I am working on some exercises for Improper Integrals (not homework). The question is 1.c in Folland Advanced Calculus :

$$\int_0^\infty x^2 e^{-x^2 } \, dx$$

It asks whether the above Improper Integral is convergent. Folland's answer to this is that it is convergent. I however cant see how that could be. If I use the theorem (4.55 in text) that states:

$$0\le f\left( x \right) \le g(x) \text{ for all sufficiently large }x. \\\text{If }\int_0^\infty g(x) \, dx \text{ converges so does }\int_0^\infty f(x) \, dx.$$

then I just can't think of a $g(x)$ that would satisfy this.

  • $\begingroup$ Have you thought about using any other methods than this one theorem? $\endgroup$ – anon271828 Mar 15 '13 at 23:41
  • $\begingroup$ I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. $\endgroup$ – apnorton Mar 15 '13 at 23:42
  • $\begingroup$ @L.F is e^x larger than x^2 ? $\endgroup$ – Kj Tada Mar 15 '13 at 23:48
  • 2
    $\begingroup$ @KjTada For large $x$, yes! In fact, here's a better suggestion: $g(x)=e^{-x^2/2}$ (replacing $x^2$ with $e^{x^2/2}$; the inequality is true over the entire real line) $\endgroup$ – L. F. Mar 15 '13 at 23:49
  • 1
    $\begingroup$ I see: The original poster did it that way. I'm sorry "anorton". $\endgroup$ – Michael Hardy Mar 15 '13 at 23:52

Take $g(x) = e^{-x} $.Than $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}=0$. So there is $K$ that $g(x)\geq f(x)$ for $x\in (K,\infty)$.

$\int_K^\infty e^{-x} = e^{-K} < \infty$.

Therefore $\int _{ K }^{ \infty }{ { x }^{ 2 } } { e }^{ { -x }^{ 2 } }dx < \infty$.


$$ \int_{0}^{\infty} x^2 e^{-x^2} \, dx < \int_0^\infty e^{-x^2+x} \, dx = e^{1/4}\int_0^\infty e^{-(x-1/2)^2}\,dx < \infty. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.