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How can I prove that $$\underset{k\geq1}{\sum}\left(\underset{m=-\infty}{\overset{\infty}{\sum}}\frac{\left(-1\right)^{m}}{\left(2k-1\right)^{2}+m^{2}}\right)=\frac{\pi\log\left(2\right)}{8}\,?$$I tried possion summation but it seems doesn't work.

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  • $\begingroup$ Do you know of Fourier series? $\endgroup$ – mickep Jan 24 '15 at 12:35
  • $\begingroup$ yes but I'm not very able to handle Fourier series... I'm an undergraduate student :) $\endgroup$ – Logger Jan 24 '15 at 12:37
  • $\begingroup$ OK, then I suggest (out of the hat), that you calculate the Fourier series of the function $\cosh(a t)$. Here $a$ is a constant that you might want to choose in a smart way. This will take care of the $m$-sum. On the other hand, you will get a not so easy sum in $k$... Edit: You could also update your question with the tools you have and in what context/the motivation to calculate your double sum. $\endgroup$ – mickep Jan 24 '15 at 12:43
  • $\begingroup$ I tried to use poisson summation but numerical test shows that it doesn't work. And now I'm stuck. $\endgroup$ – Logger Jan 24 '15 at 12:50
  • $\begingroup$ So what summation techniques have you been taught? $\endgroup$ – Mhenni Benghorbal Jan 24 '15 at 12:50
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We can exploit $$ \int_{0}^{+\infty}\frac{\sin(at)}{a}e^{-bt}\,dt=\frac{1}{a^2+b^2} $$ giving: $$\begin{eqnarray*}\sum_{m\in\mathbb{Z}}\frac{(-1)^m}{(2k-1)^2+m^2} &=& \frac{1}{(2k-1)^2}+2\sum_{m\geq 1}\int_{0}^{+\infty}\frac{\sin((2k-1)t)}{2k-1}(-1)^m e^{-mt}\,dt\\&=&\frac{1}{(2k-1)^2}-2\int_{0}^{+\infty}\frac{\sin((2k-1)t)}{2k-1}\frac{dt}{e^t+1}.\end{eqnarray*}$$ Since: $$g(t)=\sum_{k\geq 1}\frac{\sin((2k-1)t)}{2k-1}$$ is the rectangular wave whose value is $\frac{\pi}{4}$ on $(0,\pi)$, $-\frac{\pi}{4}$ on $(\pi,2\pi)$ and so on, the original series equals: $$\frac{\pi^2}{8}-2\int_{0}^{+\infty}\frac{g(t)}{e^t+1}\,dt.$$ Since: $$ \int_{k\pi}^{(k+1)\pi}\frac{dt}{e^t+1}=\log(1+e^{-k\pi})-\log(1+e^{-(k+1)\pi})$$ the final answer can be derived from the Weierstrass product for the hyperbolic cosine function.

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  • $\begingroup$ Thank you but I don't understand how use Weierstrass product... can you help me? $\endgroup$ – Logger Jan 24 '15 at 14:26
  • $\begingroup$ @user210241: in the last step, we can see that the original sum depends on a sum of logarithms of hyperbolic cosines, that is a logarithm of a product of hyperbolic cosines. By writing the hyperbolic cosine as a product, everything simplifies nicely. $\endgroup$ – Jack D'Aurizio Jan 24 '15 at 15:36

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