Evaluate: $\displaystyle \int ^\infty _{-\infty} x^4e^{-x^2/2}dx$

If I notice this is an even function I can write this as : $2\displaystyle \int ^\infty _0 x^4e^{-x^2/2}dx$

If I then proceed with the substitution $u=\frac{x^2}{2}$ the limit of integration is $(0,\infty)$

However if I do not notice this is an even function and write $u=\frac{x^2}{2}$ can I just let the limit of integration be equal $(0,\infty)$ or is there a further step I must take aswell?

It's just I get two different answers using the above approaches.


When you make a substitution $u=f(x)$, the function $f(x)$ should be one to one on the interval of integration.

That will not be the case if $f(x)=x^2/2$ and the interval is $(-\infty,\infty)$.

  • $\begingroup$ can you post some references to your claim? I don't remember it has to be one to one at all $\endgroup$ – AnalysisStudent0414 Jun 6 '14 at 19:58
  • $\begingroup$ I don't have much handy. But I do have an old Rudin, 1953. Page 106, Theorem 6.29 (change of variable, definite integral). There he specifies that $\phi(x)$ be strictly increasing. Of course strictly decreasing is no problem. $\endgroup$ – André Nicolas Jun 6 '14 at 20:14
  • $\begingroup$ That is true if you do that on a function in $\mathbb{R}^n$, $n>1$. You absolutely don't need that in $\mathbb{R}$ (see my answer) $\endgroup$ – AnalysisStudent0414 Jun 7 '14 at 19:02

You absolutely can. When you write

$$\displaystyle \int ^\infty _{-\infty} x^4e^{-x^2/2}dx = 2\displaystyle \int ^\infty _0 x^4e^{-x^2/2}dx$$

you're saying those are equal. Then you solve the right member separately, so

$$2\displaystyle \int ^\infty _0 x^4e^{-x^2/2}dx = 2\displaystyle \int ^\infty _0 2u \sqrt{u} \, e^{-u} du =\dots$$


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