Please help me to evaluate this integral: $$\large\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx$$

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    $\begingroup$ It might help to note that $$\frac{1+\cos x-\sin x}{1+\cos x+\sin x}=\tan \left(\frac{\pi}{4}-\frac{x}{2} \right)$$ $\endgroup$ Nov 6, 2013 at 14:29

2 Answers 2


Your integral can be expressed in terms of certain special functions: $$\begin{align} \int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx&=\frac{\,\pi^2}6K+4\,\beta(4)\\&=\frac{\,\pi^2}6K-\frac{\pi^4}{24}+\frac1{192}\psi^{(3)}\left(\frac14\right)\\&=\frac{\,\pi^2}6K-\frac{119\,\pi^4}{2880}-\frac1{32}H^{(4)}_{-3/4},\end{align}$$ where $K$ is Catalan's constant, $\beta(x)$ is the Dirichlet beta function, $\psi^{(n)}(z)$ is the polygamma function and $H^{(r)}_n$ is the generalized harmonic number.

  • $\begingroup$ I got this result semi-manually, with a help from Mathematica. I'll try to write a proof later. $\endgroup$ Nov 9, 2013 at 17:55

I have been dabbling with this one off and on and thought I would post some results.

L.T., may I ask where you found this monster?.

Using Integrals idea of using $\displaystyle \frac{1+\cos(x)-\sin(x)}{1+\cos(x)+\sin(x)}=\tan(\frac{\pi}{4}-\frac{x}{2})$, one can make the sub $t=\tan(\frac{\pi}{4}-\frac{x}{2})$ and obtain:


The leftmost integral is rather famous and evaluates to $\displaystyle\frac{7\pi\zeta(3)}{4}$

The other one is the one that is the booger. It evaluates to $\displaystyle\frac{\pi^{2}}{24}G+\beta(4)-\frac{7\pi\zeta(3)}{16}$

So, using the integral form of digamma: $\displaystyle\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt=\psi(a)+\gamma$, along with the arctan series, diff this twice w.r.t 'a' in order to introduce the $log^{2}$ term. Make the sub $y=t^{2}$


$\displaystyle=\frac{1}{8}\sum_{n=0}^{\infty}\frac{(-1)^{n}\psi''(n+1)}{2n+1}$, by noting that a=1.....................................[1]

By using the identity $\psi(x+1)=\psi(x)+1/x$, one may write:


where the n=0 term was brought out of the digamma sum due to the identity not being defined there.

The middle series is equal to $\displaystyle2-\frac{\pi}{2}-log(2)+\frac{\pi^{2}}{24}-\frac{3\zeta(3)}{16}$

Thus, giving $\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{(2n+1)}+2-\frac{\pi}{2}-log(2)+\frac{\pi^{2}}{24}-\frac{7\zeta(3)}{16}$

I tried writing the sum in terms of its odd and even componets. At least, to eliminate the alternation.


This is as far as I got. If anyone knows how to evaluate an alternating tetragamma series, please, I would love to see it.

I have found little work done on these. May be a good topic for a paper. :)

EDIT: Due to the identity $\displaystyle \psi''(n+1)=-2\zeta(3)+2H_{n}^{(3)}$, we may write the digamma sum in [1] as:

$\displaystyle \frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\left[-2\zeta(3)+2H_{n}^{(3)}\right]}{2n+1}-\frac{\zeta(3)}{4}$

As a side note, The only other problem I have encountered that has the $\frac{\pi^{2}}{6}G$ term is this one:

A double series yielding Riemann's $\zeta$

Maybe this integral can some how be connected to series like this?.

    – Cody
    Dec 5, 2013 at 20:00

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