Integral of alternating series I was trying to answer a question about a random walk when I came across the integral
$$
\int_0^\infty \sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\left(1-e^{-(2m+1)^2/x^2}\right)\mathrm{d}x.
$$
For probabilistic reasons, I think it has a finite value. Is there a simple proof of this? Is there a way to compute or simplify the expression? If one could exchange the $\int$ with the $\sum$, then one could use that
$$\int_0^\infty 1-e^{-(2m + 1)^2 / x^2}\mathrm{d}x = (2m+1)\sqrt{\pi}.$$ 
Thanks.
 A: The integral converges to $\sqrt{\pi}/2$. Indeed, let
$$ F(x) = \int_{0}^{x} (1- e^{-1/t^2}) \, dt. $$
By the Abel's test, the series
$$ \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) $$
converges uniformly on $[0,\infty)$ (with the convention $e^{-\infty} = 0$). So we can switch the integration and summation in the following computation:
\begin{align*}
I_R
&:= \int_{0}^{R} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx \\
&\hspace{3em}= \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} \int_{0}^{R} (1 - e^{-(2m+1)^2/x^2}) \, dx\\
&\hspace{6em}= \sum_{m=0}^{\infty} (-1)^m F\left(\frac{R}{2m+1}\right).
\end{align*}
Proceeding,
\begin{align*}
I_R
&= \sum_{m=0}^{\infty} \left\{ F\left(\frac{R}{4m+1}\right) - F\left(\frac{R}{4m+3}\right) \right\} \\
&\hspace{3em}= \sum_{m=0}^{\infty} \int_{\frac{R}{4m+3}}^{\frac{R}{4m+1}} (1 - e^{-1/t^2}) \, dt \\
&\hspace{6em}= \sum_{m=0}^{\infty} \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx,
\end{align*}
where we applied the substitution $x = 1/t$ in the last line. (This substitution is not essential for our argument, but I adopted this step to make clear how monotonicity works.)
Now using the fact that the integrand is decreasing, we can bound $2I_R$ from below by
$$ 2I_R
\geq \sum_{m=0}^{\infty} \left( \int_{\frac{4m+1}{R}}^{\frac{4m+3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{4m+3}{R}}^{\frac{4m+5}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx \right)
= \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Similar idea shows that
$$ 2I_R \leq \int_{\frac{1}{R}}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx + \int_{\frac{1}{R}}^{\frac{3}{R}} \frac{1 - e^{-x^2}}{x^2} \, dx. $$
Finally, taking $R \to \infty$ proves
$$ \int_{0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx
= \lim_{R\to\infty} I_R = \frac{1}{2} \int_{0}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx = \frac{\sqrt{\pi}}{2}. $$
