25
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ It might help to note that $$\frac{1+\cos x-\sin x}{1+\cos x+\sin x}=\tan \left(\frac{\pi}{4}-\frac{x}{2} \right)$$ $\endgroup$ – Shobhit Bhatnagar Nov 6 '13 at 14:29
13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I got this result semi-manually, with a help from Mathematica. I'll try to write a proof later. $\endgroup$ – Vladimir Reshetnikov Nov 9 '13 at 17:55
7
$\begingroup$

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:

$\displaystyle\pi\int_{0}^{1}\frac{log^{2}(t)}{1-t^{2}}dt+4\int_{0}^{1}\frac{\tan^{-1}(t)log^{2}(t)}{1-t^{2}}dt$.

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 $y=t^{2}$

$\displaystyle1/2\int_{0}^{1}\frac{y^{n+a/2-1/2}}{1-y}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}dy$

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

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

$\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{2n+1}+\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{3}(2n+1)}-\frac{\zeta(3)}{4}$

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.

$\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{2n+1}=\frac{1}{32}\sum_{n=1}^{\infty}\frac{\psi''(2n)}{n+1/4}-\frac{1}{32}\sum_{n=1}^{\infty}\frac{\psi''(2n+1)}{n+3/4}+\frac{\zeta(3)}{12}$

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?.

$\endgroup$
  • $\begingroup$ THE PROBLEM HAS BEEN SOLVED USING CONTOURS AND THE DIGAMMA KERNEL. $\endgroup$ – Cody Dec 5 '13 at 20:00

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.