# An inequality involving Bessel functions of imaginary order

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?

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.