# Integral $\int_0^\infty x\exp(-\frac{x^2}{2}\ )\mathrm{d}x$

I was trying to solve

$$\int_0^\infty x \exp \left( { - \frac{{{x^2}}}{2}} \right)\;{\text{d}}x$$

I was trying to use $$(f(g(x)))'=f'(g(x))\cdot g'(x)$$

(So $[e^{f(x)}]'=[e^{f(x)}]\cdot f'(x)$ ?), but I got the wrong answer. (I substituted $0$ and $\infty$ to $-\exp \left(-\frac{x^2}{2}\right)$ and got 1. But the test solution says $\sqrt{e}$.)

Could you point out what's wrong with my solution?

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The book's answer is incorrect; your integral evaluates to $1$ –  David Mitra Aug 21 '12 at 0:17
Nothing wrong with your answer. In the long run you may find the notation $u$, $\frac{du}{dx}$ more useful for integration by substitution. –  André Nicolas Aug 21 '12 at 0:20
It should be noted that $*$ is often used to denote convolution of functions, rather than pointwise multiplication. If you do mean pointwise multiplication it is best to use $\cdot$ (\cdot) or remove the symbol altogether. –  Asaf Karagila Aug 21 '12 at 0:22
@AsafKaragila I see. Thank you for your advice! –  Tengu Aug 21 '12 at 0:34
@DavidMitra In that case, I may not have written the problem correctly. Actually the problem is A-3 Putnam exam in 1997. Here are the links for problem and the solution. amc.maa.org/a-activities/a7-problems/putnamindex.shtml –  Tengu Aug 21 '12 at 0:38

We know that

$$\int f'(g(x))g'(x)\, dx=f(g(x))$$

This is the basis of a basic integral substitution. If we let $g(x)=u \implies g'(x)\, dx=du$ we can change the above integral into

$$\int f'(u)\, du=f(u)=f(g(x))$$

Using this method, we let

$$f(x)=\exp(-x)$$ $$g(x)=u= \frac{x^2}{2} \implies x=\sqrt{2 u} \implies dx= \frac{1}{\sqrt{2u}}\, du$$

so the integral becomes

$$\int x\exp(-x^2/2)\, dx=\int \sqrt{2u}\cdot\exp(-u)\, \cdot\frac{1}{\sqrt{2u}}\, du=-\exp(-u)=-\exp(-\frac{x^2}{2})$$

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As $(\exp(-x^2/2))´=-x \exp(-x^2/2)$ we have \begin{eqnarray} \int_{0}^{\infty}x \exp(-x^2/2)dx &=& \lim_{\varepsilon\uparrow \infty}\int_{0}^{\varepsilon}x \exp(-x^2/2)dx \\ &=& \lim_{\varepsilon\uparrow \infty}[-\exp(-\varepsilon^2/2)]_{0}^{\varepsilon}\\ &=& \lim_{\varepsilon\uparrow \infty}( -\exp(-\varepsilon^2/2)+1 )\\ &=& 1. \end{eqnarray}

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This is a very unusual $\varepsilon$. Usually $\varepsilon \to 0$, but this one $\to \infty$. Good for it! –  marty cohen Aug 21 '12 at 3:22

If you've reproduced the problem correctly, then you are correct and the solution key is incorrect.

As you said, if we set $f(x) = \exp(-x)$ and $g(x) = x^2 /2$, then the integrand is of the form $-f'(g'(x))\cdot g'(x)$, so we have by the (inverse) chain rule

$$\int_0^\infty x \exp(-x^2 / 2) dx = \left [ -f(g(x)) \right ]_0^\infty = \left [ -\exp(x^2 /2) \right ]_0^\infty = 1$$

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@Calvin McPhail-Snyder Thank you for your answer. I started to doubt I've reproduced the problem correctly. Actually this problem is from Putnam exam in 1997. (A-2). Could you take a look at it? I still don't understand where I got it wrong. Here is a link. amc.maa.org/a-activities/a7-problems/putnamindex.shtml –  Tengu Aug 21 '12 at 0:43
@DavidMitra Thanks. I already felt bad about posting this after your comment, but I would hope that I could at least get it right... –  Calvin McPhail-Snyder Aug 21 '12 at 0:47