# Can you please explain this equality? int r/theta^2 exp(-r^2/(2theta^2))dr=-exp(-r^2/2theta^2)

Jairo Decker 2022-10-24 Answered
Can you please explain this equality?
$\int \frac{r}{{\sigma }^{2}}\mathrm{exp}\left(\frac{-{r}^{2}}{2{\sigma }^{2}}\right)\phantom{\rule{thickmathspace}{0ex}}dr=-\mathrm{exp}\left(\frac{-{r}^{2}}{2{\sigma }^{2}}\right)$
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Ramiro Sosa
To check that an indefinite integral is equal to a function
$\int f\left(x\right)\phantom{\rule{thickmathspace}{0ex}}dx=F\left(x\right)\phantom{\rule{1em}{0ex}}\left(+C\right)$
you can simply check that ${F}^{\prime }\left(x\right)=f\left(x\right)$. In your example you check that
$\frac{d}{dr}\left[-\mathrm{exp}\left(\frac{-{r}^{2}}{2{\sigma }^{2}}\right)\right]=-\mathrm{exp}\left(\frac{-{r}^{2}}{2{\sigma }^{2}}\right)\frac{d}{dr}\frac{-{r}^{2}}{2{\sigma }^{2}}\phantom{\rule{0ex}{0ex}}=\mathrm{exp}\left(\frac{-{r}^{2}}{2{\sigma }^{2}}\right)\frac{r}{{\sigma }^{2}}$
If you don't like this, then you can also find the integral using the substitution method with $u=\frac{-{r}^{2}}{2{\sigma }^{2}}$. So then
$\begin{array}{rl}du& =\frac{-2r}{2{\sigma }^{2}}dr\phantom{\rule{1em}{0ex}}⇒\\ -du& =\frac{r}{{\sigma }^{2}}dr\end{array}$
So the integral becomes
$\int -{e}^{u}\phantom{\rule{thickmathspace}{0ex}}du.$