Help with integral of a gaussian function

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Discussion Overview

The discussion revolves around the computation of the integral \int_0^\pi \exp(\frac{-x^2}{2c^2})\sin(\frac{m\pi x}{2}) dx, with a focus on techniques such as integration by parts and the potential use of polar coordinates. Participants explore the challenges associated with this integral, particularly in relation to closed-form solutions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to compute the integral using integration by parts.
  • Another participant suggests that many integrals of this form do not have solutions in closed form.
  • A different participant mentions that the integral can be expressed in terms of the error function, but this does not simplify the problem significantly.
  • There is a discussion about the potential complications introduced by including the sine function in the integral, with references to Euler's identity and the challenges of complex analysis.
  • Some participants question whether the original integral is being interpreted correctly.

Areas of Agreement / Disagreement

Participants generally agree that the integral is complex and may not have a straightforward solution, but multiple competing views remain regarding the best approach to tackle it and the implications of including the sine function.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the use of integration techniques and the implications of complex numbers in the context of the integral.

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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