I am having some trouble obtaining the correct answer to the following problem. I would appreciate it if someone could point out where I went wrong in my computations.
This problem occurs as exercise 1020 in A Collection of Problems On Complex Analysis by L. I. Volkovyskiĭ, et. al.
Problem:
Prove that
$$\sum_{n=1}^{\infty}\frac{n}{(n^2-3)\sqrt{4n^2-3}} = \int_{0}^{\sqrt{3}/2}\frac{x\cot(\pi x)}{(3-x^2)\sqrt{3-4x^2}}\mathrm{d}x+\frac{1}{6}\cot(\pi(2-\sqrt{3}))$$
Hint: Use the integral
\begin{equation} \int_C\frac{z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z \end{equation}
over the contour $C$ bounding the doubly connected domain given in figure
where the small circular arcs are centered around $z=\pm\sqrt{3}/2$ and the large circular arc is of radius $n/2$ centered at the origin. (The book doesn't actually specify the orientation of the contours.)
Attempt at a Solution:
First of all, I don't believe the book's suggested radius of the outer circle is correct if you take $n$ to be an arbitrary integer, as $\cot(\pi z)$ is not bounded on $\vert{z}\vert=k$ for $k\in\mathbb{Z}$. I'm also not sure why they write $\cot(\pi(2-\sqrt{3}))$ instead of $-\cot(\pi\sqrt{3})$.
The answer I obtain is close but off by a factor of $\pi$ for the last term and $2\pi$ for the integral term. In particular, I get
\begin{equation} \sum_{k=1}^{\infty}\frac{k}{(k^2-3)\sqrt{4k^2-3}} = 2\pi \int_{0}^{\sqrt{3}/2}\frac{t\cot(\pi t)}{(3-t^2)\sqrt{3-4t^2}}\mathrm{d}t - \frac{\pi}{6}\cot(\pi\sqrt{3}) \end{equation}
The factor of pi arises because the residues of $\cot(\pi z)$ at the integers is $1/\pi$ while the factor of two on the integral comes from the fact that integrating over the dog bone contour results in two integrals from $-\sqrt{3}/2$ to $\sqrt{3}/2$, one each from the bottom and top segments, and these simplify to a single integral over $0$ to $\sqrt{3}/2$ multiplied by 4, due to the integrand being an even function.
I would love to know where I went wrong!
Proof:
Diverging from the hint slightly, we consider the integral
\begin{equation} \int_C\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z \end{equation}
over the contour bounding the doubly connected domain. To integrate over the doubly connected domain we will use the following fact.
Proposition: If $D$ is a doubly connected domain with boundary curves $C_1$ and $C_2$ oriented in the same direction and if f is analytic in a domain containing $D$ and its boundary, then
\begin{equation} \int_{C_1}f(z)\mathrm{d}z = \int_{C_2}f(z)\mathrm{d}z \end{equation}
Before beginning with integration, we need to select a branch of $\sqrt{4z^2-3}$ which is analytic in the complex plane slit from $-\sqrt{3}/2$ to $\sqrt{3}/2$. To do this we may factor the polynomial in the square root and choose a branch for each of the square roots of linear factors. We define the branch of $\sqrt{2z-\sqrt{3}}$ to be such that $-\pi<\arg{(2z-\sqrt{3})}\leq \pi$ and the branch of $\sqrt{2z+\sqrt{3}}$ to be such that $-\pi<\arg{(2z+\sqrt{3})}\leq \pi$. Recall that the complex logarithm $\log(z)$ is defined as
