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The Fourier transform of an integral
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I learned on a book that, if $f(x)$ has Fourier transform $F[f(x)] = hatf(lambda)$, and the Fourier transform of $int_-infty^x f(xi) dxi$ exists, $hatf(0)=0$, then
$$
F[int_-infty^x f(xi) dxi]=frac1ilambdahatf(lambda)
$$
I tried this:
beginalign*
F[int_-infty^x f(xi) dxi] &= int_-infty^+infty int_-infty^x f(xi) dxi e^-ilambda x dx\
&= int_-infty^+infty dxi int_xi^+infty f(xi)e^-ilambda x dx\
&= int_-infty^+infty dxi int_0^+infty f(xi)e^-ilambda ye^-ilambda xi dy\
&= hatf(lambda) int_0^+infty e^-ilambda y dy\
&= hatf(lambda) int_0^+infty[ cos(lambda y)-isin(lambda y)] dy
endalign*
Why $int_0^+infty e^-ilambda y dy = frac1ilambda$? It seems to me $lim_y to +infty e^-ilambda y$ does not exist. And neither $cos(lambda y)$ nor $sin(lambda y)$ is integrable on $[0,+infty)$. Thank you for any help!
real-analysis fourier-analysis
asked Jul 22 at 5:36
Edward Wang
598311
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I learned on a book that, if $f(x)$ has Fourier transform $F[f(x)] = hatf(lambda)$, and the Fourier transform of $int_-infty^x f(xi) dxi$ exists, $hatf(0)=0$, then
$$
F[int_-infty^x f(xi) dxi]=frac1ilambdahatf(lambda)
$$
I tried this:
beginalign*
F[int_-infty^x f(xi) dxi] &= int_-infty^+infty int_-infty^x f(xi) dxi e^-ilambda x dx\
&= int_-infty^+infty dxi int_xi^+infty f(xi)e^-ilambda x dx\
&= int_-infty^+infty dxi int_0^+infty f(xi)e^-ilambda ye^-ilambda xi dy\
&= hatf(lambda) int_0^+infty e^-ilambda y dy\
&= hatf(lambda) int_0^+infty[ cos(lambda y)-isin(lambda y)] dy
endalign*
Why $int_0^+infty e^-ilambda y dy = frac1ilambda$? It seems to me $lim_y to +infty e^-ilambda y$ does not exist. And neither $cos(lambda y)$ nor $sin(lambda y)$ is integrable on $[0,+infty)$. Thank you for any help!
real-analysis fourier-analysis
asked Jul 22 at 5:36
Edward Wang
598311
1
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
1
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I learned on a book that, if $f(x)$ has Fourier transform $F[f(x)] = hatf(lambda)$, and the Fourier transform of $int_-infty^x f(xi) dxi$ exists, $hatf(0)=0$, then
$$
F[int_-infty^x f(xi) dxi]=frac1ilambdahatf(lambda)
$$
I tried this:
beginalign*
F[int_-infty^x f(xi) dxi] &= int_-infty^+infty int_-infty^x f(xi) dxi e^-ilambda x dx\
&= int_-infty^+infty dxi int_xi^+infty f(xi)e^-ilambda x dx\
&= int_-infty^+infty dxi int_0^+infty f(xi)e^-ilambda ye^-ilambda xi dy\
&= hatf(lambda) int_0^+infty e^-ilambda y dy\
&= hatf(lambda) int_0^+infty[ cos(lambda y)-isin(lambda y)] dy
endalign*
Why $int_0^+infty e^-ilambda y dy = frac1ilambda$? It seems to me $lim_y to +infty e^-ilambda y$ does not exist. And neither $cos(lambda y)$ nor $sin(lambda y)$ is integrable on $[0,+infty)$. Thank you for any help!
