
$$
beginaligned
hat f(t)
&=
frac1sqrt 2piint_Bbb R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2piint_-R^R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2pi
int_C
frace^itz1+z^4; dz
-
lim_Rto infty
frac1sqrt 2pi
underbrace
int_S
frace^itz1+z^4; dz_to 0
\
&qquadqquadtextwhere $C=Jsqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,
\
&=
lim_Rto infty
frac1sqrt 2pi
cdot 2pi i
sum_atext pole inside C
operatornameRes_z=a
frace^itz1+z^4
\
&=
frac1sqrt 2pi
cdot 2pi i
sum_ainz_1,z_2=(pm1+i)/sqrt 2
operatornameRes_z=a
frace^itzz^4+1
\
&qquadqquadtextand we can isolate the residues in $z_1,z_2=frac 1sqrt 2(pm1+i)$,
\
&qquadqquadtextby factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
\
&=
frac1sqrt 2pi
cdot 2pi i
left(
exp(itz_1)cdotfrac 1(z_1-z_2)(z_1-z_3)(z_1-z_4)
+
exp(itz_2)cdotfrac 1(z_2-z_1)(z_2-z_3)(z_2-z_4)
right)
\
&qquadqquadtextnow we draw the unit circle and the four roots $z_1,2,3,4$ on it to simply see the relations
\
&qquadqquadtext$z_1-z_2=sqrt2$, $z_1-z_3=sqrt2(1+i)$, $z_1-z_4=isqrt 2$, and
\
&qquadqquadtext$z_2-z_1=-sqrt2$, $z_2-z_4=-sqrt2(1-i)$, $z_2-z_3=isqrt 2$,
\
&=
frac1sqrt 2pi
cdot 2pi i
cdot
sum_pm
frac 1sqrt 2cdot isqrt 2cdot (1pm i)sqrt 2
expleft(frac tsqrt 2(ipm 1)right) .
endaligned
$$
A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...
In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).
answered Aug 3 at 0:22
dan_fulea
3,9571211
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
add a comment |Â
up vote
1
down vote
Note that
$$
fracx^4x^4+1
=
1-underbracefrac1x^4+1_f(x) ,
$$
so there is a simpler way to reduce to the function $f$
isolated above. Now it is a matter of taste to go.
Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $tge 0$.
(Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.)
$$
beginaligned
hat f(t)
&=
frac1sqrt 2piint_Bbb R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2piint_-R^R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2pi
int_C
frace^itz1+z^4; dz
-
lim_Rto infty
frac1sqrt 2pi
underbrace
int_S
frace^itz1+z^4; dz_to 0
\
&qquadqquadtextwhere $C=Jsqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,
\
&=
lim_Rto infty
frac1sqrt 2pi
cdot 2pi i
sum_atext pole inside C
operatornameRes_z=a
frace^itz1+z^4
\
&=
frac1sqrt 2pi
cdot 2pi i
sum_ainz_1,z_2=(pm1+i)/sqrt 2
operatornameRes_z=a
frace^itzz^4+1
\
&qquadqquadtextand we can isolate the residues in $z_1,z_2=frac 1sqrt 2(pm1+i)$,
\
&qquadqquadtextby factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
\
&=
frac1sqrt 2pi
cdot 2pi i
left(
exp(itz_1)cdotfrac 1(z_1-z_2)(z_1-z_3)(z_1-z_4)
+
exp(itz_2)cdotfrac 1(z_2-z_1)(z_2-z_3)(z_2-z_4)
right)
\
&qquadqquadtextnow we draw the unit circle and the four roots $z_1,2,3,4$ on it to simply see the relations
\
&qquadqquadtext$z_1-z_2=sqrt2$, $z_1-z_3=sqrt2(1+i)$, $z_1-z_4=isqrt 2$, and
\
&qquadqquadtext$z_2-z_1=-sqrt2$, $z_2-z_4=-sqrt2(1-i)$, $z_2-z_3=isqrt 2$,
\
&=
frac1sqrt 2pi
cdot 2pi i
cdot
sum_pm
frac 1sqrt 2cdot isqrt 2cdot (1pm i)sqrt 2
expleft(frac tsqrt 2(ipm 1)right) .
endaligned
$$
A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...
In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).
answered Aug 3 at 0:22
dan_fulea
3,9571211
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that
$$
fracx^4x^4+1
=
1-underbracefrac1x^4+1_f(x) ,
$$
so there is a simpler way to reduce to the function $f$
isolated above. Now it is a matter of taste to go.
Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $tge 0$.
(Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.)
$$
beginaligned
hat f(t)
&=
frac1sqrt 2piint_Bbb R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2piint_-R^R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2pi
int_C
frace^itz1+z^4; dz
-
lim_Rto infty
frac1sqrt 2pi
underbrace
int_S
frace^itz1+z^4; dz_to 0
\
&qquadqquadtextwhere $C=Jsqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,
\
&=
lim_Rto infty
frac1sqrt 2pi
cdot 2pi i
sum_atext pole inside C
operatornameRes_z=a
frace^itz1+z^4
\
&=
frac1sqrt 2pi
cdot 2pi i
sum_ainz_1,z_2=(pm1+i)/sqrt 2
operatornameRes_z=a
frace^itzz^4+1
\
&qquadqquadtextand we can isolate the residues in $z_1,z_2=frac 1sqrt 2(pm1+i)$,
\
&qquadqquadtextby factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
\
&=
frac1sqrt 2pi
cdot 2pi i
left(
exp(itz_1)cdotfrac 1(z_1-z_2)(z_1-z_3)(z_1-z_4)
+
exp(itz_2)cdotfrac 1(z_2-z_1)(z_2-z_3)(z_2-z_4)
right)
\
&qquadqquadtextnow we draw the unit circle and the four roots $z_1,2,3,4$ on it to simply see the relations
\
&qquadqquadtext$z_1-z_2=sqrt2$, $z_1-z_3=sqrt2(1+i)$, $z_1-z_4=isqrt 2$, and
\
&qquadqquadtext$z_2-z_1=-sqrt2$, $z_2-z_4=-sqrt2(1-i)$, $z_2-z_3=isqrt 2$,
\
&=
frac1sqrt 2pi
cdot 2pi i
cdot
sum_pm
frac 1sqrt 2cdot isqrt 2cdot (1pm i)sqrt 2
expleft(frac tsqrt 2(ipm 1)right) .
