Integrals involving Dirichlet kernel and sinc function The Next CEO of Stack OverflowHelp evaluating or upper-bounding integral $int_-infty^infty fracaboperatornamesinc^2(cx)a+boperatornamesinc^2(cx);dx$Integrating sinc / gaussian function with 2nd order polynomials as argumentsIntegral involving $operatornamesinc$ and exponentialCalculate $int_-T^T operatornamesincbig(tau-lambdabig) operatornamesincbig(tau-nubig)dtau$.Fourier transform of a product of two rect functionsImpossible definite integral!evaluate a Fraunhofer diffraction integralDefinite integral $int_y^infty$ involving two Meijer's G functionIntegral involving 2-dimensional Gaussian functionFourier transform of the convolution of a Dirac comb with the product of a complex exponential function and a rect function
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Integrals involving Dirichlet kernel and sinc function
The Next CEO of Stack OverflowHelp evaluating or upper-bounding integral $int_-infty^infty fracaboperatornamesinc^2(cx)a+boperatornamesinc^2(cx);dx$Integrating sinc / gaussian function with 2nd order polynomials as argumentsIntegral involving $operatornamesinc$ and exponentialCalculate $int_-T^T operatornamesincbig(tau-lambdabig) operatornamesincbig(tau-nubig)dtau$.Fourier transform of a product of two rect functionsImpossible definite integral!evaluate a Fraunhofer diffraction integralDefinite integral $int_y^infty$ involving two Meijer's G functionIntegral involving 2-dimensional Gaussian functionFourier transform of the convolution of a Dirac comb with the product of a complex exponential function and a rect function
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I am looking to evaluate the following definite integral involving Dirichlet kernel and sinc function with phase.
beginequationI(d) = int_-infty^infty fracsin[N(k^"-k^')]sin(k^"-k^') fracsin(c(k^"-k^'))c(k^"-k^') fracsin(Nk^')sin(k^') fracsin(ck^')ck^' exp(ik^'d+iphi(k^')) rect Big(frack^'k_0Big) dk^' endequation
Seems to be difficult due to Dirichlet kernel. Thank you in advance.
definite-integrals convolution
asked Mar 25 at 21:02
user16409user16409
85111
$endgroup$
add a comment |
$begingroup$
I am looking to evaluate the following definite integral involving Dirichlet kernel and sinc function with phase.
beginequationI(d) = int_-infty^infty fracsin[N(k^"-k^')]sin(k^"-k^') fracsin(c(k^"-k^'))c(k^"-k^') fracsin(Nk^')sin(k^') fracsin(ck^')ck^' exp(ik^'d+iphi(k^')) rect Big(frack^'k_0Big) dk^' endequation
Seems to be difficult due to Dirichlet kernel. Thank you in advance.
definite-integrals convolution
asked Mar 25 at 21:02
user16409user16409
85111
$endgroup$
$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06
add a comment |
$begingroup$
I am looking to evaluate the following definite integral involving Dirichlet kernel and sinc function with phase.
beginequationI(d) = int_-infty^infty fracsin[N(k^"-k^')]sin(k^"-k^') fracsin(c(k^"-k^'))c(k^"-k^') fracsin(Nk^')sin(k^') fracsin(ck^')ck^' exp(ik^'d+iphi(k^')) rect Big(frack^'k_0Big) dk^' endequation
Seems to be difficult due to Dirichlet kernel. Thank you in advance.
definite-integrals convolution
asked Mar 25 at 21:02
user16409user16409
85111
$endgroup$
I am looking to evaluate the following definite integral involving Dirichlet kernel and sinc function with phase.
beginequationI(d) = int_-infty^infty fracsin[N(k^"-k^')]sin(k^"-k^') fracsin(c(k^"-k^'))c(k^"-k^') fracsin(Nk^')sin(k^') fracsin(ck^')ck^' exp(ik^'d+iphi(k^')) rect Big(frack^'k_0Big) dk^' endequation
Seems to be difficult due to Dirichlet kernel. Thank you in advance.
definite-integrals convolution
definite-integrals convolution
asked Mar 25 at 21:02
user16409user16409
85111
asked Mar 25 at 21:02
user16409user16409
85111
edited Mar 28 at 9:25
user16409
asked Mar 25 at 21:02
user16409user16409
85111
asked Mar 25 at 21:02
user16409user16409
85111
asked Mar 25 at 21:02
user16409user16409
85111
85111
$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06
add a comment |
$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06
$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06
add a comment |
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$begingroup$
No, the Dirichlet kernel cancel and with $int_-infty^infty$ instead of $int_0^k_0$ it would have a closed form.
$endgroup$
– reuns
Mar 25 at 22:13
$begingroup$
Kindly explain how they cancel? btw, I corrected for missing constant in the sinc argument.
$endgroup$
– user16409
Mar 25 at 22:52
$begingroup$
With your $c$ they don't anymore. Why would you want to evaluate $int_0^k_0$ ? With $int_-infty^infty$ it still has a closed form
$endgroup$
– reuns
Mar 25 at 22:54
$begingroup$
$int_0^k_0$ ? --> I have optical low pass filter that limits spatial frequencies.
$endgroup$
– user16409
Mar 25 at 23:06