# How to evaluate this infinite sum using Fourier series?

Is there any way to solve the seris $S=\displaystyle\sum_{n=0}^{+\infty}\frac{(-1)^n}{a^2+(2n+1)^2}$ where $a\neq 0$?
I tried using Fourier series but I could only evaluate $\displaystyle\sum_{n=0}^{+\infty}\frac{(-1)^n}{a^2+n^2}$ and $\displaystyle\sum_{n=0}^{+\infty}\frac{(-1)^n}{a^2+(2n)^2}.$

I need hints or suggestions if there are other methods.

• en.wikipedia.org/wiki/Mittag-Leffler%27s_theorem Apr 16, 2018 at 23:01
• how you can evaluate this $$\displaystyle\sum_{n=0}^{\infty^{+}}\frac{(-1)^n}{a^2+(2n)^2}$$? Your series is kinda similar, probably from the series above you can guess what kind of function produces such Fourier coefficients. Apr 16, 2018 at 23:44

If you are able to show that $$\sum_{n\geq 0}\frac{(-1)^n}{a^2+n^2} = \frac{1}{2a^2}+\frac{\pi}{2a\sinh(\pi a)} \tag{1}$$ $$\sum_{n\geq 0}\frac{(-1)^n}{a^2+(2n)^2} = \frac{1}{2a^2}+\frac{\pi}{4a\sinh(\pi a/2)} \tag{2}$$ for instance through the Fourier sine series of $\sinh$ over $(-\pi,\pi)$, then by considering the difference between $(1)$ and $(2)$: $$\sum_{n\geq 0}\frac{\color{red}{1}}{a^2+(2n+1)^2} = \frac{\pi\tanh(\pi a/2)}{4a} \tag{3}$$ but if the last $\color{red}{1}$ is replaced by $(-1)^n$ the outcome is not as elementary. We have $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(s)-\psi(1-s)=-\pi\cot(\pi s)\tag{4}$$ where $\psi(s)=\frac{d}{ds}\log\Gamma(s)$, hence $$\sum_{n\geq 0}\frac{\color{red}{(-1)^n}}{a^2+(2n+1)^2}=\frac{1}{4a}\,\text{Im}\left[\psi\left(\tfrac{1+ia}{4}\right)-\psi\left(\tfrac{3+ia}{4}\right)\right]\tag{5}$$ which essentially has a nice closed form (Catalan's constant $G$) only for $a\to 0^+$.
Note that the digamma function can be written as $$\psi(x)=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag1$$ where $\gamma$ is the Euler-Mascheroni constant. Thus, \begin{align} \sum_{n=0}^\infty\frac{(-1)^n}{a^2+(2n+1)^2} &=\frac1{2ia}\sum_{n=0}^\infty(-1)^n\left(\frac1{2n+1-ia}-\frac1{2n+1+ia}\right)\tag2\\ &=\frac1{4ia}\sum_{n=0}^\infty(-1)^n\left(\frac1{n+\frac{1-ia}2}-\frac1{n+\frac{1+ia}2}\right)\tag3\\ &=\frac1{4ia}\sum_{n=0}^\infty\left(\color{#C00}{\frac1{n+\frac{1-ia}4}-\frac1{n+\frac{1+ia}4}}\color{#090}{-\frac1{n+\frac{1-ia}2}+\frac1{n+\frac{1+ia}2}}\right)\tag4\\[6pt] &=\frac1{4ia}\left(\psi\left(\frac{1+ia}4\right)-\psi\left(\frac{1-ia}4\right)+\psi\left(\frac{1-ia}2\right)-\psi\left(\frac{1+ia}2\right)\right)\tag5 \end{align} Explanation:
$(2)$: partial fractions
$(3)$: put $(2)$ into the form of $(1)$
$(4)$: an alternating sum is $\color{#C00}{\text{twice the even terms}}$ $\color{#090}{\text{minus all the terms}}$
$(5)$: apply $(1)$