interesting square of log sin integral
Solution 1:
As I remarked in a comment, I have spent a while trying to derive this identity by means of contour integration. Though I have not succeeded at that, (I take great pleasure in striking that out; see my other answer below!) I have found some related identities in my search which surprised me quite a bit, and I think they're worth sharing. For example, using the integral you asked about, I have shown that
$$
\begin{equation}
\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{5}{4}\cdot \frac{\pi^4}{90} = \frac{5}{4} \zeta(4). \tag{1}
\end{equation}
$$
Here
$$
H_n = \sum_{k = 1}^n \frac{1}{k}
$$
is the $n$th harmonic number. Here are a couple more examples. I will use the first to derive $(1)$.
- With the method from this answer, one can show that $$ \int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{5}\left(\frac{\pi}{2}\right)^5 + \pi \int_0^\infty y \log^2(1-e^{-2y})\,dy, $$ and it then follows from the result of your question that $$ \pi \int_0^\infty y \log^2(1- e^{-2y})\,dy = \frac{11}{45}\left(\frac{\pi}{2}\right)^5 - \frac{1}{5}\left(\frac{\pi}{2}\right)^5 = \frac{\pi}{8} \cdot \frac{\pi^4}{90}. $$
- I also shared this example in the other answer, but for completeness, let me mention that $$ \int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{30}\left(\frac{\pi}{2}\right)^5 + \frac{1}{6}\int_0^{\pi/2} \log^4(2\cos{x})\,dx, $$ hence $$ \int_0^{\pi/2} \log^4(2\cos{x})\,dx = \frac{19}{15}\left(\frac{\pi}{2}\right)^5. $$
More details about the derivations of 1. and 2. are contained in the answer I referenced above (near the bottom).
Let me now turn to deriving $(1)$. Taking $x = - e^{-2y}$ in the series expansion $$ \log{(1+x)} = \sum_{n = 1}^\infty \frac{(-1)^{n+1}x^n}{n} $$ and inserting the result into the integral evaluated in 1. gives $$ \begin{align} \frac{1}{8}\cdot\frac{\pi^4}{90} &= \int_0^\infty y \log^2(1 - e^{-2y})\,dy \\ & = \int_0^\infty y \left(\sum_{n = 1}^\infty \frac{e^{-2ny}}{n}\right)^2\,dy \\ & = \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{1}{nm} \int_0^\infty y e^{-2(n + m)y}\,dy \\ & = \frac{1}{4}\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2}. \tag{2} \end{align} $$ Now put $r = m+n$ and $s=n$ in order to write $$ \begin{align} \sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2} & = \sum_{r = 2}^\infty \frac{1}{r^2}\sum_{s = 1}^{r-1} \frac{1}{s(r-s)} = 2 \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s}, \tag{3} \end{align} $$ with the last equation a consequence of the identity $\frac{1}{s(r-s)} = \frac{1}{r}\left(\frac{1}{s}+ \frac{1}{r-s}\right)$. Insert $(3)$ into $(2)$ and multiply both sides by $2$ to get $$ \frac{1}{4} \cdot \frac{\pi^4}{90} = \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s} = \sum_{r=2}^\infty \frac{H_{r-1}}{r^3}. $$ Since $H_r - H_{r-1} = 1/r$, the identity $(1)$ now follows from the equations $$ \frac{\pi^4}{90} = \sum_{r=1}^\infty \frac{1}{r^4} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \sum_{r=2}^\infty \frac{H_{r-1}}{r^3} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \frac{1}{4}\cdot\frac{\pi^4}{90}. $$
Solution 2:
With a little time on my hands these past couple of days, I have finally succeeded in computing this integral through contour integration; the method is very similar in spirit to my answer here. The only background needed is knowledge of Cauchy's integral theorem and the fact that $\zeta(4) = \pi^4/90$. I've made a few remarks on generalization towards the bottom.
The strategy is as follows. We integrate the principal branch of $f(z) = z\log^3{(1+e^{2i z})}$ over an appropriately chosen contour in order to prove $$ \begin{align} \int_{-\pi/2}^{\pi/2}x^2\log^2{(2\cos{x})}\,dx & = \int_{-\pi/2}^{\pi/2}x^4\,dx-\frac{\pi}{3}\int_0^\infty \log^3{(1-e^{-2y})}\,dy. \tag{1} \end{align} $$ This leaves us in good shape; due to the fact that the integrand is even, the left-hand side is twice the integral we wish to evaluate. The first integral on the right is easily computed as $\pi^5/80$. To evaluate the second integral, put $e^{-v} = 1-e^{-2y}$; mild computations give $-(e^v - 1)^{-1}\,dv = dy$, and then $$ \begin{align} -\frac{\pi}{3}\int_0^\infty \log^3{(1-e^{-2y})}\,dy & = \frac{\pi}{6}\int_0^\infty\frac{v^3}{e^v - 1}\,dv = \pi\zeta(4). \tag{2} \end{align} $$ The last equation follows from the well-known identity $$ \Gamma(s)\zeta(s) = \int_0^\infty \frac{v^{s-1}}{e^{v}-1}\,dv $$ which holds for $\text{Re}\,s > 1$ and can be derived easily by developing the integrand in a geometric series. Upon inserting $(2)$ into $(1)$ and simplifying we arrive at the desired result.
