# On the integral $\int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x$

Recently I run into this integral

$$\mathcal{J} = \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x$$

I don't know to what it evaluates. I tried several approaches.

1st: Differentiation under the integral sign

Consider the function $\displaystyle f(\alpha)= \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x$. Hence

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \frac{\mathrm{d} }{\mathrm{d} \alpha} \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\ &= \int_{0}^{\pi/2} \frac{\partial }{\partial \alpha} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\ &= -\int_{0}^{\pi/2} \frac{x \sin x}{\sin x \left ( 1- \alpha \sin x \right )} \, \mathrm{d}x\\ &=- \int_{0}^{\pi/2} \frac{x}{1- \alpha \sin x} \, \mathrm{d}x \end{align*}

And the last integral equals?

2nd: Taylor series expansion

Lemma: It holds that

$$x \sin^n x = \left\{\begin{matrix} 2^{1-n}\displaystyle\mathop{\sum}\limits_{k=0}^{\frac{n-1}{2}}(-1)^{\frac{n-1}{2}-k}\binom{n}{k}\,x\sin\big((n-2k)x\big) & , & n \;\; \text{odd} \\\\ 2^{-n}\displaystyle\binom{n}{\frac{n}{2}}\,x+2^{1-n}\mathop{\sum}\limits_{k=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-k}\binom{n}{k}\,x\cos\big((n-2k)x\big) & , & n \;\; \text{even} \end{matrix}\right.$$

Hence,

\begin{align*} \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x &= -\int_{0}^{\pi/2} \frac{x}{\sin x} \sum_{n=1}^{\infty} \frac{\sin^n x}{n} \, \mathrm{d}x \\ &=-\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x \sin^{n-1} x \, \mathrm{d}x \end{align*}

However the lemma does not help at all. In fact, if someone substitutes the RHS what it seems to be in there is an $\arcsin$ Taylor expansion. The series that remains to be evaluated is very daunting.

To sum up, I don't know to what this integral evaluates. I don't even know if a nice closed form exists neither do I expect one. But , I still hope.

• Nice closed form exists: $\int_0^{\frac{\pi }{2}} \frac{x \log (1-\sin (x))}{\sin (x)} \, dx=-\frac{\pi ^3}{8}$ found by lookup: isc.carma.newcastle.edu.au/advancedCalc Jun 30, 2018 at 8:53
• Sorry, try:isc.carma.newcastle.edu.au Jun 30, 2018 at 9:00
• @MariuszIwaniuk I'm curious to see where that number comes from. It's interesting. Jun 30, 2018 at 9:06
• By numeric integration of yours integral.I used CAS. Jun 30, 2018 at 9:07

## 2 Answers

The given problem is equivalent to the evaluation of $$\int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\cdot\frac{\log(1-x)}{x}\,dx =\sum_{n\geq 1}\frac{4^n}{2n\binom{2n}{n}}\int_{0}^{1}x^{2n-2}\log(1-x)\,dx=\sum_{n\geq 1}\frac{4^n H_{2n-1}}{2n\binom{2n}{n}(1-2n)}$$ which is a twisted hypergeometric series. On the other hand $$\mathcal{J}= 2\int_{0}^{\pi/4}\frac{2x \log(1-\sin(2x))}{\sin(2x)}\,dx=2\int_{0}^{1}\frac{\arctan(t)}{t}\log\left(\frac{(1-t)^2}{1+t^2}\right)\,dt$$ appears to be manageable through the polylogarithms machinery.
Indeed $\arctan t=\text{Im}\log(1+it)$ and the integrals $$\int \frac{\log(1+it)\log(1\pm it)}{t}\,dt, \qquad \int \frac{\log(1+it)\log(1-t)}{t}\,dt$$ have closed forms in terms of $\text{Li}_2$ and $\text{Li}_3$. However the simplest way to recover $\mathcal{J}=-\frac{\pi^3}{8}$ might be to exploit complex analysis and contour integration, as it often happens when integrating multiples of $\frac{x}{\sin x}$.

Through the Fourier series of $\log\sin$ we have $$\log(1-\cos x)=-\log(2)-2\sum_{n\geq 1}\frac{\cos(nx)}{n}$$ pointwise on $(0,\pi/2)$. We have that $\int_{0}^{\pi/2}\frac{x}{\sin x}\,dx$ equals $2K$, with $K$ being Catalan's constant, and by induction

$$\int_{0}^{\pi/2}\frac{x}{\sin x}\cos\left[n\left(\frac{\pi}{2}-x\right)\right]\,dx$$ up to the sign, equals $\sum_{m>n/2}\frac{2(-1)^m}{(2m+1)^2}$ or $\sum_{m> n/2}\frac{1}{(2m+1)^2}$, according to the parity of $n$. This allows to write the original twisted sum in terms of standard Euler sums. $K$ disappears from the outcome after some simplification and $$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32}$$ is well-known.

• (+1) Yes, real analysis seems to stuck in Euler sums and plenty of other twisted things!! Let's see what you or other members can additionally come up with. Jun 30, 2018 at 20:17

After the substitution $$t=\tan\frac x2$$, it can be shown that the integral reduces to

\begin{align} & \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, {d}x\\ =&\> 4\int_0^1 \frac{\tan^{-1}t\ln t}tdt = -2\int_0^1 \frac{\ln^2t}{1+t^2}dt = -\frac{\pi^3}8 \end{align}