Deduce an integral I came across

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Discussion Overview

The discussion revolves around the evaluation of the integral \(\int_0^\infty x^3e^{-x^2}dx\), which is encountered in the context of thermodynamics. Participants are exploring methods to deduce the value of this integral, including substitution and integration by parts.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in evaluating the integral and mentions previous attempts using partial integration.
  • Another participant suggests using the substitution \(u = x^2\) as a potential method to simplify the integral.
  • A later reply acknowledges the initial difficulty but indicates that the substitution worked upon re-evaluation, suggesting a possible earlier mistake.
  • Further clarification is provided on rewriting the integral in a specific form to facilitate integration by parts, leading to a standard integral result.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the evaluation method, as multiple approaches are suggested and discussed without a definitive resolution.

Contextual Notes

Some participants reference standard integrals and methods, but there are indications of potential mistakes in earlier attempts, highlighting the complexity of the evaluation process.

Funzies
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Hey guys, I'm trying to deduce an integral I came across whilst studying some thermodynamics, but I can't seem to evaluate it:
[tex]\int_0^\infty x^3e^{-x^2}dx[/tex]
I've tried partial integration numerous times, but I can't seem to get it right. Can you help me?
I consider
[tex]\int_{-\infty}^\infty e^{-x^2}dx = \frac{1}{2}\sqrt{\pi}[/tex]
as a standard integral.
 
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Try substitution [tex]u=x^2[/tex].
 


Thanks, I did try that substitution before, but now I tried it again and it came out, I must have made a mistake. Thanks!
 


Specically, write the integral as
[tex]\int_0^\infty x^2e^{-x^2}(x dx)[/tex]
and then set [itex]u= x^2[/itex]. You will get an integral of the form
[tex]\frac{1}{2}\int_0^\infty ue^{-u}du[/tex]
that you can integrate by parts.
 

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