I have an improper integral as follows:
$$\int_0^\infty r^2 e^{-a\cdot r}dr.$$
I try to evaluate it by parts and get $ [ -(\dfrac{r^2}{a} + \dfrac{2r}{a^2} + \dfrac{2}{a^3}) \cdot e^{-a \cdot r} ]^\infty_0 $, which leads me to nowhere.
According to the given solution, this integral can be solved like this:
$\int_0^\infty r^2 e^{-a\cdot r} = \dfrac{\delta^2}{\delta r^2} \int_0^\infty e^{-a\cdot r} = ... = \dfrac{2}{a^3} $
Which rule did they apply here (for the derivative)?
On
Hint...if you evaluate the expression you found, rather than leading you "nowhere" you get the same answer as given in the alternative solution. This is because $$x^ne^{-x}\rightarrow0$$ as $x\rightarrow\infty$
On
Method 1. One may integrate by parts twice as follows, $$ \begin{align} \int_0^\infty r^2 e^{-a\cdot r}dr&=\left[ r^2\:\frac{e^{-a\cdot r}}{-a}\right]_0^\infty+\frac1a\int_0^\infty (2r)\: e^{-a\cdot r}dr \\\\&=0+\frac{2}a\int_0^\infty r\: e^{-a\cdot r}dr \\\\&=\frac{2}a\left(\left[ r\:\frac{e^{-a\cdot r}}{-a}\right]_0^\infty+\frac1a\int_0^\infty e^{-a\cdot r}dr\right) \\\\&=\frac{2}a\left(0+\frac1a\cdot \frac1a\right) \\\\&=\frac{2}{a^3} \end{align} $$ where we have used that, for any fixed real numbers $n$ and $a>0$, $$ \lim_{r \to \infty}\left(r^n \cdot e^{-a\cdot r}\right)=0. $$
Method 2. One may apply the Leibniz integral rule twice (all conditions are OK here): $$\frac{d}{da} \left ( \int_\alpha^\beta f(a,x) \,\mathrm{d}x \right )= \int_\alpha^\beta \frac{\partial}{\partial a}f(a,x) \,\mathrm{d}x,$$ on the one hand, $$\frac{d^2}{da^2} \left (\int_0^\infty e^{-a\cdot x} \,\mathrm{d}x \right )= \int_0^\infty \frac{\partial^2}{\partial a^2}e^{-a\cdot x} \,\mathrm{d}x=\int_0^\infty x^2 \cdot e^{-a\cdot x} \,\mathrm{d}x$$ on the other hand, $$\frac{d^2}{da^2} \left (\int_0^\infty e^{-a\cdot x} \,\mathrm{d}x \right )= \frac{d^2}{da^2} \left (\frac1a\right )=\frac{2}{a^3}.$$
OK, made a mistake for the second method (apologies): For the second method you have to consider $$I(a) = \int_0^\infty e^{-a\cdot r} \, dr$$ Then if you look at $\frac{\partial^2 }{\partial \, a^2} I(a)$ this is equal to the required integral. But you have $I(a)= \frac{1}{a}$ and you can easily work out $\frac{\partial^2 }{\partial \, a^2} \frac{1}{a}$. As an additional point, you can't really look at $\frac{\partial^2 }{\partial \, r^2} \int_0^\infty r^2 e^{-a\cdot r} \, dr$, since $r$ is an integration variable.