How can the error function be used to solve this integral?

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Discussion Overview

The discussion revolves around solving the integral \(\int_{0}^{\infty} e^{-(x - t)^2/2 \sigma^2} x^n\ dx\), with a focus on its relationship to the error function and potential methods for evaluation. Participants explore techniques for integrating this expression, considering both odd and even values of \(n\) and the implications of the Gaussian convolution.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that if \(n\) is odd, a substitution \(u= -(x-\mu)^2/(2\sigma^2)\) could be useful, while for even \(n\), integration by parts might be employed.
  • Another participant proposes expressing the integral in terms of the error function as long as \(t\) is not zero.
  • A later reply indicates a method involving the substitution \(u=x-t\), expanding the polynomial, and using successive integration by parts to relate the result to the error function.

Areas of Agreement / Disagreement

Participants express a shared interest in using the error function for the integral, but there is no consensus on the specific steps or methods to achieve this. Multiple approaches are suggested without agreement on a definitive solution.

Contextual Notes

Participants do not clarify the assumptions regarding the parameters involved, such as the values of \(n\) or \(t\), nor do they resolve the mathematical steps necessary for the integration.

astralmeme
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Greetings,

I am a computer scientist revisiting integration after a long time. I am stuck with this simple-looking integral that's turning out to be quite painful (to me). I was wondering if one of you could help.

The goal is to solve the integral

<br /> \int_{0}^{\infty} e^{-(x - t)^2/2 \sigma^2} x^n\ dx . <br />

Note that this is the convolution of the Gaussian centered around 0 with the function that equals $x^n$ for $x > 0$, and 0 elsewhere (modulo scaling).

In particular, I would be interested in seeing any relationship with the integral

<br /> \int_{-\infty}^{\infty} e^{-(x - t)^2/2 \sigma^2} x^n\ dx . <br />

which I have worked out.

Any suggestions?

Thanks in advance,
Swar
 
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If n is odd, that can be done by letting u= -(x-\mu)^2/(2\sigma^2). If x is even, try integration by parts, letting u= x^{n-1}, dv= xe^{-(x-\mu)^2/(2\sigma^2)} to reduce it to n odd.
 
astralmeme said:
Greetings,

I am a computer scientist revisiting integration after a long time. I am stuck with this simple-looking integral that's turning out to be quite painful (to me). I was wondering if one of you could help.

The goal is to solve the integral

<br /> \int_{0}^{\infty} e^{-(x - t)^2/2 \sigma^2} x^n\ dx . <br />

Note that this is the convolution of the Gaussian centered around 0 with the function that equals $x^n$ for $x > 0$, and 0 elsewhere (modulo scaling).

In particular, I would be interested in seeing any relationship with the integral

<br /> \int_{-\infty}^{\infty} e^{-(x - t)^2/2 \sigma^2} x^n\ dx . <br />

which I have worked out.

Any suggestions?

Thanks in advance,
Swar

As long as t is not 0, the best you can do is express the integral in terms of the error function.
 
Regarding error function, that is my guess too, but can you tell me what exactly would need be done?

I apologize if the question is obvious.


Swar
 
astralmeme said:
Regarding error function, that is my guess too, but can you tell me what exactly would need be done?

I apologize if the question is obvious.


Swar
What I would do is first let u=x-t. Then xn becomes (u+t)n.
Expand the polynomial in u and then by succesive itegration by parts, get all the terms to a 0 exponent for u, which will be proportional to erf(t).
 

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