Showing $\int_{-\infty}^\infty e^{-tu}\frac{\sinh(\frac{u}2(\pi-2\gamma))}{\sinh(\frac{u}2\pi)}du=\frac{\sin2\gamma}{\sin(\gamma+t)\sin(\gamma-t)}$

Question

For $-\gamma<t<\gamma$ and $0<\gamma<\frac{\pi}{2}$, I want to verify the identity

$$\int_{-\infty}^{\infty} e^{-t \lambda} \frac{\sinh\left(\frac{\lambda}{2}(\pi-2\gamma)\right)}{\sinh\left(\frac{\lambda}{2}\pi\right)} \mathrm{d} \lambda = \frac{\sin(2\gamma)}{\sin(\gamma+t)\sin(\gamma-t)}$$ using elementary methods if possible, and I am also curious if this can be done using bilateral Laplace Transform?

To that end, I want to know if we can use the inverse Laplace Transform to solve the problem in the reverse direction, that is given the function $g(t,\gamma):=\frac{\sin(2\gamma)}{\sin(\gamma+t)\sin(\gamma-t)}$, determine $f(\lambda, \gamma)$ such that $$\int_{-\infty}^{\infty} e^{-t \lambda} f(\lambda,\gamma) \mathrm{d} \lambda = g(t,\gamma)$$ Thanks for your help in advance.

Added later Regarding the first part of my question: Notice that the requirements on $\gamma$ and $t$ ensure that the integrand $e^{-t \lambda} \frac{\sinh\left(\frac{\lambda}{2}(\pi-2\gamma)\right)}{\sinh\left(\frac{\lambda}{2}\pi\right)}$ has exponential decay as $\lambda \to \pm \infty$, ensuring the convergence of the integral.

Answer 1

The given integral is $\int_{-\infty}^\infty e^{-t\lambda}f(\gamma,\lambda)\,d\lambda=\int_0^\infty(e^{t\lambda}+e^{-t\lambda})f(\gamma,\lambda)\,d\lambda$ where, for $\lambda>0$, $$f(\gamma,\lambda)=\frac{\sinh\frac\lambda2(\pi-2\gamma)}{\sinh\frac\lambda2\pi}=\frac{e^{-\gamma\lambda}-e^{-(\pi-\gamma)\lambda}}{1-e^{-\pi\lambda}}=e^{-\gamma\lambda}-\sum_{n=1}^\infty(e^{-(n\pi-\gamma)\lambda}-e^{-(n\pi+\gamma)\lambda}),$$ and termwise integration of the sum is justified by positivity of its terms, so that $$\int_{-\infty}^\infty e^{-t\lambda}f(\gamma,\lambda)\,d\lambda=\frac{2\gamma}{\gamma^2-t^2}-2\sum_{n=1}^\infty\left(\frac{\gamma-t}{n^2\pi^2-(\gamma-t)^2}+\frac{\gamma+t}{n^2\pi^2-(\gamma+t)^2}\right),$$ which is $\cot(\gamma-t)+\cot(\gamma+t)$ thanks to $\cot z=1/z-2\sum_{n=1}^\infty z/(n^2\pi^2-z^2)$.

Answer 2

There is an antiderivative $$I=\int e^{-t \lambda}\, \frac{\sinh\left(\frac{\lambda}{2}(\pi-2\gamma)\right)}{\sinh\left(\frac{\lambda}{2}\pi\right)} \,d\lambda $$

$$\lambda=\frac{2 \log (x)}{\pi }\implies I=\frac 2\pi \int \frac{x^{-\frac{2 t}{\pi }} \left(x^{\frac{\pi -2 \gamma }{\pi }}-x^{-\frac{\pi -2 \gamma }{\pi }}\right)}{(x^2-1)} \,dx$$

$$J=\int \frac{x^a}{(x^2-1)} \,dx=-\frac{x^{a+1} }{a+1}\,\, _2F_1\left(1,\frac{a+1}{2};\frac{a+3}{2};x^2\right)$$ So, there is no problem to compute in terms of the Gaussian hypergeometric function

$$K=\int_{-a}^{+a} e^{-t \lambda}\, \frac{\sinh\left(\frac{\lambda}{2}(\pi-2\gamma)\right)}{\sinh\left(\frac{\lambda}{2}\pi\right)} \,d\lambda $$ Using the conditions, if $a\to \infty$, the result is $$\int_{-\infty}^{+\infty} e^{-t \lambda}\, \frac{\sinh\left(\frac{\lambda}{2}(\pi-2\gamma)\right)}{\sinh\left(\frac{\lambda}{2}\pi\right)} \,d\lambda=\cot (t+\gamma )-\cot (t-\gamma ) $$ which is your expected result.