A way to justify interchanging a summation and integration

Question

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 $a > 0$?

Answer 1

Note that:

$$\sum_{n = 1}^k (-1)^n n e^{-2nx}=\frac{-e^{2x}}{(1+e^{2x})^2}+\frac{(-1)^k}{e^{2(k-1)x}(1+e^{2x})^2}+\frac{(-1)^kk}{e^{2kx}(1+e^{2x})}$$ Since $k\ge1$ and $\displaystyle\frac k{e^{2kx}}\le\frac k{2kx}=\frac1{2x}$, hence

$$\left|\sum_{n = 1}^k (-1)^n n e^{-2nx}\right|\le\frac{e^{2x}}{(1+e^{2x})^2}+\frac{1}{(1+e^{2x})^2}+\frac{1}{2x(1+e^{2x})}$$

further,

$$\left|x^a\cdot\sum_{n = 1}^k (-1)^n n e^{-2nx}\right|\le\frac{x^ae^{2x}}{(1+e^{2x})^2}+\frac{x^a}{(1+e^{2x})^2}+\frac{x^{a-1}}{2(1+e^{2x})}$$

Apparently, the RHS is convergent for $a>0$, hence we can switch the summation and integration.

Answer 2

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: $$\int_0^\infty x^a \Big(\lim_{N\to\infty}S_N(x)\Big)dx=\lim_{N\to\infty}\int_0^\infty x^a \Big(S_N(x)\Big)dx$$