Integrate X²e^-x²: Solving a Tricky Integral

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Homework Help Overview

The discussion revolves around the integration of the function x²e^-x² over the interval from 0 to infinity. Participants are exploring various methods to approach this integral, including the use of integration by parts and referencing known results related to the Gaussian integral.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the possibility of using integration by parts and question the existence of a closed form for the integral. There are references to known results and methods, including double integrals and polar coordinates, as potential avenues for evaluation.

Discussion Status

The discussion is active, with participants sharing their thoughts on the integral and its evaluation. Some express uncertainty about the methods discussed, while others suggest that there are multiple ways to approach the problem. The original poster has provided a known result, which has prompted further exploration of the integral's evaluation.

Contextual Notes

Participants note that the integral is part of a study on improper integrals in an advanced math course, indicating a level of complexity and the potential for varied approaches. There is mention of prior discussions on other forums, suggesting a broader context for the inquiry.

arsmath
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I am working on an integral I am finding tricky, and I think I'm missing something.
I need to integrate on the interval 0 to infinity, x²e^-x².
We have proved that on the interval of -∞ to ∞, e^-x²=√∏ so from o to ∞, it equals √∏/2. I can use this in my proof, but I don't see how. When I try integrating by parts I have trouble getting a finite answer. I would love some help,
 
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Try integrating by parts the

I=\int_0^\infty e^{-x^2}\,d\,x
 
Rainbow Child said:
Try integrating by parts the

I=\int_0^\infty e^{-x^2}\,d\,x
HOw can one integrate this by parts, i do not think this has any closed form does it?
 
and this question is already on another forum!lol
 
HOw can one integrate this by parts, i do not think this has any closed form does it?

Like this:

I=\int_0^\infty e^{-x^2}\,d\,x=\int_0^\infty (x)'\,e^{-x^2}\,d\,x

and

I=\frac{\sqrt{\pi}}{2}

by OP
 
Rainbow Child said:
Like this:

I=\int_0^\infty e^{-x^2}\,d\,x=\int_0^\infty (x)'\,e^{-x^2}\,d\,x

and

I=\frac{\sqrt{\pi}}{2}

by OP

At what level is one supposed to learn how to integrate this?? I mean where is it covered?
Becasue this is my first time seeing such a trick!
 
Did you read the original post?
They gave him the result I=\frac{\sqrt{\pi}}{2}.

As for the actual calculation, there are many ways to calulate I. The simplest one is by double integrals.
 
my fault

sutupidmath said:
and this question is already on another forum!lol

i posted in the other forum before finding this one which I think may be more appropriate.
 
Rainbow Child said:
Did you read the original post?
They gave him the result I=\frac{\sqrt{\pi}}{2}.

As for the actual calculation, there are many ways to calulate I. The simplest one is by double integrals.

Well, i did not read the op's post!
and as for evaluating that integral, i think i should wait a few more months.
 
  • #10
arsmath said:
i posted in the other forum before finding this one which I think may be more appropriate.
ok then, like Hurky said, show what u did so far?
 
  • #11
sutupidmath said:
At what level is one supposed to learn how to integrate this?? I mean where is it covered?
Becasue this is my first time seeing such a trick!
I'm studying improper integrals for a "topics in advanced math" course. . .the proof is lengthy and involves double integrals and polar coords
 
  • #12
arsmath said:
I'm studying improper integrals for a "topics in advanced math" course. . .the proof is lengthy and involves double integrals and polar coords
AH, i have never worked with double integrals, so i guess i cannot be of any further help, but surely the other guys will give u enough hints to get it right!
good luck!
 

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