A calculus problem by rahul singh

Relevant wiki: Riemann Zeta Function

Using the half-angle tangent substitution , and let $t = \tan \frac x2$ , then we have $\dfrac {1-\cos x}{1+\cos x} = \dfrac {1-\frac {1-t^2}{1+t^2}}{1+\frac {1-t^2}{1+t^2}} = t^2$ . And $\dfrac {(1-\cos x)^{n-2}}{(1+\cos x)^{n+2}} = \left(\dfrac {1-\cos x}{1+\cos x}\right)^n \left(\dfrac 1{(1-\cos x)(1+\cos x)}\right)^2 = t^{2n}\left(\dfrac 1{1-\cos^2 x}\right)^2 = \dfrac {t^{2n}}{\sin^4 x} = \dfrac {t^{2n-4}(1+t^2)^4}{16}$ . Also $dt = \dfrac 12 \sec \frac x2 \ dx$ . Therefore,

$\begin{aligned} I & = \sum_{n=2}^\infty \int_0^\frac \pi 2 \sqrt {\frac {(1-\cos x)^{n-2}}{(1+\cos x)^{n+2}}}\log\left(\frac {1-\cos x}{1+\cos x}\right) dx \\ & = \sum_{n=2}^\infty \int_0^1 \sqrt {\frac {t^{2n-4}(1+t^2)^4}{16}}\cdot 2 \log t \cdot \frac 2{1+t^2} dt \\ & = \sum_{n=2}^\infty \int_0^1 \left(t^{n-2} + t^n\right)\log t\ dt & \small \color{#3D99F6} \text{By integration by parts} \\ & = \sum_{n=2}^\infty \left[ \left(\frac {t^{n-1}}{n-1} + \frac {t^{n+1}}{n+1} \right) \log t - \int \left(\frac {t^{n-2}}{n-1} + \frac {t^n}{n+1} \right)\ dt\right]_0^1 \\ & = \sum_{n=2}^\infty \left[ 0 - \left(\frac {t^{n-1}}{(n-1)^2} + \frac {t^{n+1}}{(n+1)^2}\right) \right]_0^1 \\ & = - \sum_{n=2}^\infty \frac 1{(n-1)^2} - \sum_{n=2}^\infty \frac 1{(n+1)^2} \\ & = - \sum_{n=1}^\infty \frac 1{n^2} - \sum_{n=3}^\infty \frac 1{n^2} & \small \color{#3D99F6} \text{Riemann zeta function }\zeta (n) = \sum_{k=1}^\infty \frac 1{k^n} \\ & = - 2 \zeta(2) + \frac 1{1^2} + \frac 1{2^2} & \small \color{#3D99F6} \text{and }\zeta (2) = \frac {\pi^2}6 \\ & = - 2\left(\frac {\pi^2}6\right) + \frac 1{1^2} + \frac 1{2^2} \\ & \approx -2.0399 \end{aligned}$

Therefore, $\lfloor I \rfloor = \boxed{-3}$ .