<s>[INST] Double summation with improper integral So my friend sent me this really interesting problem. It goes:
Evaluate the following expression: 
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$
Here is my approach:
First evaluate the integral:
$$ \frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$
This can be done using integration by parts and we get:
$$ \frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$
We can do this $ b $ times until we get:
$$ \frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$
and hence we end up with:
$$ \frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$
Now we can apply the sum of GP to infinity formula and we get:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}}  = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$
This is a telescoping series and we end up with $$ \frac{1}{a-1} = \frac{1}{2-1} = 1.$$
Do you guys have any other ways of solving this problem? Please do share it here.
 [/INST] since $\frac{x^b}{e^{ax} b!}$  is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
\begin{align} 
&\sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx \\
&=\int_{0}^\infty \sum_{a=2}^\infty e^{-ax} \sum_{b=1}^\infty \frac{x^{b}}{ \ b!} \ dx \\
&=  \int_{0}^\infty \underbrace{\left(\sum_{a=2}^\infty (e^{-x})^a\right)}_{\text{geometric series}} \overbrace{\left(\sum_{b=0}^\infty \frac{x^{b}}{ \ b!}-1\right)}^{\text{series definition of $e^x$}} \ dx \\
&=    \int_{0}^\infty \frac{1}{e^x(e^x-1)}(e^{x}-1)dx \\
&= \int_0^\infty e^{-x} dx \\&= 1.\end{align}
</s>