For all real $a>0$ consider
$$\int_0^\infty x^a \left (\sum_{n = 1}^\infty (-1)^n n e^{-2nx} \right ) \, dx.$$
I would like to interchange the summation with the integration in order to find its value but I am having trouble justifying such a change.
Fubini's theorem is not strong enough to justify the interchange for all $a > 0$. If we put absolute values on the terms, we have
$$\left | \sum_{n = 1}^\infty (-1)^n n e^{-2nx} \right | \leqslant \sum_{n = 1}^\infty n e^{-2nx} = \frac{e^{2x}}{(e^{2x} – 1)^2}$$
but $\displaystyle\int_0^\infty \frac{x^a e^{2x}}{(e^{2x} – 1)^2}~dx$ only converges for $a > 1$ and not for all $a > 0$.
So my question is, how can the interchange between the summation and integration be justified so that it holds for all $a > 0$?
Best Answer
We can use the dominated convergence theorem to justify the interchange.
Let's denote $S_N(x)=\sum_{n=0}^N(-1)^nne^{-2nx}$.
As $N$ is fixed, $$S_N(x)=-\frac 12\frac d{dx}\sum_{n=0}^N(-1)^ne^{-2nx}$$ $$=\frac{\big(1+(N+2)(-1)^Ne^{-2x(N+1)}\big)}{1+e^{2x}}-\frac{1+(-1)^Ne^{-2x(N+1)}}{(1+e^{2x})^2}$$ It is straightforward to see that for $N>3$ $$|S_N(x)|<e^{-2x}\Big(1+(N+3)e^{-2Nx}\Big)<e^{-2x}+2Ne^{-2Nx}e^{-2x}$$ Given that $2Nx<e^{2Nx}=1+2Nx+...$ $$|S_N(x)|<(1+1/x)\,e^{-2x}\tag{1}$$ Then $$|I|=\int_0^\infty x^a|S_N(x)|dx<\int_0^\infty x^a\Big(1+\frac 1x\Big)e^{-2x}dx\tag{2}$$ It means that we constructed the dominating function $S(x)>x^a|S_N(x)|$ for all $N>3,\, a>0$
According to the DCT, we are allowed to take the limit under the integral sign: $$\int_0^\infty x^a \Big(\lim_{N\to\infty}S_N(x)\Big)dx=\lim_{N\to\infty}\int_0^\infty x^a \Big(S_N(x)\Big)dx$$