The value of $\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}2}\frac{\ln\left({1+\frac{\sin x}2}\right)}{\sin x}dx$ is equal to:

Question

The value of $\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}2}\frac{\ln\left({1+\frac{\sin x}2}\right)}{\sin x}dx$ is equal to:

A. $\frac{\pi}6$

B. $\frac{\pi^2}6$

C. $\frac{\pi}3$

D. $\frac{\pi^2}3$

Let $I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}2}\frac{\ln\left({1+\frac{\sin x}2}\right)}{\sin x}dx$

\begin{align} I &= \int\limits_{0}^{\frac{\pi}2}\left(\frac{\ln\left(1+\frac{\sin x}2\right)}{\sin x}+\frac{\ln\left(1+\frac{\sin(-x)}2\right)}{\sin({-x})}\right)dx \\ &= \int\limits_{0}^{\frac{\pi}2}\left(\frac{\ln\left(\frac{1+\frac{\sin x}2}{1-\frac{\sin x}2}\right)}{\sin x}\right)dx \\ \end{align}

I could not proceed after this point. King's property doesn't seem very useful here.

Answer 1

We can do this with Feynman's trick and the Weierstrass substitution of $u=\tan(x/2)$.

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Consider the more general integral,

$$\mathcal{I}(\alpha) = \int_{-\pi/2}^{\pi/2} \frac{1}{\sin(x)} \ln \Big( 1 + \alpha \sin(x) \Big) \, \mathrm{d}x \newcommand{\II}{\mathcal{I}} \newcommand{\dd}{\mathrm{d}} \newcommand{\a}{\alpha}$$

Note that you seek $\II(1/2)$.

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On the assumption we may do so: differentiate w.r.t. $\a$ and interchange the integral and derivative to find

$$\II'(\a) = \int_{-\pi/2}^{\pi/2} \frac{1}{\sin(x)} \frac{1}{1 + \a \sin(x)} \sin(x) \, \dd x = \int_{-\pi/2}^{\pi/2} \frac{\dd x}{1 + \a \sin(x)}$$

Let $u = \tan(x/2)$. Then

$$\II'(\a) = \int_{-1}^1 \frac{1}{1 + \a \frac{2u}{u^2+1}} \cdot \frac{2}{u^2+1} \, \dd u = 2\int_{-1}^1 \frac{\dd u}{u^2 + 2 \a u + 1} $$

Complete the square in the denominator; we get

$$\II'(\a) = 2 \int_{-1}^1 \frac{\dd u}{(u+\a)^2 + 1 - \a^2}$$

We then factor out $1-\a^2$:

$$\II'(\a) = \frac{2}{1-\a^2} \int_{-1}^1 \frac{\dd u}{1 + \left( \frac{u+\a}{\sqrt{1-\a^2}} \right)^2} $$

A second substitution, letting $v$ be our parenthetical, gives

$$\II'(\a) =\frac{2}{\sqrt{1 - \a^2} } \int_{v(-1)}^{v(1)} \frac{\dd v}{1 + v^2} $$

Recall that

$$\int \frac{1}{1+ \xi^2} \, \dd \xi = \arctan(\xi) + C$$

so

$$\II'(\a) = \frac{2}{\sqrt{1 - \a^2} } \arctan\left( \frac{u+\a}{\sqrt{1-\a^2}} \right) \bigg|_{u=-1}^{u=1}$$

With some algebraic manipulation and using that $\arctan(\cdot)$ is an odd function, we see that

\begin{align*} &\arctan\left( \frac{u+\a}{\sqrt{1-\a^2}} \right) \bigg|_{u=-1}^{u=1}\\ &= \arctan\left( \frac{1+\a}{\sqrt{1-\a^2}} \right) - \arctan\left( \frac{-1+\a}{\sqrt{1-\a^2}} \right) \\ &= \arctan\left( \frac{1+\a}{(1-\a)^{1/2} (1+\a)^{1/2}} \right) - \arctan\left( \frac{-1+\a}{(1-\a)^{1/2} (1+\a)^{1/2}} \right) \\ &= \arctan\left( \frac{(1+\a)^{1/2}}{(1-\a)^{1/2} } \right) - \arctan\left( \frac{-1+\a}{(1-\a)^{1/2} (1+\a)^{1/2}} \right) \\ &= \arctan\left( \frac{(1+\a)^{1/2}}{(1-\a)^{1/2} } \right) + \arctan\left( \frac{1-\a}{(1-\a)^{1/2} (1+\a)^{1/2}} \right) \\ &= \arctan\left( \frac{(1+\a)^{1/2}}{(1-\a)^{1/2} } \right) + \arctan\left( \frac{(1-\a)^{1/2} }{(1+\a)^{1/2}} \right) \\ &= \arctan\left( \sqrt{ \frac{1+\a}{1-\a} } \right) + \arctan\left( \sqrt{ \frac{1-\a}{1+\a} } \right) \end{align*}

