Closed from of $int_0^{infty} frac{e^{iax}}{x^{n}+1}dx$?
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I've been trying to find the general form of a certain group of integrals of the form$$I(a,n)=int_0^{infty} frac{e^{iax}}{x^{n}+1}dx$$
I know that the real part of $I(a,2)$ can be calculated using Fourier Transform or residues, and $I(a,1)$ reduces to a form of the exponential integral.
I thought about approaching the integral via Fourier Transform but I did not know how to apply it to this integral. It might be able to be calculated with residues but I am not that great at complex analysis. I'm very interested in a closed form for this integral so any help would be appreciated.
improper-integrals contour-integration fourier-transform
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
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add a comment |
$begingroup$
I've been trying to find the general form of a certain group of integrals of the form$$I(a,n)=int_0^{infty} frac{e^{iax}}{x^{n}+1}dx$$
I know that the real part of $I(a,2)$ can be calculated using Fourier Transform or residues, and $I(a,1)$ reduces to a form of the exponential integral.
I thought about approaching the integral via Fourier Transform but I did not know how to apply it to this integral. It might be able to be calculated with residues but I am not that great at complex analysis. I'm very interested in a closed form for this integral so any help would be appreciated.
improper-integrals contour-integration fourier-transform
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
$endgroup$
$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28
add a comment |
$begingroup$
I've been trying to find the general form of a certain group of integrals of the form$$I(a,n)=int_0^{infty} frac{e^{iax}}{x^{n}+1}dx$$
I know that the real part of $I(a,2)$ can be calculated using Fourier Transform or residues, and $I(a,1)$ reduces to a form of the exponential integral.
I thought about approaching the integral via Fourier Transform but I did not know how to apply it to this integral. It might be able to be calculated with residues but I am not that great at complex analysis. I'm very interested in a closed form for this integral so any help would be appreciated.
improper-integrals contour-integration fourier-transform
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
$endgroup$
I've been trying to find the general form of a certain group of integrals of the form$$I(a,n)=int_0^{infty} frac{e^{iax}}{x^{n}+1}dx$$
I know that the real part of $I(a,2)$ can be calculated using Fourier Transform or residues, and $I(a,1)$ reduces to a form of the exponential integral.
I thought about approaching the integral via Fourier Transform but I did not know how to apply it to this integral. It might be able to be calculated with residues but I am not that great at complex analysis. I'm very interested in a closed form for this integral so any help would be appreciated.
improper-integrals contour-integration fourier-transform
improper-integrals contour-integration fourier-transform
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
asked Dec 13 '18 at 20:55
aledenaleden
2,362511
2,362511
$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28
add a comment |
$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28
$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28
$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28
add a comment |
1 Answer
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Since the integrand is a product of two Meijer G-functions and the integration range is $[0, infty)$, there is a closed form, but it involves the Fox H-function:
$$int_0^infty frac {e^{i a x}} {x^n + 1} dx =
int_0^infty
G_{0, 1}^{1, 0} {left(- i a x middle| { - atop 0} right)}
G_{1, 1}^{1, 1} {left(x^n middle| { 0 atop 0} right)} dx =
frac i a H_{2, 1}^{1, 2}
{left(
left( frac i a right)^{!n} middle| {(0, 1), (0, n) atop (0, 1)}
right)}.$$
This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since the integrand is a product of two Meijer G-functions and the integration range is $[0, infty)$, there is a closed form, but it involves the Fox H-function:
$$int_0^infty frac {e^{i a x}} {x^n + 1} dx =
int_0^infty
G_{0, 1}^{1, 0} {left(- i a x middle| { - atop 0} right)}
G_{1, 1}^{1, 1} {left(x^n middle| { 0 atop 0} right)} dx =
frac i a H_{2, 1}^{1, 2}
{left(
left( frac i a right)^{!n} middle| {(0, 1), (0, n) atop (0, 1)}
right)}.$$
This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
$endgroup$
add a comment |
$begingroup$
Since the integrand is a product of two Meijer G-functions and the integration range is $[0, infty)$, there is a closed form, but it involves the Fox H-function:
$$int_0^infty frac {e^{i a x}} {x^n + 1} dx =
int_0^infty
G_{0, 1}^{1, 0} {left(- i a x middle| { - atop 0} right)}
G_{1, 1}^{1, 1} {left(x^n middle| { 0 atop 0} right)} dx =
frac i a H_{2, 1}^{1, 2}
{left(
left( frac i a right)^{!n} middle| {(0, 1), (0, n) atop (0, 1)}
right)}.$$
This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
$endgroup$
add a comment |
$begingroup$
Since the integrand is a product of two Meijer G-functions and the integration range is $[0, infty)$, there is a closed form, but it involves the Fox H-function:
$$int_0^infty frac {e^{i a x}} {x^n + 1} dx =
int_0^infty
G_{0, 1}^{1, 0} {left(- i a x middle| { - atop 0} right)}
G_{1, 1}^{1, 1} {left(x^n middle| { 0 atop 0} right)} dx =
frac i a H_{2, 1}^{1, 2}
{left(
left( frac i a right)^{!n} middle| {(0, 1), (0, n) atop (0, 1)}
right)}.$$
This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
$endgroup$
Since the integrand is a product of two Meijer G-functions and the integration range is $[0, infty)$, there is a closed form, but it involves the Fox H-function:
$$int_0^infty frac {e^{i a x}} {x^n + 1} dx =
int_0^infty
G_{0, 1}^{1, 0} {left(- i a x middle| { - atop 0} right)}
G_{1, 1}^{1, 1} {left(x^n middle| { 0 atop 0} right)} dx =
frac i a H_{2, 1}^{1, 2}
{left(
left( frac i a right)^{!n} middle| {(0, 1), (0, n) atop (0, 1)}
right)}.$$
This becomes a G-function if $n$ is rational, but a rational $n$ produces an infinite number of double poles. This gives an infinite sum of polygamma terms instead of gamma terms when the H-function is evaluated by applying the residue theorem. Such a sum may have a closed form in terms of simpler functions in some special cases, which happens for $n = 1$ and $n = 2$.
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
answered Dec 17 '18 at 14:14
MaximMaxim
5,5981220
5,5981220
add a comment |
add a comment |
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$begingroup$
You may or may not be able to use the solution as per my question here - math.stackexchange.com/questions/3045895/…
$endgroup$
– DavidG
Dec 28 '18 at 1:28