real-analysis fourier-analysis
asked Jul 22 at 5:36
Edward Wang
598311
I learned on a book that, if $f(x)$ has Fourier transform $F[f(x)] = hatf(lambda)$, and the Fourier transform of $int_-infty^x f(xi) dxi$ exists, $hatf(0)=0$, then
$$
F[int_-infty^x f(xi) dxi]=frac1ilambdahatf(lambda)
$$
I tried this:
beginalign*
F[int_-infty^x f(xi) dxi] &= int_-infty^+infty int_-infty^x f(xi) dxi e^-ilambda x dx\
&= int_-infty^+infty dxi int_xi^+infty f(xi)e^-ilambda x dx\
&= int_-infty^+infty dxi int_0^+infty f(xi)e^-ilambda ye^-ilambda xi dy\
&= hatf(lambda) int_0^+infty e^-ilambda y dy\
&= hatf(lambda) int_0^+infty[ cos(lambda y)-isin(lambda y)] dy
endalign*
Why $int_0^+infty e^-ilambda y dy = frac1ilambda$? It seems to me $lim_y to +infty e^-ilambda y$ does not exist. And neither $cos(lambda y)$ nor $sin(lambda y)$ is integrable on $[0,+infty)$. Thank you for any help!
real-analysis fourier-analysis
asked Jul 22 at 5:36
Edward Wang
598311
edited Jul 22 at 5:51
asked Jul 22 at 5:36
Edward Wang
598311
asked Jul 22 at 5:36
Edward Wang
598311
asked Jul 22 at 5:36
Edward Wang
598311
598311
1
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
1
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12
add a comment |Â
1
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
1
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12
1
1
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
1
1
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12
add a comment |Â
1 Answer
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up vote
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Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain:
$$int_-infty^+inftyint_-infty^xf(xi)e^-ilambda xdxi dx = int_-infty^+inftyint_xi^+inftyf(xi)e^-ilambda xdx dxi.$$
In fact, if $|f|_1neq0$, this double integral isn't absolutely convergent, because:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)|dxi dx=int_-infty^+infty |f|_L^1((-infty,x]) dx = +infty,$$
and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$xmapsto e^x.$$
First, let's state explicitly the hypothesis: assume that $fin L^1(mathbbR)$ and $xmapsto int_-infty^x f(xi)dxi in L^1(mathbbR)$.
Here the idea: by Lebesgue dominated convergence theorem:
$$int_-infty^+inftyleft(int_-infty^xf(xi)dxiright) e^-ilambda x dx = lim_varepsilonrightarrow0^+int_-infty^+inftyleft(int_-infty^xf(xi)dxi right) e^-ilambda x e^ dx$$
and now, for $varepsilon>0$:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)e^-varepsilon x|dxi dx<+infty,$$
so we can use Fubini theorem to switch integration order.
Here the details for the case $lambdaneq 0$:
$$mathcalFleft(xmapstoint_-infty^xf(xi)dxiright)(lambda) = int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x dx = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x e^xdx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdxi dx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdx dxi \ = lim_varepsilonrightarrow 0^+ int_-infty^+infty f(xi) int_-infty^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi = (*). $$
Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+infty)$ and then in $(-infty,0)$.
We have:
$$int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_xi^+inftye^-(varepsilon+ilambda) x dx dxi \ = frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi.$$
On the other hand:
$$int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^(varepsilon-ilambda) x dx dxi \ = int_-infty^0 f(xi) int_xi^0 e^(varepsilon-ilambda) x dx dxi = int_-infty^0 f(xi) frac1-e^(varepsilon-ilambda)xivarepsilon-ilambda dxi \ = frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxi.$$
So, again using Lebesgue dominated convergence theorem, we get:
$$(*) = lim_varepsilonrightarrow 0^+ left(frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi + frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxiright) \ = frac1ilambda int_-infty^0 f(xi) dxi + frac1ilambda int_0^+infty f(xi)e^-ilambda xi dxi + frac1-ilambda int_-infty^0 f(xi) dxi - frac1-ilambda int_-infty^0 f(xi) e^-ilambdaxi dxi \ = frac1ilambda int_-infty^+infty f(xi) e^-ilambdaxi dxi = frac1ilambda mathcalF(f)(lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.
answered Jul 22 at 15:42
Bob
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Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain:
$$int_-infty^+inftyint_-infty^xf(xi)e^-ilambda xdxi dx = int_-infty^+inftyint_xi^+inftyf(xi)e^-ilambda xdx dxi.$$
In fact, if $|f|_1neq0$, this double integral isn't absolutely convergent, because:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)|dxi dx=int_-infty^+infty |f|_L^1((-infty,x]) dx = +infty,$$
and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$xmapsto e^x.$$
First, let's state explicitly the hypothesis: assume that $fin L^1(mathbbR)$ and $xmapsto int_-infty^x f(xi)dxi in L^1(mathbbR)$.