\begin{equation} \log(z)=\log(\vert{z}\vert)+i\arg(z) \end{equation}
Thus the complex square root function is given by
\begin{equation} \sqrt{z}=e^{\frac{1}{2}(\log(\vert{z}\vert)+i\arg(z))}=\sqrt{\vert{z}\vert}e^{i\frac{1}{2}\arg(z)} \end{equation}
This shows that
\begin{equation} \sqrt{4z^2-3} = \sqrt{\vert{4z^2-3}\vert}e^{i\frac{1}{2}(\arg{(2z-\sqrt{3})}+\arg{(2z+\sqrt{3})})} \end{equation}
where $-2\pi < \arg{(2z-\sqrt{3})}+\arg{(2z+\sqrt{3})} \leq 2\pi$. We would like to show that $\sqrt{4z^2-3}$ is continuous across the branch cut below $z=-\sqrt{3}/2$. To do this let $z = -t$ where $t>\sqrt{3}/2$ and consider the points $-t+i\varepsilon$ and $-t-i\varepsilon$ above and below the real axis for $\varepsilon$ real and tending to $0$. For the point above we see that
\begin{align} \lim_{\varepsilon\to 0}\sqrt{4(-t+i\varepsilon)^2-3} &= \lim_{\varepsilon\to 0}\sqrt{\vert{4(-t+i\varepsilon)^2-3}\vert}e^{i\frac{1}{2}(\arg{(2(-t+i\varepsilon)-\sqrt{3})}+\arg{(2(-t+i\varepsilon)+\sqrt{3})})} \\ &= \sqrt{4t^2-3}e^{i\frac{1}{2}(\pi+\pi)} \\ &= -\sqrt{4t^2-3} \end{align}
For the point below we have
\begin{align} \lim_{\varepsilon\to 0}\sqrt{4(-t-i\varepsilon)^2-3} &= \lim_{\varepsilon\to 0}\sqrt{\vert{4(-t-i\varepsilon)^2-3}\vert}e^{i\frac{1}{2}(\arg{(2(-t-i\varepsilon)-\sqrt{3})}+\arg{(2(-t-i\varepsilon)+\sqrt{3})})} \\ &= \sqrt{4t^2-3}e^{i\frac{1}{2}(-\pi-\pi)} \\ &= -\sqrt{4t^2-3} \end{align}
Thus $\sqrt{4z^2-3}$ is in fact continuous across the branch cut. To show that $1/\sqrt{4z^2-3}$ is holomorphic in $D$ we integrate along any circle in $D$, splitting it in half along the real line. By continuity the integrals along the segments on the branch cut annihilate each other, and since the integrals along the half circles are 0 the integral along the whole circle vanishes as well. Thus, by Morera's Theorem $1/\sqrt{4z^2-3}$ is holomorphic in $D$.
Let us denote the circular contour of radius $N+1/2$ as $C_N$ and denote the interior dog bone contour by $D$. The the above proposition shows that
\begin{equation} \int_{C_N}f(z)\mathrm{d}z = \int_{D}f(z)\mathrm{d}z \end{equation}
if we traverse each contour in the counterclockwise direction. Beginning with the circular contour, by parameterizing we see that
\begin{equation} \int_{C_N}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z = \pi i(N+1/2)^2\int_{0}^{2\pi}\frac{e^{2it}\cot(\pi (N+1/2)e^{it})}{((N+1/2)^2e^{2it}-3)\sqrt{4(N+1/2)^2e^{2it}-3}}\mathrm{d}t \end{equation}
Taking the absolute value of both sides we have
\begin{align} \bigg\vert{\int_{C_N}\frac{\pi z\cot(z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z}\bigg\vert &\leq \pi (n+1/2)^2\bigg\vert{\int_{0}^{2\pi}\frac{e^{2it}\cot(\pi (N+1/2)e^{it})}{((N+1/2)^2e^{2it}-3)\sqrt{4(N+1/2)^2e^{2it}-3}}\mathrm{d}t}\bigg\vert \\ &\leq \pi(n+1/2)^2\coth^2(\pi/2)\int_{0}^{2\pi}\bigg\vert{\frac{1}{((N+1/2)^2e^{2it}-3)\sqrt{4(N+1/2)^2e^{2it}-3}}}\bigg\vert\mathrm{d}t \\ &\leq \frac{\pi^2\coth^2(\pi/2)}{n+1/2} \end{align}
Thus
\begin{equation} \lim_{N\to\infty}\int_{C_N}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z = 0 \end{equation}
