Help with integral of a gaussian function

AI Thread Summary
The integral of the Gaussian function combined with a sine function, specifically \(\int_0^\pi \exp(-\frac{x^2}{2c^2})\sin(\frac{m\pi x}{2}) dx\), poses significant challenges for computation. Integration by parts is suggested but may not yield a closed-form solution. The integral without the sine can be expressed in terms of the error function, but adding the sine complicates matters further. Techniques such as using Euler's identity and complex analysis may be necessary, but they introduce complexities that require advanced knowledge. Ultimately, numerical methods or software tools like Mathematica or MATLAB are recommended for practical evaluation.
maxtor101
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Hi all!

I'm curious as to how one would go about actually computing this integral

\int_0^\pi \exp(\frac{-x^2}{2c^2})\sin(\frac{m\pi x}{2}) dx

I start off by using integration by parts but I am unsure how to solve this integral

v = \int_0^\pi \exp(\frac{-x^2}{2c^2}) dx

Any help would be greatly appreciated
Max
 
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Note that \intexp(-ax^{2})dx x \intexp(-ay^{2})dy = \int\intexp{-a[x^{2}+y^{2}]}dxdy.

Then change to polar coordinates.
 
Note that a lot of integrals of that form (including the normal probability density function) don't have solutions in closed form.
 
grzz said:
Note that \intexp(-ax^{2})dx x \intexp(-ay^{2})dy = \int\intexp{-a[x^{2}+y^{2}]}dxdy.

Then change to polar coordinates.

I don't think the OP is doing the integral you think he's doing.

maxtor101 said:
Hi all!

I'm curious as to how one would go about actually computing this integral

\int_0^\pi \exp(\frac{-x^2}{2c^2})\sin(\frac{m\pi x}{2}) dx

I start off by using integration by parts but I am unsure how to solve this integral

v = \int_0^\pi \exp(\frac{-x^2}{2c^2}) dx

Any help would be greatly appreciated
Max

You're not really going to have much luck with that integral. Without the sine, the integral can only be expressed in terms of the error function, but the error function is defined by

\mbox{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x dt~e^{-t^2},

so you haven't gained much except that you know you can express it as the error function, which mathematica, matlab, etc., have code to compute.

If you through the sine back in there, that just makes it worse. wolframalpha calculates the integral in terms of functions related to the error function. I imagine the way it does this is to use Euler's identity and write

\sin x = \frac{e^{ix}-e^{-ix}}{2i},

and then complete the square in the exponential, but there's some trickiness associated with that because you'll be introducing imaginary numbers into the limits of the integral. That's not really a problem if you know contour integrals and complex analysis, but if you don't writing the integral that way isn't really going to help you much.
 
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