The following inequality: $$\frac{\pi k}{\sinh{(\pi k)}}\;|J_{ik}(\tau)|^2\le 1,\;\;\;k,\tau\ge 0,$$ for Bessel function $J_{ik}(\tau)$, I found in http://link.springer.com/article/10.1134%2F1.558677 (Rindler solutions and their physical interpretation, by A.I. Nikishov and V.I. Ritus), where it is stated without a proof and it is said that the inequality "was not found in mathematical literature". How this inequality can be proved? What other papers it was considered in?

Here is a proof:

Use, e.g. formula 10.9.4 in DLMF, to write $$ |J_{i k}(\tau)|^2 = \frac{4}{\pi} \left|\exp\left\{i k \ln \frac{\tau}{2}\right\}\right|^2 \frac{1}{\left|\Gamma\left(i k + \frac{1}{2}\right)\right|^2}\left|\int_{0}^{1} dt (1-t^2)^{i k -\frac{1}{2}} \cos(\tau t)\right|^2. $$ Then use, e.g. 5.4.4 in DLMF, to simplify the $\Gamma$-function $$ |J_{i k}(\tau)|^2 = \frac{4}{\pi^2} \cosh(\pi k)\left|\int_{0}^{1} dt \exp\left\{i k \ln(1-t^2)\right\}(1-t^2)^{-\frac{1}{2}} \cos(\tau t)\right|^2. $$ Using the upper bound 1 for $\cos(\tau t)$ (this is the only approximation), we find after the substitution $t = \sqrt{1-e^{-x}}$ $$ |J_{i k}(\tau)|^2 \leq \frac{1}{\pi^2} \cosh(\pi k)\left|\int_{0}^{\infty} dx\ e^{-i k x} e^{-x/2}(1-e^{-x})^{-1/2}\right|^2. $$ The integral can be found in Gradshteyn and Ryzhik (7th Edition, formula 3.312). The solution is then $$ |J_{i k}(\tau)|^2 \leq \frac{1}{\pi^2} \cosh(\pi k)\left|B\left(\frac{1}{2}+i k,\frac{1}{2}\right)\right|^2. $$ The Beta function can be evaluated by using 5.4.4 , 5.4.3 and $\Gamma(1+x)=x\Gamma(x)$ $$ \left|B\left(\frac{1}{2}+i k,\frac{1}{2}\right)\right|^2 = \left|\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}+i k\right)}{\Gamma\left(1+i k\right)}\right|^2=\frac{\pi}{k}\tanh(\pi k). $$ Inserting in the above equation finishes the proof.

**Edit**: Of course, the much simpler
$$
\int_{0}^{1} dt \ t^{x-1}(1-t^2)^{y-1}= \frac{1}{2} B\left(\frac{x}{2},y\right)
$$
could have been used as well, instead of the transformation of the integrand.