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## An arctan integral

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### An arctan integral

Tue Sep 06, 2016 5:32 pm

Posts: 50
I was playing around with this integral and thought it may be of interest to some:
$$\int^\infty_0\frac{x^{s-1}\arctan{x}}{\sqrt{1+x^2}}\ dx=\frac{1}{2\cos\left(\frac{\pi s}{2}\right)}\int^1_0\frac{x^{-s}\ln\left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^2}}\ dx=\frac{\sqrt\pi}{2\cos\left(\frac{\pi s}{2}\right)}\frac{\Gamma\left(1-\tfrac s2\right)}{\Gamma\left(\tfrac{3-s}2\right)}{}_3F_2\left(\tfrac12, 1-\tfrac s2,1;\tfrac 32,\tfrac{3-s}2;1\right)$$
Edit: After a series of further manipulations I managed to obtain the identity
$$\int^\infty_0\frac{J_0(x)\left[\left(C\left(\sqrt{\frac{2x}\pi}\right)-\frac12\right)^2+\left(S\left(\sqrt{\frac{2x}\pi}\right)-\frac12\right)^2\right]}{\sqrt{x}}\ dx=\frac{\Gamma^2\left(\frac{1}{4}\right)}{8\pi}$$
where $J_0(x)$ is the zeroth-order Bessel function of the first kind and $C$ and $S$ are Fresnel integrals.