Convincing myself the answer key was correct
I’ve been working through Likharev’s Classical Electrodynamics rather slowly. I came across something in the answer key as I working through the electric field of an exponentially decaying, spherically symmetric charge distribution. He claimed that the integral in the problem was actually the second partial derivative of an elementary integral.
$$ \int_{0}^{r} r'^{2} e^{-\lambda r'} \, dr' = \frac{\partial^2 }{\partial \lambda^2} \int_{0}^{r} e^{-\lambda r'} dr' $$I was integrating by parts, which is a tried and true way to solve this problem, but I wanted to see why this was true. I’d remembered hearing about some similar technique for integrals years ago, and this seemed like a good time to see if it actually worked. Plus this seemed like one of those things that I didn’t want to take on faith
I started by defining a function of lambda
$$ F(\lambda) = \int_{0}^{r} e^{-\lambda r'} dr' = \frac{1}{\lambda} (1 - e^{-\lambda r}) $$This result was fairly straightforward and elementary. Then as a sanity check I began took the first partial derivative of this with respect to $ \lambda $.
$$ \frac{\partial F}{\partial \lambda} = - \frac{1}{\lambda^{2}} (1-e^{-\lambda r}) + \frac{1}{\lambda} (r e^{-\lambda r}) $$I sanity checked myself with by integrating $ r e^{-\lambda r} $ over the same bounds and was stumped by the fact that I was getting a negative sign. Then I remembered the sign in the exponential function and it all clicked.
It made sense why this particular trick worked. The integral was over $r$ not $\lambda$ and when I took its partial derivatives inside the integral I was just pumping up the power of $r$ in front of the exponential function. Because I knew the result of the integration, it was really just as simple as taking the partial derivative of the result.
I didn’t prove this out in full generality because the problem didn’t require it but there does appear to be a caveat to this rule for the odd powers of $n$. This is what I expect the general result is. I’m going to have to keep my eye out for more patterns like this.
$$ \int_{0}^{r} r'^{n} e^{-\lambda r'} \, dr' = (-1)^n \frac{\partial^{n}}{\partial \lambda^n} [ \frac{1}{\lambda} (1 - e^{-\lambda r}) ] $$