endaligned
$$
A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...
In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).
answered Aug 3 at 0:22
dan_fulea
3,9571211
Note that
$$
fracx^4x^4+1
=
1-underbracefrac1x^4+1_f(x) ,
$$
so there is a simpler way to reduce to the function $f$
isolated above. Now it is a matter of taste to go.
Personally, i would opt for the residue theorem, so that the integral is painless solved in some lines. We may assume first $tge 0$.
(Else we substitute $x$ by $-x$ in the integral below, to reduce ourselves to $|t|$.)
$$
beginaligned
hat f(t)
&=
frac1sqrt 2piint_Bbb R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2piint_-R^R
frace^itx1+x^4; dx
\
&=
lim_Rto infty
frac1sqrt 2pi
int_C
frace^itz1+z^4; dz
-
lim_Rto infty
frac1sqrt 2pi
underbrace
int_S
frace^itz1+z^4; dz_to 0
\
&qquadqquadtextwhere $C=Jsqcup S$ is the contour with $J=[-R,R]$, and $S$ is the semicircle with diameter $J$ in upper half plane,
\
&=
lim_Rto infty
frac1sqrt 2pi
cdot 2pi i
sum_atext pole inside C
operatornameRes_z=a
frace^itz1+z^4
\
&=
frac1sqrt 2pi
cdot 2pi i
sum_ainz_1,z_2=(pm1+i)/sqrt 2
operatornameRes_z=a
frace^itzz^4+1
\
&qquadqquadtextand we can isolate the residues in $z_1,z_2=frac 1sqrt 2(pm1+i)$,
\
&qquadqquadtextby factoring $z^4+1=(z-z_1)(z-z_2)(z-z_3)(z-z_4)$
\
&=
frac1sqrt 2pi
cdot 2pi i
left(
exp(itz_1)cdotfrac 1(z_1-z_2)(z_1-z_3)(z_1-z_4)
+
exp(itz_2)cdotfrac 1(z_2-z_1)(z_2-z_3)(z_2-z_4)
right)
\
&qquadqquadtextnow we draw the unit circle and the four roots $z_1,2,3,4$ on it to simply see the relations
\
&qquadqquadtext$z_1-z_2=sqrt2$, $z_1-z_3=sqrt2(1+i)$, $z_1-z_4=isqrt 2$, and
\
&qquadqquadtext$z_2-z_1=-sqrt2$, $z_2-z_4=-sqrt2(1-i)$, $z_2-z_3=isqrt 2$,
\
&=
frac1sqrt 2pi
cdot 2pi i
cdot
sum_pm
frac 1sqrt 2cdot isqrt 2cdot (1pm i)sqrt 2
expleft(frac tsqrt 2(ipm 1)right) .
endaligned
$$
A Galois theorist would stop here. I also have to stop here, since there is an other answer along the same lines. What i wanted to say...
In the original post we have a separation, instead of working with $1/(z^4+1)$ there is a split and we have to work with the two fractions $1/(z^2pm i)$ (up to a multiplicative constant). Each one has one root in the upper half plane, so this splitting does not sensibly simplify algebra (if we finally again pass to residues).
answered Aug 3 at 0:22
dan_fulea
3,9571211
answered Aug 3 at 0:22
dan_fulea
3,9571211
answered Aug 3 at 0:22
dan_fulea
3,9571211
answered Aug 3 at 0:22
dan_fulea
3,9571211
3,9571211
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
add a comment |Â
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
Thank you very much! I know that splitting the fractions is not going to simplify the algebra if we're going on the residue route. My goal was to go to a known Fourier transform! As I said, for me at least, this is the transform of $$frac1x^2+a^2$$ Your answer is still really appreciated, all resolutions matter to me!
– Davide Morgante
Aug 3 at 8:12
add a comment |Â
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The Fourier transform of your function involves a Delta-Dirac function, and some step functions, I doubt very much the utility of standard tricks to facilitate its computation.
– uniquesolution
Aug 2 at 20:38
1
Why would you like to avoid the complex analysis?
– Botond
Aug 2 at 20:41
@uniquesolution Would that decomposition work? Sure at the end I have to take a fourth derivative of some quite horrible function but leaving constants aside, it's nothing so terrible
– Davide Morgante
Aug 2 at 20:41
@Botond I love complex analysis but knowing my professor I know for certain that that transform can be done with some tricks!
– Davide Morgante
Aug 2 at 20:43
1
You can avoid the differentiation with noting that $fracx^4x^4+1=1-frac11+x^4$, and the FT of $1$ is just the dirac delta (maybe a scaled dirac delta, it depends on your definition), but the $frac11+x^4$ is still left to compute.
– Botond
Aug 2 at 20:48