Open Problem (Calculus - 1)

The third one is equal to $\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right)$, where csch is the hyperbolic cosecant. The first step is to take $\frac{(-1)^n}{n^2+\gamma^2} = \sum_{k=0}^{\infty} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}}$ We can then switch the order to $\lim_{a\to\infty}\,\sum_{k=0}^{\infty} \sum_{n=-a}^{a} \frac{(-1)^{n+k} \gamma^{2k}}{n^{2k+2}}$ to get $\sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k}$ This of course neglects the $\frac{(-1)^0}{0^2+\gamma^2}$ term (which is undefined for $n = 0$ in the denominator of the summand), so we must add $\frac{1}{\gamma^2}$ to our previous sum to get $\frac{1}{\gamma^2} + \sum_{k=0}^{\infty} (-1)^{k-1} \left(2-4^{-k}\right) \zeta (2 k+2)\gamma^{2k}$ We can see the relation between these coefficients and the coefficients of the series for the hyperbolic cosecant to get that our answer should be $\boxed{\frac{\pi}{\gamma} \text{csch}\left(\gamma\pi\right)}$.

Quick sketch : the first two can be done using complex analysis just consider the integral over a half circle from $R$ to $-R$ counterclockwise and $(-R,R)$.

For the sum, consider the integral of $z\mapsto \frac{\pi \csc(\pi z)}{z^2+x^2}$ over a rectangle with vertices $(n+1/2)\left\{1+i,1-i,-1+i,-1-i\right\}$.

I can't write full solutions now, but I'll write when have time.

Problem No 2

Answer is $\displaystyle \boxed{\frac{\pi e^{-\beta}}{6} + \frac{\pi}{\sqrt3}e^{-\beta/2}\left(\sin(\frac{\sqrt3 \beta}{2})+\cos(\frac{\sqrt3 \beta}{2})\right)}$

please anybody solve the integral using the residue theorem

Problem No 1

Answer is

$\displaystyle \boxed{\frac{2\pi}{\sqrt3}e^{-\frac{\sqrt3 \alpha}{2}}\sin(\frac{\alpha}{2})}$