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How can one show that

$$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$

assume $\alpha\ge1$.

I can't see how to simplify $\tan(x)\tan(x/2)$ and $\sin(x)\tan(x/2)$.

Michael Hardy
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    Another post where the community should be embarassed. This question should have been closed without any answer. –  May 23 '18 at 13:10

5 Answers5

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Use properties of trig functions to get \begin{equation*} \sin(x)\tan\left(\frac{x}{2}\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=2\sin^2\left(\frac{x}{2}\right). \end{equation*} Now we can simplify the integral to \begin{equation*} \int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)\left(\ln \tan(x)\tan\left(\frac{x}{2}\right)-\ln\alpha\right) dx. \end{equation*} It is easy to verify that \begin{equation*} \int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)(-\ln\alpha)dx=-\frac{\ln \alpha}{\sqrt{2}}. \end{equation*} which is your answer. Now we only need to show \begin{equation}\label{x} \int_0^{\pi/2} \sin(x) \ln \tan(x)\tan\left(\frac{x}{2}\right) dx=0. \end{equation} Let $t=\cos(x)$, we can draw a right triangle with sides $1,t,\sqrt{1-t^2}$ and draw a bisector of the angle $x$. Then \begin{equation*} \tan(x)=\frac{\sqrt{1-t^2}}{t},\quad\tan\left(\frac{x}{2}\right)=\frac{\sqrt{1-t^2}}{1+t} \end{equation*} and the integral becomes \begin{equation*} \int_0^1 \ln \left(\frac{1-t}{t}\right)dt=\int_0^1 \ln(1-t)dt-\int_0^1 \ln tdt=0 \end{equation*} by substitution $u=1-t$.

Haoran Chen
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This integral becomes deceptively easy when you simplify the integrand and use Feynman's Trick to differentiate with respect to its parameter $a$. First off, use the double angle identity for $\sin x$$$\sin x=2\sin\frac x2\cos\frac x2$$and call our integral $I$. What's left becomes$$\begin{align*}I(a) & =\int\limits_0^{\pi/2}dx\,\cos\frac x2\sqrt{2\sin^2\frac x2}\log\left(\frac 1a\tan x\tan\frac x2\right)\\ & =\sqrt2\int\limits_0^{\pi/2}dx\,\sin\frac x2\cos\frac x2\log\left(\frac 1a\tan x\tan\frac x2\right)\\ & =\frac 1{\sqrt2}\int\limits_0^{\pi/2}dx\,\sin x\log\left(\frac 1a\tan x\tan\frac x2\right)\end{align*}$$

Now differentiate with respect to $a$. The natural log then becomes $-1/a$ and the resulting integral becomes trivial

$$I'(a)=-\frac 1{a\sqrt2}\int\limits_0^{\pi/2}dx\,\sin x=-\frac 1{a\sqrt2}$$

Integrate with respect to $a$ and we get

$$I(a)=-\frac {\log a}{\sqrt2}+C$$

When $a=0$, then $I(a)=0$ and the right-hand side also equals zero. Therefore, $C=0$ and the identity has been proven$$\int\limits_0^{\pi/2}dx\,\cos\frac x2\log\left(\frac 1a\tan x\tan\frac x2\right)\sqrt{\sin x\tan\frac x2}\color{blue}{=-\frac {\log a}{\sqrt2}}$$

Frank W
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Using $$\tan(x) \, \tan(x/2) = \frac{2 \, \sin^{2}(x/2)}{\cos(x)}$$ and $$\sin(x) \, \tan(x/2) = 2 \, \sin^{2}(x/2)$$ then \begin{align} I &= \int_{0}^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha}\tan(x)\tan\left(\frac{x}{2}\right)\right]\sqrt{\sin(x)\tan\left(\frac{x}{2}\right)}\mathrm dx \\ &= 2\sqrt{2} \, \int_{0}^{\pi/2} \sin(x/2) \, \cos(x/2) \, \ln\left(\frac{2}{\alpha} \, \frac{\sin^{2}(x/2)}{\cos(x)} \right) \, dx \end{align} Let $u = \sin(x/2)$ to obtain \begin{align} I &= 4 \sqrt{2} \, \int_{0}^{1/\sqrt{2}} u \, \ln u \, du - 2\sqrt{2} \, \ln\left(\frac{\alpha}{2}\right) \, \int_{0}^{1/\sqrt{2}} u \, du - \sqrt{2} \,\int_{0}^{\pi/2} \sin(x) \, \ln(\cos(x)) \, dx \\ &= - \frac{1 + \ln 2}{\sqrt{2}} - \frac{1}{\sqrt{2}} \, \ln\left(\frac{\alpha}{2}\right) + \frac{1}{\sqrt{2}} \\ &= - \frac{\ln \alpha}{\sqrt{2}}. \end{align}

Leucippus
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Try to apply $\frac{d}{d \alpha}$ on both sides observing that $\sin x = 2 \sin \left(\frac x2\right) \cos \left(\frac x 2\right)$, the integral is then solvable.

edmz
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Rodrigo Fontana
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Let $u=\frac{1}{2}x \implies \int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sqrt{2\sin(u)\cos(u)\tan(u)}du$ $$\int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sqrt{2\sin^2(u)}du$$ $$\sqrt{2}\int_0^{\pi/4} \cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]\sin(u)du$$ $$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{1}{\alpha}\tan(2u)\tan(u)]du$$ $$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{1}{\alpha}\frac{2\sin(u)\cos(u)}{cos(2u)}tan(u)]du$$ $$\sqrt{2}\int_0^{\pi/4} \sin(u)\cos(u)\ln[\frac{2}{\alpha}\frac{\sin^2(u)}{\cos(2u)}]du$$ $$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + \int_0^{\pi/4} \sin(u)\cos(u)ln[\frac{\sin^2(u)}{\cos(2u)}]du)$$ $$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + \frac{1}{2} \int_0^{\pi/4} \sin(2u)ln[\frac{\sin^2(u)}{\cos(2u)}]du)$$ $$\sqrt{2}(\ln[\frac{2}{\alpha}] \int_0^{\pi/4} \sin(u)\cos(u)du + \frac{-1}{4} \int_1^{0} ln[\frac{1-\frac{1}{2}(1+a)}{a}]da)$$ Now is almost immediate.

tobiasbriones
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