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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.

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Have you thought about using any other methods than this one theorem? – anon271828 Mar 15 '13 at 23:41
I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. – apnorton Mar 15 '13 at 23:42
@L.F is e^x larger than x^2 ? – Kj Tada Mar 15 '13 at 23:48
@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) – L. F. Mar 15 '13 at 23:49
I see: The original poster did it that way. I'm sorry "anorton". – Michael Hardy Mar 15 '13 at 23:52
up vote 2 down vote accepted

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

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

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