We can then use the famous identity

$$\arctan(\xi) + \arctan \left( \frac 1 \xi \right) = \begin{cases} \pi/2 & \xi > 0 \\ -\pi/2 & \xi < 0 \end{cases}$$

The radical (our $\xi$) in our case is never negative, so we use that and finally have

$$\II'(\a) = \frac{\pi}{\sqrt{1-\a^2}}$$

Recall that

$$\int \frac{\dd \xi}{\sqrt{1- \xi^2}} = \arcsin(\xi)+C$$

Integration w.r.t. $\a$ gives us, clearly,

$$\II(\a) = \pi \arcsin(\a) + C$$

Finally, note that $\II(0)$ is obvious:

$$\II(0) =\int_{-\pi/2}^{\pi/2} \frac{1}{\sin(x)} \ln \Big( 1 + 0 \cdot \sin(x) \Big) \, \mathrm{d}x = \int_{-\pi/2}^{\pi/2} \frac{\ln(1)}{\sin(x)} \, \mathrm{d}x = 0$$

as $\ln(1) = 0$. Thus we easily find $C$:

$$\II(0) = 0 = \pi \arcsin(0) + C = C \implies C = 0$$

hence,

$$\II(\a) = \pi \arcsin(\a) \implies \boxed{\II \left( \frac 1 2 \right) = \pi \arcsin \left( \frac 1 2 \right) = \frac{\pi^2}{6}}$$

Answer 2

You made a good job arriving at $$I=\int_0^{\frac \pi 2} \csc (x) \log \left(\frac{1+\frac{\sin (x)}{2}}{1-\frac{\sin (x)}{2}}\right)\,dx$$ There is an antiderivative ( a monster); so consider the Taylor expansion $$\frac{\log \left(\frac{1+\frac{a}{2}}{1-\frac{a}{2}}\right)}{a}=\sum_{n=0}^\infty \frac{4^{-n}}{2 n+1} a^{2n}$$

$$I=\sum_{n=0}^\infty\frac{4^{-n}}{2 n+1}\int_0^{\frac \pi 2}\sin^{2n}(x) \,dx$$ $$\int_0^{\frac \pi 2}\sin^{2n}(x) \,dx=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ $$I=\sqrt{\pi }\sum_{n=0}^\infty\frac{2^{-(2 n+1)}}{2 n+1} \,\,\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)}$$ Computing the partial sums, you generate the sequence $$\left\{\frac{\pi }{2},\frac{25 \pi }{48},\frac{2009 \pi }{3840},\frac{112579 \pi }{215040},\cdots\right\}$$ Even the first term is sufficient to conclude.

Edit

Another way to do it. Using the tangent half-angle substitution, we have $$I=\int_0^1 \frac 1 t \log \left(1+\frac{2 t}{t^2-t+1}\right)\,dt$$ $$\frac 1 t \log \left(1+\frac{2 t}{t^2-t+1}\right)=2\sum_{n=0}^\infty(-1)^n \frac{\sqrt{3} \sin \left(\frac{\pi n}{3}\right)+\cos \left(\frac{\pi n}{3}\right)}{2 n+1}\,t^{2n}$$ $$I=2\sum_{n=0}^\infty(-1)^n \frac{\sqrt{3} \sin \left(\frac{\pi n}{3}\right)+\cos \left(\frac{\pi n}{3}\right)}{(2 n+1)^2}$$ Computing the partial sums, you generate the sequence $$\left\{2,\frac{14}{9},\frac{368}{225},\frac{18482}{11025},\frac{161438}{99225},\cdots\right\}$$ and the second term is more than sufficient to conclude.