The function $tanh(\pi z)$ has its poles at the points $i(n+\frac{1}{2})$ for $n \in \mathbb{Z}$. Now I want to take an $\epsilon$ circle around such a pole and contour integrate the function $z\text{ }tanh(\pi z) log(z^2 + a^2)$ around it in the limit $\epsilon \rightarrow 0$ (and $a >0$)
Basicaly I want to compute $lim_{\epsilon \rightarrow 0} [ \int _{\phi = -\pi} ^{\pi} z\text{ }tanh(\pi z) log(z^2 + a^2) dz]$ for $z = i(n+\frac{1}{2}) + \epsilon e^{i\phi}$
Is this a meaningful calculation?
I tried computing this quantity on Wolfram alpha but I couldn't get any answer,
Series [ (Ix + yexp(Ip))(tanh (pi*(Ix + yexp(Ip))) )(log (( Ix + yexp(Ip) )^2 + a^2))(Iyexp(I*p)), {y,0,2}, {Assuming x>0,y>0,a>0, p real}]
- If I understand Mathematica correctly it is evaluating,
$$lim_{\epsilon \rightarrow 0} [ \int _{\phi = -\pi} ^{\pi} z\text{ }tanh(\pi z) log(z^2 + a^2) dz] \text{ for }z = i(n+\frac{1}{2}) + \epsilon e^{i\phi} = (1+2n)log[a^2 - (n +\frac{1}{2})^2]$$
Am I using Mathematica correctly?
Let's look at $i/2.$ I assume $a>1/2$ and that $\log z$ is the principal branch. Then $\log (a^2+z^2)$ is analytic in a deleted neighborhood of $i/2.$ By the residue theorem, for small $\epsilon$ your integral is $2\pi i$ times the residue of the integrand at $i/2.$ Finding the residue is not too bad because $\cosh z$ has a zero of order $1$ at $i/2.$ The residue equals $$\lim_{z\to i/2} \frac{(z-i/2)z\sinh \pi z \log (a^2+z^2)}{\cosh \pi z}.$$ Now $(z-i/2)/\cosh \pi z \to 1/[\pi \sinh (\pi i/2)].$ Thus the above limit is $(i/2)\log (a^2-1/4))/\pi.$ Multiply that by $2\pi i$ to get $-\pi \log (a^2 - 1/4).$ My answer is different from the one you posted.