It remains to prove $(1)$; as mentioned, the argument is entirely analogous to one I gave in this answer to another of Cody's questions. For completeness, I include it here. Consider the region obtained from $\mathbb C$ by removing the half-lines on which $\text{Re}\,z$ is an integer multiple of $\pi/2$ and $\text{Im}\,z \leq 0$. On this region, we can define a branch of $\log{(1+e^{2iz})} = \log{(2e^{iz}\cos{z})}$; we choose the branch with imaginary part between $-\pi$ and $\pi$. Having done this, let $f(z) = z\log^3{(1+e^{2iz})}$. We wish to integrate $f(z)$ over the contour obtained by removing the corners from the rectangle determined by $-\pi/2$ and $\pi/2 + iR$, where $R > 0$. To complete the contour we replace the bottom corners with quarter-circles of radius $\delta >0$. The intention is to let $R \to \infty$ and $\delta \to 0$.
${}$
For fixed values of $\delta$ and $R$, Cauchy's theorem says that $f(z)$ integrates to zero over the contour. As $R \to \infty$, the contribution from the upper horizontal side tends to $0$ because $f(x+iR) \to 0$ uniformly for $-\pi/2 \leq x \leq \pi/2$. By writing $1+e^{2i z} = 1- e^{2i(z-\pi/2)}$ one sees that $1+e^{2i z} = O(z-\pi/2)$, and therefore that $\log{(1+e^{2iz})} = O(\log{|z-\pi/2|})$ as $z \to \pi/2$. It follows that the contribution from the left quarter-circle is $O(\delta^2\log^3{\delta})$, hence that it vanishes in the limit $\delta \to 0$. The same argument applies to the other quarter-circle, and we are left with the contributions from the vertical sides and the lower horizontal side.
After taking limits, the contribution from the vertical sides is $$ \begin{align} i\int_0^\infty f(iy+\pi/2)\,dy -i\int_0^\infty f(iy - \pi/2)\,dy = i\pi\int_0^\infty\log^3{(1-e^{-2y})}\,dy. \tag{3} \end{align} $$ Now for $x$ between $-\pi/2$ and $\pi/2$, the quantity $2\cos{x}$ is positive. This means that the unique value of $\arg{(2e^{ix}\cos{x})}$ which lies between $-\pi$ and $\pi$ is simply $x$. As we have chosen the principle branch, we obtain $\log{(2e^{ix}\cos{x})} = \log{(2\cos{x})}+ix$. Therefore the contribution from the lower horizontal side may be written $$ \int_{-\pi/2}^{\pi/2} f(x)\,dx = \int_{-\pi/2}^{\pi/2}x\left(\log{(2\cos{x})} + ix\right)^3\,dx. \tag{4} $$ By the preceding analysis, $(3)$ and $(4)$ sum to zero. Since $(3)$ is purely imaginary, this means in particular that the imaginary part of $(4)$ is equal to the negative of $(3)$. The last statement is equivalent to $(1)$.
Because I have spent quite some time with this question, I would like to make a few remarks in the direction of a generalization. By taking $g(z) = p(z) \log^m(1+e^{2iz})$ in place of $f(z)$ above, where $p(z)$ is a polynomial and $m \in \mathbb N$, one finds by repeating the same arguments that $$ \begin{align} \int_{-\pi/2}^{\pi/2} p(x)&\left(\log{(2\cos{x})} + ix\right)^m\,dx \\ &= -i \int_0^\infty \left(p(iy + \pi/2)-p(iy - \pi/2)\right)\log^m{(1-e^{-2y})}\,dy. \tag{5} \end{align} $$ Clever choices of $p$ and $m$ then furnish a number of integral and series identities. Moreover, the identities $$ \int_0^\infty y^n \log{(1-e^{-2y})}\,dy = -\frac{1}{2^{n+1}}\Gamma(n+1)\zeta(n+2) \tag{6} $$ and $$ \int_0^\infty \log^m{(1-e^{-2y})}\,dy = \frac{(-1)^{m}}{2}\int_0^\infty \frac{y^m}{e^y - 1}\,dy = \frac{(-1)^m}{2}\Gamma(m+1)\zeta(m+1) \tag{7} $$ provide a connection between the integrals on the left-hand side of $(5)$ and the values taken by $\zeta$ at positive integers $n\geq2$. Here $(6)$ is derived by expanding $\log{(1-e^{-2y})}$ in powers of $e^{-2y}$ while $(7)$ is derived in the same way as $(2)$.
Solution 3:
I found the solution to this problem. If anyone is interested, see here:
http://de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R%28x,log,cos%29#0.2