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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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    I understand that the series sums to $\sqrt{e}$ but I don't really know why I need to think about the series and how I can find the series, $x^{2n+1} exp(−\frac{x^2}{2})$2012-08-21

3 Answers 3

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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

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

1

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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    @DavidMitra Thanks. I already felt bad about posting this after your comment, but I would hope that I could at least get it right...2012-08-21
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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!2012-08-21