On the other hand the contour encloses the poles at $z=\pm \sqrt{3}$ as well as the poles of cotangent at $z\in \mathbb{Z}$ except for $z = 0$. Therefore the residue theorem shows that
\begin{equation} \frac{1}{2\pi i}\int_{C_N}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z = \sum_{k=-N}^{N}\underset{{z=k}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) + \underset{{z=\pm \sqrt{3}}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) \end{equation}
With this in mind we first evaluate the residues at $\pm \sqrt{3}$, finding that
\begin{align} \underset{{z=\sqrt{3}}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) &= \lim_{z\to\sqrt{3}}(z-\sqrt{3})\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}} \\ &= \lim_{z\to\sqrt{3}}\frac{\pi z\cot(\pi z)}{(z+\sqrt{3})\sqrt{\vert{4z^2-3}\vert}e^{i\frac{1}{2}(\arg{(2z-\sqrt{3})}+\arg{(2z+\sqrt{3})})}} \\ &= \frac{\pi}{6}\frac{\cot(\pi\sqrt{3})}{e^{i\frac{1}{2}(\arg{(\sqrt{3})}+\arg{(3\sqrt{3})})}} \\ &= \frac{\pi}{6}\frac{\cot(\pi\sqrt{3})}{e^{i\frac{1}{2}(0+0)}} \\ &= \frac{\pi}{6}\cot(\pi\sqrt{3}) \end{align}
Likewise the residue at $z=-\sqrt{3}$ is given by
\begin{align} \underset{{z=-\sqrt{3}}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) &= \lim_{z\to-\sqrt{3}}(z+\sqrt{3})\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}} \\ &= \lim_{z\to-\sqrt{3}}\frac{\pi z\cot(\pi z)}{(z-\sqrt{3})\sqrt{\vert{4z^2-3}\vert}e^{i\frac{1}{2}(\arg{(2z-\sqrt{3})}+\arg{(2z+\sqrt{3})})}} \\ &= \frac{\pi}{6}\frac{\cot(-\pi\sqrt{3})}{e^{i\frac{1}{2}(\arg{(-3\sqrt{3})}+\arg{(-\sqrt{3})})}} \\ &= -\frac{\pi}{6}\frac{\cot(\pi\sqrt{3})}{e^{i\frac{1}{2}(\pi+\pi)}} \\ &= \frac{\pi}{6}\cot(\pi\sqrt{3}) \end{align}
Similarly evaluating the residues at $z = \pm k$ for $k \in \mathbb{Z}_{>0}$ we find that
\begin{equation} \underset{{z=k}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) = \frac{k}{(k^2-3)\sqrt{4k^2-3}} \end{equation}
and
\begin{equation} \underset{{z=-k}}{\mathrm{res}}\left(\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\right) = \frac{k}{(k^2-3)\sqrt{4k^2-3}} \end{equation}
Therefore, we find that
\begin{equation} \frac{1}{2\pi i}\int_{C_N}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z = 2\sum_{k=1}^{N}\frac{k}{(k^2-3)\sqrt{4k^2-3}} + \frac{\pi}{3}\cot(\pi\sqrt{3}) \end{equation}
Taking the limit as $N\to \infty$ and using Cauchy's Theorem for multiply connected domains gives
\begin{equation} 2\sum_{k=1}^{\infty}\frac{k}{(k^2-3)\sqrt{4k^2-3}} + \frac{\pi}{3}\cot(\pi\sqrt{3}) = \int_{D}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z \end{equation}
Next we break up the dog bone contour into four parts
\begin{equation} \int_{D}=\int_{C_{\varepsilon}^{-}}+\int_{\Gamma_{\varepsilon}^{-}}+\int_{\Gamma_{\varepsilon}^{+}}+\int_{C_{\varepsilon}^{+}} \end{equation}
Where $C_{\varepsilon}^{-}$ and $C_{\varepsilon}^{+}$ denotes the circular arcs of radius $\varepsilon$ traversed in the counter clockwise direction and centered at $-\sqrt{3}/2$ and $\sqrt{3}/2$ respectively, and $\Gamma_{\varepsilon}^{-}$ and $\Gamma_{\varepsilon}^{+}$ are the lines a distance of $\varepsilon$ below and above the real line respectively.
