# Evaluation of integral $\int_{0}^{\infty} \frac{\sqrt{k}}{(k+\frac{1}{2})^{n+2}} J_{0}(2k\sin\theta) \sin{(\tau\sqrt{k})}\mathrm{d}k$

How to compute the integral that showed in the title. $J_0$ is the Bessel function. Parameters $\tau$ is a positive real number and $\theta\in[0,2\pi]$, n is a positive integer. Any help on it is much appreciated. Many thanks in advance. $$\int_{0}^{\infty} \frac{\sqrt{k}}{(k+\frac{1}{2})^{n+2}} J_{0}(2k\sin\theta) \sin{(\tau\sqrt{k})}\mathrm{d}k = \int_{0}^{\infty} \frac{2k^2}{(k^2+\frac{1}{2})^{n+2}} J_{0}(2k^2\sin\theta) \sin{(\tau k)}\mathrm{d}k$$

• Have you tried replacing $k$ with $x^2$ then applying the residue theorem? – Jack D'Aurizio Mar 12 '15 at 13:32
• @JackD'Aurizio While $k=\pm i \frac{\sqrt 2}{2}$ are high order singularities. Can residue theorem solve it? – Peng Mar 13 '15 at 1:07
• yes, as well as it is possible to compute $\int_{\mathbb{R}}\frac{dx}{(1+x^2)^n}$ with the residue theorem. – Jack D'Aurizio Mar 13 '15 at 12:24
• May I have the details when the Bessel functions included? Much appreciated. – Peng Mar 15 '15 at 2:34