Here the idea: by Lebesgue dominated convergence theorem:
$$int_-infty^+inftyleft(int_-infty^xf(xi)dxiright) e^-ilambda x dx = lim_varepsilonrightarrow0^+int_-infty^+inftyleft(int_-infty^xf(xi)dxi right) e^-ilambda x e^ dx$$
and now, for $varepsilon>0$:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)e^-varepsilon x|dxi dx<+infty,$$
so we can use Fubini theorem to switch integration order.
Here the details for the case $lambdaneq 0$:
$$mathcalFleft(xmapstoint_-infty^xf(xi)dxiright)(lambda) = int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x dx = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x e^xdx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdxi dx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdx dxi \ = lim_varepsilonrightarrow 0^+ int_-infty^+infty f(xi) int_-infty^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi = (*). $$
Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+infty)$ and then in $(-infty,0)$.
We have:
$$int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_xi^+inftye^-(varepsilon+ilambda) x dx dxi \ = frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi.$$
On the other hand:
$$int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^(varepsilon-ilambda) x dx dxi \ = int_-infty^0 f(xi) int_xi^0 e^(varepsilon-ilambda) x dx dxi = int_-infty^0 f(xi) frac1-e^(varepsilon-ilambda)xivarepsilon-ilambda dxi \ = frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxi.$$
So, again using Lebesgue dominated convergence theorem, we get:
$$(*) = lim_varepsilonrightarrow 0^+ left(frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi + frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxiright) \ = frac1ilambda int_-infty^0 f(xi) dxi + frac1ilambda int_0^+infty f(xi)e^-ilambda xi dxi + frac1-ilambda int_-infty^0 f(xi) dxi - frac1-ilambda int_-infty^0 f(xi) e^-ilambdaxi dxi \ = frac1ilambda int_-infty^+infty f(xi) e^-ilambdaxi dxi = frac1ilambda mathcalF(f)(lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.
answered Jul 22 at 15:42
Bob
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up vote
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Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain:
$$int_-infty^+inftyint_-infty^xf(xi)e^-ilambda xdxi dx = int_-infty^+inftyint_xi^+inftyf(xi)e^-ilambda xdx dxi.$$
In fact, if $|f|_1neq0$, this double integral isn't absolutely convergent, because:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)|dxi dx=int_-infty^+infty |f|_L^1((-infty,x]) dx = +infty,$$
and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$xmapsto e^x.$$
First, let's state explicitly the hypothesis: assume that $fin L^1(mathbbR)$ and $xmapsto int_-infty^x f(xi)dxi in L^1(mathbbR)$.
Here the idea: by Lebesgue dominated convergence theorem:
$$int_-infty^+inftyleft(int_-infty^xf(xi)dxiright) e^-ilambda x dx = lim_varepsilonrightarrow0^+int_-infty^+inftyleft(int_-infty^xf(xi)dxi right) e^-ilambda x e^ dx$$
and now, for $varepsilon>0$:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)e^-varepsilon x|dxi dx<+infty,$$
so we can use Fubini theorem to switch integration order.
Here the details for the case $lambdaneq 0$:
$$mathcalFleft(xmapstoint_-infty^xf(xi)dxiright)(lambda) = int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x dx = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x e^xdx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdxi dx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdx dxi \ = lim_varepsilonrightarrow 0^+ int_-infty^+infty f(xi) int_-infty^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi = (*). $$
Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+infty)$ and then in $(-infty,0)$.