It is not difficult to show that the integrals along the small circular arcs $C_{\varepsilon}^{-}$ and $C_{\varepsilon}^{+}$ vanish as $\varepsilon \to 0$. On the other hand the integrals along the lines above and below the real axis can be evaluated as
\begin{align} \lim_{\varepsilon \to 0}\int_{\Gamma_{\varepsilon}^{+}}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z &= \lim_{\varepsilon \to 0}\int_{\sqrt{3}/2}^{-\sqrt{3}/2}\frac{\pi (t+i\varepsilon)\cot(\pi (t+i\varepsilon))}{((t+i\varepsilon)^2-3)\sqrt{4(t+i\varepsilon)^2-3}}\mathrm{d}t \\ &= -2i\int_{0}^{\sqrt{3}/2}\frac{\pi t\cot(\pi t)}{(t^2-3)\sqrt{4t^2-3}}\mathrm{d}t \end{align}
and
\begin{align} \lim_{\varepsilon \to 0}\int_{\Gamma_{\varepsilon}^{-}}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z &= \lim_{\varepsilon \to 0}\int_{-\sqrt{3}/2}^{\sqrt{3}/2}\frac{\pi (t-i\varepsilon)\cot(\pi (t-i\varepsilon))}{((t-i\varepsilon)^2-3)\sqrt{4(t-i\varepsilon)^2-3}}\mathrm{d}t \\ &= -2i\int_{0}^{\sqrt{3}/2}\frac{\pi t\cot(\pi t)}{(t^2-3)\sqrt{4t^2-3}}\mathrm{d}t \end{align}
Where we have used the fact that for $-\sqrt{3}/2<x<\sqrt{3}/2$ we have on the line above the real axis
\begin{align} \lim_{\varepsilon \to 0}\sqrt{\vert{4(-x+i\varepsilon)^2-3}\vert}e^{i\frac{1}{2}(\arg{(2(-x+i\varepsilon)-\sqrt{3})}+\arg{(2(-x+i\varepsilon)+\sqrt{3})})} &= \sqrt{4x^2-3}e^{i\frac{1}{2}(\pi+0)} \\ &= i\sqrt{4x^2-3} \end{align}
Likewise below the real axis we find that
\begin{align} \lim_{\varepsilon \to 0}\sqrt{\vert{4(-x-i\varepsilon)^2-3}\vert}e^{i\frac{1}{2}(\arg{(2(-x-i\varepsilon)-\sqrt{3})}+\arg{(2(-x-i\varepsilon)+\sqrt{3})})} &= \sqrt{4x^2-3}e^{i\frac{1}{2}(-\pi+0)} \\ &= -i\sqrt{4x^2-3} \end{align}
Putting this all together the integral over the contour $D$ becomes
\begin{align} \int_{D}\frac{\pi z\cot(\pi z)}{(z^2-3)\sqrt{4z^2-3}}\mathrm{d}z = -4i\int_{0}^{\sqrt{3}/2}\frac{\pi t\cot(\pi t)}{(t^2-3)\sqrt{4t^2-3}}\mathrm{d}t \end{align}
giving in the end
\begin{equation} \sum_{k=1}^{\infty}\frac{k}{(k^2-3)\sqrt{4k^2-3}} = 2\int_{0}^{\sqrt{3}/2}\frac{\pi t\cot(\pi t)}{(3-t^2)\sqrt{3-4t^2}}\mathrm{d}t - \frac{\pi}{6}\cot(\pi\sqrt{3}) \end{equation}