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

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


\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.

  • 1
    $\begingroup$ 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 $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 8:53
  • $\begingroup$ Sorry, try:isc.carma.newcastle.edu.au $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 9:00
  • $\begingroup$ @MariuszIwaniuk I'm curious to see where that number comes from. It's interesting. $\endgroup$ – Tolaso Jun 30 '18 at 9:06
  • $\begingroup$ By numeric integration of yours integral.I used CAS. $\endgroup$ – Mariusz Iwaniuk Jun 30 '18 at 9:07

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.

  • $\begingroup$ (+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. $\endgroup$ – Tolaso Jun 30 '18 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.