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How to integrate $int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega$
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I would like to evaluate the integral:
$$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega,$$
where $i$ is the imaginary number and $ainmathbbR$ and $ninmathbbN$ are constants.
I have tried to perform a binomial expansion of the denominator where the even indices will represent the real part and the odd indices the imaginary part. From that I can calculate the real and imaginary parts of the fraction but then I have also to calculate the squared term which makes the integrand very messy and I cannot see anyway of evaluating the integral.
I have also tried to use partial fractions, such as,
$$left|frac1(a+iomega)^n+1right|^2 = frac1(a+iomega)^n+1(a-iomega)^n+1 = fracA(a+iomega)^1+fracBomega+C(a+iomega)^2+...$$
but again things start to get super-messy and very difficult to solve.
Question:
Is there are any trick to solve this integral?
Or, should I still follow one of the two methods I have described?
complex-analysis complex-integration partial-fractions
asked Jul 25 at 9:36
DOMiguel
16411
add a comment |Â
up vote
2
down vote
favorite
I would like to evaluate the integral:
$$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega,$$
where $i$ is the imaginary number and $ainmathbbR$ and $ninmathbbN$ are constants.
I have tried to perform a binomial expansion of the denominator where the even indices will represent the real part and the odd indices the imaginary part. From that I can calculate the real and imaginary parts of the fraction but then I have also to calculate the squared term which makes the integrand very messy and I cannot see anyway of evaluating the integral.
I have also tried to use partial fractions, such as,
$$left|frac1(a+iomega)^n+1right|^2 = frac1(a+iomega)^n+1(a-iomega)^n+1 = fracA(a+iomega)^1+fracBomega+C(a+iomega)^2+...$$
but again things start to get super-messy and very difficult to solve.
Question:
Is there are any trick to solve this integral?
Or, should I still follow one of the two methods I have described?
complex-analysis complex-integration partial-fractions
asked Jul 25 at 9:36
DOMiguel
16411
2
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
1
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to evaluate the integral:
$$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega,$$
where $i$ is the imaginary number and $ainmathbbR$ and $ninmathbbN$ are constants.
I have tried to perform a binomial expansion of the denominator where the even indices will represent the real part and the odd indices the imaginary part. From that I can calculate the real and imaginary parts of the fraction but then I have also to calculate the squared term which makes the integrand very messy and I cannot see anyway of evaluating the integral.
I have also tried to use partial fractions, such as,
$$left|frac1(a+iomega)^n+1right|^2 = frac1(a+iomega)^n+1(a-iomega)^n+1 = fracA(a+iomega)^1+fracBomega+C(a+iomega)^2+...$$
but again things start to get super-messy and very difficult to solve.
Question:
Is there are any trick to solve this integral?
Or, should I still follow one of the two methods I have described?
complex-analysis complex-integration partial-fractions
asked Jul 25 at 9:36
DOMiguel
16411
I would like to evaluate the integral:
$$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega,$$
where $i$ is the imaginary number and $ainmathbbR$ and $ninmathbbN$ are constants.
I have tried to perform a binomial expansion of the denominator where the even indices will represent the real part and the odd indices the imaginary part. From that I can calculate the real and imaginary parts of the fraction but then I have also to calculate the squared term which makes the integrand very messy and I cannot see anyway of evaluating the integral.
I have also tried to use partial fractions, such as,
$$left|frac1(a+iomega)^n+1right|^2 = frac1(a+iomega)^n+1(a-iomega)^n+1 = fracA(a+iomega)^1+fracBomega+C(a+iomega)^2+...$$
but again things start to get super-messy and very difficult to solve.
Question:
Is there are any trick to solve this integral?
Or, should I still follow one of the two methods I have described?
complex-analysis complex-integration partial-fractions
asked Jul 25 at 9:36
DOMiguel
16411
edited Jul 25 at 11:23
asked Jul 25 at 9:36
DOMiguel
16411
asked Jul 25 at 9:36
DOMiguel
16411
asked Jul 25 at 9:36
DOMiguel
16411
16411
2
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
1
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14
add a comment |Â
2
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
1
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14
2
2
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
1
1
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The Fourier Transform of $f(t)=e^-aomegau(t)$ is $F(omega)=dfrac1a+jomega$. Also we know that $$FT((-jt)^nf(t))=F^(n)(omega)$$from the other side$$F^(n)(omega)=dfrac(-j)^nn!(a+jomega)^n+1$$therefore according to Parseval's identity we have $$int_Bbb R|f(t)|^2dt=dfrac12piint_Bbb R|F(omega)|^2domega$$hence by substitution we obtain:$$int_0^inftyt^2ne^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$which means that$$int_Bbb R|dfrac1(a+jomega)^n+1|^2domega=dfrac2piGamma(2n+1)(2a)^2n+1Gamma^2(n+1)$$here is a sketch of integral for $a=1$
where red square markers indicate on when $ninBbb Z$
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses ofLARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.
– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The Fourier Transform of $f(t)=e^-aomegau(t)$ is $F(omega)=dfrac1a+jomega$. Also we know that $$FT((-jt)^nf(t))=F^(n)(omega)$$from the other side$$F^(n)(omega)=dfrac(-j)^nn!(a+jomega)^n+1$$therefore according to Parseval's identity we have $$int_Bbb R|f(t)|^2dt=dfrac12piint_Bbb R|F(omega)|^2domega$$hence by substitution we obtain:$$int_0^inftyt^2ne^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$which means that$$int_Bbb R|dfrac1(a+jomega)^n+1|^2domega=dfrac2piGamma(2n+1)(2a)^2n+1Gamma^2(n+1)$$here is a sketch of integral for $a=1$
where red square markers indicate on when $ninBbb Z$
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses ofLARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.
– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
 |Â
show 5 more comments
up vote
2
down vote
accepted
The Fourier Transform of $f(t)=e^-aomegau(t)$ is $F(omega)=dfrac1a+jomega$. Also we know that $$FT((-jt)^nf(t))=F^(n)(omega)$$from the other side$$F^(n)(omega)=dfrac(-j)^nn!(a+jomega)^n+1$$therefore according to Parseval's identity we have $$int_Bbb R|f(t)|^2dt=dfrac12piint_Bbb R|F(omega)|^2domega$$hence by substitution we obtain:$$int_0^inftyt^2ne^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$which means that$$int_Bbb R|dfrac1(a+jomega)^n+1|^2domega=dfrac2piGamma(2n+1)(2a)^2n+1Gamma^2(n+1)$$here is a sketch of integral for $a=1$
where red square markers indicate on when $ninBbb Z$
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses ofLARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.
– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
 |Â
show 5 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The Fourier Transform of $f(t)=e^-aomegau(t)$ is $F(omega)=dfrac1a+jomega$. Also we know that $$FT((-jt)^nf(t))=F^(n)(omega)$$from the other side$$F^(n)(omega)=dfrac(-j)^nn!(a+jomega)^n+1$$therefore according to Parseval's identity we have $$int_Bbb R|f(t)|^2dt=dfrac12piint_Bbb R|F(omega)|^2domega$$hence by substitution we obtain:$$int_0^inftyt^2ne^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$which means that$$int_Bbb R|dfrac1(a+jomega)^n+1|^2domega=dfrac2piGamma(2n+1)(2a)^2n+1Gamma^2(n+1)$$here is a sketch of integral for $a=1$
where red square markers indicate on when $ninBbb Z$
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
The Fourier Transform of $f(t)=e^-aomegau(t)$ is $F(omega)=dfrac1a+jomega$. Also we know that $$FT((-jt)^nf(t))=F^(n)(omega)$$from the other side$$F^(n)(omega)=dfrac(-j)^nn!(a+jomega)^n+1$$therefore according to Parseval's identity we have $$int_Bbb R|f(t)|^2dt=dfrac12piint_Bbb R|F(omega)|^2domega$$hence by substitution we obtain:$$int_0^inftyt^2ne^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$which means that$$int_Bbb R|dfrac1(a+jomega)^n+1|^2domega=dfrac2piGamma(2n+1)(2a)^2n+1Gamma^2(n+1)$$here is a sketch of integral for $a=1$
where red square markers indicate on when $ninBbb Z$
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
edited Jul 25 at 11:52
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
answered Jul 25 at 10:13
Mostafa Ayaz
8,5373630
8,5373630
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses ofLARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.
– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
 |Â
show 5 more comments
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses ofLARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.
– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
$n$ may be not an integer.
– Szeto
Jul 25 at 11:16
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
@Szeto, it is true but in my example it is an integer. I will edit it.
– DOMiguel
Jul 25 at 11:22
Please avoid the unnecessary uses of
LARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.– Did
Jul 25 at 11:31
Please avoid the unnecessary uses of
LARGE, they only make your answer look childish. Oh, and the power of $a$ in your formula is incorrect, already for homogeneity reasons.– Did
Jul 25 at 11:31
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
Now, the computation and the final result itself are probably also off by a factor depending on $n$, maybe something similar to $(2n)!/4^n$.
– Did
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
I do not understand why equality $$int_0^inftye^-2atdt=dfrac12piint_Bbb R|F(omega)|^2domega=dfrac(n!)^22piint_Bbb R|dfrac1(a+jomega)^n+1|^2domega$$ holds. Where is $t^n$?
– DOMiguel
Jul 25 at 11:37
 |Â
show 5 more comments
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2
The change of variable $$omega=atan t$$ yields $$int_-infty^inftyleft|frac1(a+iomega)^n+1right|^2,mathrmdomega=frac1aint_-pi/2^pi/2cos^2nt,dt$$ and you might recognize the integral on the RHS as $2W_2n$ where $(W_n)$ denotes the sequence of Wallis integrals, whose values are known.
– Did
Jul 25 at 9:46
or just realize that $|z|^q=|z^q|$.
– G Cab
Jul 25 at 10:04
@Did Thank you for your answer. Nevertheless, I still cannot see how do you get rid of the complex number $1+itan t$. By the way, the factor should be $1/(|a|^2n+1)$, no?
– DOMiguel
Jul 25 at 11:54
1
Re the factor $1/|a|^2n+1$: indeed, well spotted, silly me! Re the complex number, simply note that $|1+itan t|^2=1/cos^2t$ and that $domega=|a|dt/cos^2t$.
– Did
Jul 25 at 12:05
Yes, sure. I forgot about the absolute value. Thank you.
– DOMiguel
Jul 25 at 12:14