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 $$\int_0^\infty 1-e^{-(2m + 1)^2 / x^2}\mathrm{d}x = (2m+1)\sqrt{\pi}.$$


  • $\begingroup$ 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

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*}


\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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.