This is the sequel of my previous question
$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.
Any hint?
This is the sequel of my previous question
$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.
Any hint?
Thanks for the nice question.
The answer is $ I(a) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}} $ The sketch of the proof: expand $\arctan$ in series, and integrate term-wise (can do this for small enough $a$, since the sine is bounded): $ \arctan\left(a \sin^2(x)\right) = \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \sin^{4n+2}(x) $ This gives $ \int_0^\infty \frac{\sin^{4n+2}(x)}{x^2} \mathrm{d} x = \frac{1}{\binom{2n}{\tfrac{1}{2}}} = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma(2n+\frac{1}{2})}{(2n)!} $ The summation is easy, since the summand is a hypergeometric term: $ I(a) = \frac{\sqrt{\pi}}{2} \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \frac{\Gamma(2n+\frac{1}{2})}{(2n)!} = \frac{\pi a}{2} \cdot {}_2F_1\left(\frac{1}{4}, \frac{3}{4}; \frac{3}{2}; -a^2\right) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}} $