Explanation of $\int\frac{r}{\sigma^2}\exp\big(\frac{-r^2}{2\sigma^2}\big)\; dr=-\exp\big(\frac{-r^2}{2\sigma^2}\big)$

Question

Can you please explain this equality?

$$\int\frac{r}{\sigma^2}\exp\left(\frac{-r^2}{2\sigma^2}\right)\; dr=-\exp\left(\frac{-r^2}{2\sigma^2}\right)$$

Thanks a lot! :)

Answer 1

To check that an indefinite integral is equal to a function $$ \int f(x) \; dx= F(x) \quad (+C) $$ you can simply check that $F'(x) = f(x)$. In your example you check that $$ \frac{d}{dr} [-\exp\left(\frac{-r^2}{2\sigma^2}\right)] = -\exp\left(\frac{-r^2}{2\sigma^2}\right)\frac{d}{dr} \frac{-r^2}{2\sigma^2} \\ = \exp\left(\frac{-r^2}{2\sigma^2}\right)\frac{r}{\sigma^2} $$

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If you don't like this, then you can also find the integral using the substitution method with $\displaystyle{u = \frac{-r^2}{2\sigma^2}}$. So then $$\begin{align} du &= \frac{-2r}{2\sigma^2}dr \quad \Rightarrow\\ -du &= \frac{r}{\sigma^2}dr \end{align} $$ So the integral becomes $$ \int -e^u \; du. $$ This you can probably find.

Answer 2

A primitive function of $\frac{r}{\sigma^2}\exp\left(\frac{-r^2}{2\sigma^2}\right)$ is $-\exp\left(\frac{-r^2}{2\sigma^2}\right)$.