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Just wondering how I can integrate $\displaystyle xe^{ \large {-x^2/(2\sigma^2)}}$

Tried using substitution where $U(x) = x^2$ but I kept getting a $x^2$ at the denominator which is incorrect.

I understand that $\displaystyle \int e^{f(x)} = e^{\large \frac{f(x)}{f'(x)}}$ if $f(x)$ is linear, however, how do we handle this situation when $f(x)$ is not linear?

Step by step answer will be awesome!!!! thanks :D

M

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you shouldn't get a x^2 in the denominator... –  Eleven-Eleven Apr 25 '13 at 12:03

2 Answers 2

up vote 2 down vote accepted

A close relative of your substitution, namely $u=-\dfrac{x^2}{2\sigma^2}$, works.

In "differential" notation, we get $du=-\frac{1}{\sigma^2} x\,dx$, so $x\,dx=-\sigma^2\,du$.

Remark: Or more informally, let's guess that the answer is $e^{-x^2/(2\sigma^2)}$. Differentiate, using the Chain Rule. We get $-\frac{x}{\sigma^2}e^{-x^2/(2\sigma^2)}$, so wrong guess. Too bad.

But we got close, and there is an easy fix by multiplying by a suitable constant to get rid of the $-\frac{1}{\sigma^2}$ in front of the derivative of our wrong guess. The indefinite integral is $-\sigma^2e^{-x^2/(2\sigma^2)}+C$.

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$$\int{xe^{\frac{-x^2}{2\sigma^2}}dx}$$

Let $u=x^2$. Then $\frac{du}{2}=xdx$.

$$\frac{1}{2}\int{e^\frac{-u}{2\sigma^2}}dx=-\sigma^2e^{\frac{-x^2}{2\sigma^2}}$$

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