What is $int_{-infty}^{infty}exp(mathrm{i} n cosh{x}) , mathrm{d}x$?
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I'm hoping to determine the value of the following integral:
$$int_{-infty}^{infty}exp(mathrm{i} n cosh{x}) , mathrm{d}x$$
Here is a plot of the integrand as a function of $x$ with parameter $n$ varying from 0 to 10.
The integral appears to not converge. However, it is known that
$$int_{-infty}^{infty}exp(mathrm{i} n x) , mathrm{d}x = 2pi delta(n)$$
where $delta(x)$ is the Dirac delta function. Is it possible for the first integral to be expressed similarly using Dirac delta notation?
integration definite-integrals dirac-delta
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
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add a comment |
$begingroup$
I'm hoping to determine the value of the following integral:
$$int_{-infty}^{infty}exp(mathrm{i} n cosh{x}) , mathrm{d}x$$
Here is a plot of the integrand as a function of $x$ with parameter $n$ varying from 0 to 10.
The integral appears to not converge. However, it is known that
$$int_{-infty}^{infty}exp(mathrm{i} n x) , mathrm{d}x = 2pi delta(n)$$
where $delta(x)$ is the Dirac delta function. Is it possible for the first integral to be expressed similarly using Dirac delta notation?
integration definite-integrals dirac-delta
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
$endgroup$
1
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
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– Paul Enta
Dec 14 '18 at 17:09
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@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
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@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
2
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18
add a comment |
$begingroup$
I'm hoping to determine the value of the following integral:
$$int_{-infty}^{infty}exp(mathrm{i} n cosh{x}) , mathrm{d}x$$
Here is a plot of the integrand as a function of $x$ with parameter $n$ varying from 0 to 10.
The integral appears to not converge. However, it is known that
$$int_{-infty}^{infty}exp(mathrm{i} n x) , mathrm{d}x = 2pi delta(n)$$
where $delta(x)$ is the Dirac delta function. Is it possible for the first integral to be expressed similarly using Dirac delta notation?
integration definite-integrals dirac-delta
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
$endgroup$
I'm hoping to determine the value of the following integral:
$$int_{-infty}^{infty}exp(mathrm{i} n cosh{x}) , mathrm{d}x$$
Here is a plot of the integrand as a function of $x$ with parameter $n$ varying from 0 to 10.
The integral appears to not converge. However, it is known that
$$int_{-infty}^{infty}exp(mathrm{i} n x) , mathrm{d}x = 2pi delta(n)$$
where $delta(x)$ is the Dirac delta function. Is it possible for the first integral to be expressed similarly using Dirac delta notation?
integration definite-integrals dirac-delta
integration definite-integrals dirac-delta
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
asked Dec 14 '18 at 16:43
SprocketsAreNotGearsSprocketsAreNotGears
1736
1736
1
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:09
$begingroup$
@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
$begingroup$
@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
2
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18
add a comment |
1
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:09
$begingroup$
@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
$begingroup$
@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
2
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18
1
1
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:09
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:09
$begingroup$
@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
$begingroup$
@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
$begingroup$
@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
$begingroup$
@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
2
2
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18
add a comment |
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1
$begingroup$
This integral is related to the Hankel function for which an integral representation is $${H^{(1)}_{nu}}left(zright)=frac{e^{-frac{1}{2}nupi i}}{pi i}int_{-infty}^{infty}e^{izcosh t-nu t}mathrm{d}t$$ Here, $nu=0$.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:09
$begingroup$
@PaulEnta, then for real $z$ the integral doesn't exist in the usual sense, does it? Or is the function analytically continued somehow?
$endgroup$
– Yuriy S
Dec 14 '18 at 17:10
$begingroup$
@YuriyS The principal branch of the function has a branch cut along the negative reals. See here for example.
$endgroup$
– Paul Enta
Dec 14 '18 at 17:15
2
$begingroup$
The representation $$Y_{0}left(xright)=-frac{2}{pi}int_{0}^{infty}cosleft(xcosh tright), mathrm{d}t$$ from here may be also helpful.
$endgroup$
– Sangchul Lee
Dec 14 '18 at 17:18