We have:
$$int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_xi^+inftye^-(varepsilon+ilambda) x dx dxi \ = frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi.$$
On the other hand:
$$int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^(varepsilon-ilambda) x dx dxi \ = int_-infty^0 f(xi) int_xi^0 e^(varepsilon-ilambda) x dx dxi = int_-infty^0 f(xi) frac1-e^(varepsilon-ilambda)xivarepsilon-ilambda dxi \ = frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxi.$$
So, again using Lebesgue dominated convergence theorem, we get:
$$(*) = lim_varepsilonrightarrow 0^+ left(frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi + frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxiright) \ = frac1ilambda int_-infty^0 f(xi) dxi + frac1ilambda int_0^+infty f(xi)e^-ilambda xi dxi + frac1-ilambda int_-infty^0 f(xi) dxi - frac1-ilambda int_-infty^0 f(xi) e^-ilambdaxi dxi \ = frac1ilambda int_-infty^+infty f(xi) e^-ilambdaxi dxi = frac1ilambda mathcalF(f)(lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.
answered Jul 22 at 15:42
Bob
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Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain:
$$int_-infty^+inftyint_-infty^xf(xi)e^-ilambda xdxi dx = int_-infty^+inftyint_xi^+inftyf(xi)e^-ilambda xdx dxi.$$
In fact, if $|f|_1neq0$, this double integral isn't absolutely convergent, because:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)|dxi dx=int_-infty^+infty |f|_L^1((-infty,x]) dx = +infty,$$
and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$xmapsto e^x.$$
First, let's state explicitly the hypothesis: assume that $fin L^1(mathbbR)$ and $xmapsto int_-infty^x f(xi)dxi in L^1(mathbbR)$.
Here the idea: by Lebesgue dominated convergence theorem:
$$int_-infty^+inftyleft(int_-infty^xf(xi)dxiright) e^-ilambda x dx = lim_varepsilonrightarrow0^+int_-infty^+inftyleft(int_-infty^xf(xi)dxi right) e^-ilambda x e^ dx$$
and now, for $varepsilon>0$:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)e^-varepsilon x|dxi dx<+infty,$$
so we can use Fubini theorem to switch integration order.
Here the details for the case $lambdaneq 0$:
$$mathcalFleft(xmapstoint_-infty^xf(xi)dxiright)(lambda) = int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x dx = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x e^xdx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdxi dx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdx dxi \ = lim_varepsilonrightarrow 0^+ int_-infty^+infty f(xi) int_-infty^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi = (*). $$
Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+infty)$ and then in $(-infty,0)$.
We have:
$$int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_xi^+inftye^-(varepsilon+ilambda) x dx dxi \ = frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi.$$
On the other hand:
$$int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^(varepsilon-ilambda) x dx dxi \ = int_-infty^0 f(xi) int_xi^0 e^(varepsilon-ilambda) x dx dxi = int_-infty^0 f(xi) frac1-e^(varepsilon-ilambda)xivarepsilon-ilambda dxi \ = frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxi.$$
So, again using Lebesgue dominated convergence theorem, we get:
$$(*) = lim_varepsilonrightarrow 0^+ left(frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi + frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxiright) \ = frac1ilambda int_-infty^0 f(xi) dxi + frac1ilambda int_0^+infty f(xi)e^-ilambda xi dxi + frac1-ilambda int_-infty^0 f(xi) dxi - frac1-ilambda int_-infty^0 f(xi) e^-ilambdaxi dxi \ = frac1ilambda int_-infty^+infty f(xi) e^-ilambdaxi dxi = frac1ilambda mathcalF(f)(lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.
answered Jul 22 at 15:42
Bob
1,477522
Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain:
$$int_-infty^+inftyint_-infty^xf(xi)e^-ilambda xdxi dx = int_-infty^+inftyint_xi^+inftyf(xi)e^-ilambda xdx dxi.$$
In fact, if $|f|_1neq0$, this double integral isn't absolutely convergent, because:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)|dxi dx=int_-infty^+infty |f|_L^1((-infty,x]) dx = +infty,$$
and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$xmapsto e^x.$$
First, let's state explicitly the hypothesis: assume that $fin L^1(mathbbR)$ and $xmapsto int_-infty^x f(xi)dxi in L^1(mathbbR)$.
