# Daunting series of integrals: $\sum_{n=2}^\infty\int_0^{\pi/2}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}\log(\frac{1-\sin x}{1+\sin x})dx$

My coleague showed me the following integral yesterday

\begin{equation} I=\sum_{n=2}^{\infty}\int_0^{\pi/2}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}\log\left(\!\frac{1-\sin x}{1+\sin x}\!\right)\ dx=\frac{5}{4}-\frac{\pi^2}{3}\tag1 \end{equation}

He also claimed the following closed-form:

\begin{equation} J=\int_{2}^{\infty}\int_0^{\pi/2}\sqrt{\frac{(1-\sin x)^{y-2}}{(1+\sin x)^{y+2}}}\log\left(\!\frac{1-\sin x}{1+\sin x}\!\right)\ dx\ dy=-\frac{4}{3}\tag2 \end{equation}

$(1)$ and $(2)$ seem difficult to deal with, but I believe there are some tricks that I can use but I'm not able to spot it yet. Using substitution $x\mapsto\frac\pi2-x$, one gets \begin{equation} I=\sum_{n=2}^{\infty}\int_0^{\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\tag3 \end{equation} and \begin{equation} J=\int_{2}^{\infty}\int_0^{\pi/2}\sqrt{\frac{(1-\cos x)^{y-2}}{(1+\cos x)^{y+2}}}\log\left(\!\frac{1-\cos x}{1+\cos x}\!\right)\ dx\ dy\tag4 \end{equation} but I don't know how to use $(3)$ and $(4)$ to evaluate $(1)$ and $(2)$. I'm quite sure that the main problem here is to evaluate \begin{equation} K=\int_0^{\pi/2}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}\log\left(\!\frac{1-\sin x}{1+\sin x}\!\right)\ dx \end{equation} How does one prove $(1)$ and $(2)$?

• Have you already tried to set $\frac{1-\sin x}{1+\sin x}=u$ then deal with the logarithm through differentiation under the integral sign? – Jack D'Aurizio Jun 26 '16 at 6:30
• Concerning $(1)$ it seems that if you first sum the series then you get an integral which you can handle with derivatives of the Euler beta function. – Olivier Oloa Jun 26 '16 at 6:33
• @JackD'Aurizio I did, but the different power in the square root make it difficult – Sophie Agnesi Jun 26 '16 at 6:52
• @OlivierOloa I'm not so sure with that. – Sophie Agnesi Jun 26 '16 at 6:55
• @SophieAgnesi Not really. – Marco Cantarini Jun 26 '16 at 8:44

## 2 Answers

Hint.

1. One may evaluate $(1)$ with the following steps. From the geometric series evaluation $$\sum_{n=2}^{\infty}\sqrt{\frac{(1-\sin x)^{n-2}}{(1+\sin x)^{n+2}}}=\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}},\quad 0<x<\frac{\pi}2,$$ one may write $$I=\int_0^{\pi/2}\frac{1}{(1+\sin x)^2}\frac1{1-\sqrt{\frac{1-\sin x}{1+\sin x}}}\log\left( \frac{1-\sin x}{1+\sin x}\right)dx.$$ By the change of variable $u=\frac{1-\tan (x/2)}{1+\tan (x/2)}$ we obtain a standard integral:

$$I=\int_0^1\frac{1+u^2}{1-u}\:\log u \:du=\frac{5}{4}-\frac{\pi^2}{3}.$$

2. One may evaluate $(2)$ by first integrating with respect to $y$ : $$J=\int_{2}^{\infty}\!\sqrt{\frac{(1-\sin x)^{y-2}}{(1+\sin x)^{y+2}}} dy=-\frac2{(1+\sin x)^2\log\left( \frac{1-\sin x}{1+\sin x}\right)},\quad 0<x<\frac{\pi}2,$$ then integrating with respect to $x$ one gets a standard integral: $$J=\int_{2}^{\infty}\!\!\int_0^{\pi/2}\sqrt{\frac{(1-\sin x)^{y-2}}{(1+\sin x)^{y+2}}}\log\left(\!\frac{1-\sin x}{1+\sin x}\!\right)\ dx\ dy=-\int_0^{\pi/2}\frac{2\:dx}{(1+\sin x)^2}$$ which gives

$$J=-\frac43.$$

• Let me check your answer later. (+1) – Sophie Agnesi Jun 26 '16 at 15:39
• You turned out to be correct that $J=-4/3$. Merci beaucoup Olivier. – Sophie Agnesi Jun 27 '16 at 17:26
• De rien, @SophieAgnesi. – Olivier Oloa Jun 27 '16 at 17:29

Let's employ the weirdo substitution taught by my brother: $\sin x=\tanh t$. Doing so, one will get \begin{align} K&=\int_0^{\infty}\sqrt{\frac{(1-\tanh t)^{n-2}}{(1+\tanh t)^{n+2}}}\ln\left(\frac{1-\tanh t}{1+\tanh t}\right)\ \frac{dt}{\cosh t}\\[10pt] &=\int_0^{\infty}\sqrt{\left(\frac{\cosh t-\sinh t}{\cosh t+\sinh t}\right)^{n-2}}\frac{\cosh t}{(\cosh t+\sinh t)^2}\ \ln\left(\frac{\cosh t-\sinh t}{\cosh t+\sinh t}\right)\ dt\\[10pt] &=-\int_0^{\infty} e^{-(n-2)t}\left(e^{-t}+e^{-3t}\right)\ t\ dt\\[10pt] &=-\frac{1}{n^2+1}-\frac{1}{n^2-1} \end{align} Thus, evaluating $I$ and $J$ are easy-peasy-lemon-squeezy.

• Out of curiosity, what would compel someone to try out that substitution? What advantage does it offer (aside from making the problem EPLS)? – user170231 Jun 27 '16 at 15:09
• "EPLS" hahahaha +1 – The Count Jun 27 '16 at 15:11
• @user170231 The obvious fact: $\cosh t\pm\sinh t=e^{\pm t}$ – Anastasiya-Romanova 秀 Jun 27 '16 at 15:12
• @user170231 I can't speak for Anastasiya's brother, but in my own experience, most unintuitive ('magical') substitutions are just a retrospective shorthand, and were initially a very intuitive chain of substitutions. – nospoon Jun 27 '16 at 15:17
• @TheCount Why on earth did you put the bound in my answer since $K$ must be in the form of $n$?? – Anastasiya-Romanova 秀 Jun 27 '16 at 15:39