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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.

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  • $\begingroup$ Do you know how to determine whether it's absolutely convergent or not? $\endgroup$ – Michael McGovern Apr 16 '17 at 19:35
  • $\begingroup$ Wait if you do switch the integral and summation, don't you then end up with $\sum (-1)^m\sqrt{\pi}$ which obviously diverges. $\endgroup$ – mathworker21 Apr 16 '17 at 19:36
  • $\begingroup$ This is sort of like a theta function. Perhaps splitting the integral at 1 and applying the x to 1/x theta function transformation might work. $\endgroup$ – marty cohen Apr 16 '17 at 19:46
  • $\begingroup$ @MichaelMcGovern I think it's not absolutely convergent. $\endgroup$ – Ben Derrett Apr 16 '17 at 20:19
  • $\begingroup$ @mathworker21 Yes, but what justifies switching the integral and summation? $\endgroup$ – Ben Derrett Apr 16 '17 at 20:20
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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

$$ \int_{0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} (1 - e^{-(2m+1)^2/x^2}) \, dx = \lim_{R\to\infty} I_R = \frac{1}{2} \int_{0}^{\infty} \frac{1 - e^{-x^2}}{x^2} \, dx = \frac{\sqrt{\pi}}{2}. $$

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