Here the idea: by Lebesgue dominated convergence theorem:
$$int_-infty^+inftyleft(int_-infty^xf(xi)dxiright) e^-ilambda x dx = lim_varepsilonrightarrow0^+int_-infty^+inftyleft(int_-infty^xf(xi)dxi right) e^-ilambda x e^ dx$$
and now, for $varepsilon>0$:
$$int_-infty^+inftyint_-infty^+infty|chi_(-infty,x](xi)f(xi)e^-varepsilon x|dxi dx<+infty,$$
so we can use Fubini theorem to switch integration order.
Here the details for the case $lambdaneq 0$:
$$mathcalFleft(xmapstoint_-infty^xf(xi)dxiright)(lambda) = int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x dx = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^xf(xi)dxi e^-ilambda x e^xdx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdxi dx \ = lim_varepsilonrightarrow 0^+ int_-infty^+inftyint_-infty^+infty chi_(-infty,x](xi) f(xi) e^-ilambda x e^xdx dxi \ = lim_varepsilonrightarrow 0^+ int_-infty^+infty f(xi) int_-infty^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi = (*). $$
Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+infty)$ and then in $(-infty,0)$.
We have:
$$int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_0^+infty chi_[xi, +infty)(x) e^-(varepsilon+ilambda) x dx dxi \ = int_-infty^0 f(xi) int_0^+infty e^-(varepsilon+ilambda) x dx dxi + int_0^+infty f(xi) int_xi^+inftye^-(varepsilon+ilambda) x dx dxi \ = frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi.$$
On the other hand:
$$int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^-ilambda x e^xdx dxi \ = int_-infty^+infty f(xi) int_-infty^0 chi_[xi, +infty)(x) e^(varepsilon-ilambda) x dx dxi \ = int_-infty^0 f(xi) int_xi^0 e^(varepsilon-ilambda) x dx dxi = int_-infty^0 f(xi) frac1-e^(varepsilon-ilambda)xivarepsilon-ilambda dxi \ = frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxi.$$
So, again using Lebesgue dominated convergence theorem, we get:
$$(*) = lim_varepsilonrightarrow 0^+ left(frac1varepsilon+ilambda int_-infty^0 f(xi) dxi + frac1varepsilon+ilambda int_0^+infty f(xi)e^-ilambda xi e^-varepsilonxi dxi + frac1varepsilon-ilambda int_-infty^0 f(xi) dxi - frac1varepsilon-ilambda int_-infty^0 f(xi) e^-ilambdaxi e^varepsilonxi dxiright) \ = frac1ilambda int_-infty^0 f(xi) dxi + frac1ilambda int_0^+infty f(xi)e^-ilambda xi dxi + frac1-ilambda int_-infty^0 f(xi) dxi - frac1-ilambda int_-infty^0 f(xi) e^-ilambdaxi dxi \ = frac1ilambda int_-infty^+infty f(xi) e^-ilambdaxi dxi = frac1ilambda mathcalF(f)(lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.
answered Jul 22 at 15:42
Bob
1,477522
edited Jul 22 at 17:58
answered Jul 22 at 15:42
Bob
1,477522
answered Jul 22 at 15:42
Bob
1,477522
answered Jul 22 at 15:42
Bob
1,477522
1,477522
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1
Try taking the Fourier Transform of the derivative of your function (by applying the Fundamental Theorem of Calculus). The use the relationship of the Fourier Transform of the derivative to the Fourier Transform of the original function.
– NicNic8
Jul 22 at 6:17
1
The problem in your reasoning is that the hypothesis of Fubini's theorem don't hold. Anyway you can still get the result you're looking for via this path, using a convergence factor in order to guarantee that the hypothesis of Fubini's theorem are satisfied. If you want a detailed explanation, tag me in a comment and I will publish an answer.
– Bob
Jul 22 at 12:59
@Bob I see... Would you like to add the answer please? Thank you so much!
– Edward Wang
Jul 22 at 13:12