Solve equation about infinite integral. Find out all solutions that exist

Question

The equation of $x$ says: $$\frac{1}{2(x-1)}=\int_0^\infty\frac{\sin (x\arctan y)}{(1+y^2)^{x/2}(e^{y\pi}+1)}\,dy$$ I've tried using variable substitution $y=\tan\theta$, and it becomes: $$\frac{1}{2(x-1)}=\int_0^{\pi/2}\frac{\sin(\theta\cdot x)\sec^2\theta}{(\sec^2\theta)^{x/2}(1+e^{\pi\cdot\tan\theta})}d\theta$$ After that I got stuck. There may be another approach to solve this problem by introducing some kind of complex variable functions, but it seems really tough to deal with the exponential term. So can anyone help me out? Thanks!

Answer 1

A partial answer. The integral can be written as the imaginary part of

$$ \int_{0}^{+\infty}\frac{1}{\left(e^{\pi y}+1\right)(1-iy)^x}\,dy $$ and we may notice that assuming $x>1$ $$ \frac{1}{2(x-1)}=\text{Im}\int_{0}^{+\infty}\frac{1}{2(1-iy)^x}\,dy $$ hence the given identity occurs at the values of $x>1$ such that $$ \text{Im}\int_{0}^{+\infty}\frac{\tanh\left(\frac{\pi y}{2}\right)}{(1-iy)^x}\,dy = 0\quad \Longleftrightarrow \quad \int_{0}^{+\infty}\frac{\tanh\left(\frac{\pi y}{2}\right)}{(1-iy)^x}\,dy\in\mathbb{R}$$ or $$ \sum_{n\geq 0}\int_{\mathbb{R}}\frac{y}{(2n+1)^2+y^2}\left(\frac{1}{(1-iy)^x}-\frac{1}{(1+iy)^x}\right)\,dy\in\mathbb{R}, $$ but the computation of the residues of $\frac{y}{(2n+1)^2+y^2}\left(\frac{1}{(1-iy)^x}-\frac{1}{(1+iy)^x}\right)$ at $y=i$ and $y=(2n+1)i$ easily contradicts the previous line. This proves we have no solutions for $x>1$.