On Parseval's identity and computing a special sum

Question

In Vretblad's Fourier Analysis and its Applications, I've encountered the following sum

Prove the formula $$\sum _{k=0}^{\infty }\frac{\left(-1\right)^k}{\left(2k+1\right)\left(\left(2k+1\right)^2-\alpha ^2\right)}=\frac{\pi }{4\alpha ^2}\left(\frac{1}{\cos \left(\frac{\alpha \pi \ }{2}\right)}-1\right),$$ where $\alpha\notin \mathbb Z$. (Hint: study the series on $(0,\pi/2)$ established in proving the expansion into partial fractions of the cotangent.)

The hint refers to the following problem:

Determine the the Fourier series of $\cos{(\alpha t)}$ ($|t|\leq\pi$), where $\alpha$ is a complex number but not an integer. Use this to verify $$\pi\cot{( \alpha \pi)}=\sum_{k=-\infty}^\infty \frac{1}{\alpha-k}.$$

I have solved this problem. The Fourier coefficients of $\cos{(\alpha t)}$ ($|t|\leq\pi$) are \begin{align} a_0&=\frac{2\sin(\alpha\pi)}{\alpha\pi},\\ a_n&=\frac{2\alpha\sin(\alpha\pi)(-1)^{n+1}}{\pi(k^2-\alpha^2)} \quad n\geq 1.\end{align} So $$\cos{(\alpha t)}=\frac{\sin(\alpha \pi)}{\alpha \pi }-\sum_{k=1}^\infty \frac{2\alpha \sin(\alpha \pi)(-1)^k}{\pi(k^2-\alpha^2)}\cos(kt).$$ If you plug in $t=\pi$ in the previous equation and rearrange a bit, you arrive at $$\pi\cot{( \alpha \pi)}=\sum_{k=-\infty}^\infty \frac{1}{\alpha-k}.$$ But I'm stuck at verifying the formula stated at the top. This problems appears in a section that introduces the Parseval formula, $$\frac1{\pi}\int_{\mathbb T}|f(t)|^2dt=\frac12|a_0|^2+\sum_{n=1}^\infty |a_n|^2,\tag{*}$$ so I'm pretty sure it has something to do with this formula, but the fact that $\alpha$ is complex confuses me. For instance, is it correct that $|f(t)|^2$ is simply $\cos^2(\alpha t)$? In that case \begin{align} \frac1{\pi}\int_{\mathbb T}|f(t)|^2dt&=\frac1{\pi}\int_{-\pi}^\pi\left(\frac{1+\cos{(2\alpha t)}}{2}\right) dt \\ &=1+\frac{\sin{(2\alpha \pi)}}{2\alpha \pi}. \end{align} For the RHS of (*) I get (again, not sure $|\sin(\alpha \pi)|^2=\sin^2(\alpha \pi)$) \begin{align} |a_0|^2&=\frac{4\sin^2(\alpha \pi)}{|\alpha|^2\pi} \\ |a_n|^2&=\frac{4|\alpha|^2 \sin^2(\alpha \pi)}{\pi^2|k^2-\alpha^2|^2}. \end{align} I'm not sure I'm going in the right direction. I see no pattern. Grateful for any help.

Answer 1

This is about integration, my bad. Integrate the Fourier series of $\cos{(\alpha t)}$ from $0$ to $\pi/2$ and you will obtain the desired formula at the top.

Answer 2

Too long for a comment

Given that
$$S=\sum _{k=0}^\infty\frac{\left(-1\right)^k}{\left(2k+1\right)\left(\left(2k+1\right)^2-\alpha ^2\right)}=\frac12\sum _{k=-\infty}^\infty \frac{\left(-1\right)^k}{\left(2k+1\right)\left(\left(2k+1\right)^2-\alpha ^2\right)}$$ and considering the integral in the complex plane along a big circle of the radius $R\to \infty$ $$I_R=\oint_{C_R}\frac{1}{\left(2z+1\right)\left(\left(2z+1\right)^2-\alpha ^2\right)}\frac\pi{\sin\pi z}dz$$ $$\to2\pi i\sum\operatorname{Res}\frac{1}{\left(2z+1\right)\left(\left(2z+1\right)^2-\alpha ^2\right)}\frac\pi{\sin\pi z}\,\to 0\,\,\text{at}\,\,R\to\infty$$ The residues at $k=0,\pm1,\pm2,...\,$ give $\,2S$; together with the residues at the other poles $$2\pi i\left(2S+\underset{z=-\frac12}{\operatorname{Res}}\frac{1}{\left(2z+1\right)\left(\left(2z+1\right)^2-\alpha ^2\right)}\frac\pi{\sin\pi z}+\underset{z=-\frac12\pm\frac\alpha2}{\operatorname{Res}}\frac{1}{\left(2z+1\right)\left(\left(2z+1\right)^2-\alpha ^2\right)}\frac\pi{\sin\pi z}\right)=0$$ $$\Rightarrow\,2S=-\underset{\binom{z=-\frac12\pm\frac\alpha2}{z=-\frac12}}{\operatorname{Res}}\frac{1}{\left(2z+1\right)\left(\left(2z+1\right)^2-\alpha ^2\right)}\frac\pi{\sin\pi z}$$ $$2S=-\frac\pi2\frac1{\sin\big(-\frac\pi2\big)}\frac1{-\alpha^2}-\frac\pi{\sin\pi\big(-\frac12+\frac\alpha2\big)}\frac1{2\alpha}\frac1{2\alpha}-\frac\pi{\sin\pi\big(-\frac12-\frac\alpha2\big)}\frac1{-2\alpha}\frac1{-2\alpha}$$ $$=-\frac\pi{2\alpha^2}+\frac\pi{2\alpha^2}\frac1{\cos\frac{\pi\alpha}2}$$