Finding $\int_{0}^{\pi/2} x \log(\sec{x}) \mathrm{d}x$ How do I evaluate $I=\displaystyle\int_{0}^{\pi/2} x \log(\sec{x}) \mathrm{d}x$?
We could write $I$ as $\dfrac{1}{2}\displaystyle\int_{0}^{\pi/2} x \log(1+\tan^2{x})$ and then taylor expand to get $$\begin{align}I&=\int_{0}^{\pi/2} \sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{x \tan^{2n}{x}}{n} \mathrm{d}x \\ &= \sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}\int_{0}^{\pi/2}x \tan^{2n}{x} \mathrm{d}x\end{align}$$ But I don't know what to do from here on. Please help me out. Thank you.
 A: We can use Fourier series. Over $I=\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ we have:
$$\log\sec t = \log 2 +\sum_{n\geq 1}\frac{(-1)^n}{n}\cos(2nt)\tag{1}$$
hence it follows that:
$$ I= \int_{0}^{\pi/2} t \log\sec t\,dt = \frac{\pi^2}{8}\log 2 + \sum_{n\geq 1}\frac{(-1)^n}{n}\int_{0}^{\pi/2}t\cos(2nt)\,dt \tag{2}$$
and:
$$ I = \frac{\pi^2}{8}\log 2 + \sum_{n\geq 1}\frac{(-1)^n}{n}\cdot\frac{-1+(-1)^n}{4n^2} = \frac{\pi^2}{8}\log 2 + \frac{1}{2}\sum_{n\geq 0}\frac{1}{(2n+1)^3},\tag{3}$$
so:
$$ I = \frac{\pi^2}{8}\log 2 +\frac{7}{16}\zeta(3).\tag{4}$$
A: Using $\log(\sec(x))=\log(1/\cos(x))=\underbrace{\log(1)}_{=0}-\log(\cos(x))$ and $\cos(x)=\frac{1}{2}(e^{i x}+e^{-i x})$ we can rewrite the integral as
$$
I=-\int_0^{\pi/2}dx x\log \left[\frac{1}{2}\left(e^{i x}\left(1+e^{-2 ix}\right)\right)\right]=-\int_0^{\pi/2}dx\left[\log\left(\frac{1}{2}\right)x+ix^2+x\log(1+e^{-2ix})\right]
$$
 which yields 
$$
I=-\log\left(\frac{1}{2}\right)\frac{\pi^2}{8} -i\frac{\pi^3}{24}-\underbrace{\int_0^{\pi/2}dxx\log(1+e^{-2ix})}_{J}
$$
it remains to calculate $J$. Expanding the logarithm as a Taylor Series 
$\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}x^n$ , we get
$$
J=\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\int_0^{\pi/2}dxx e^{-2 i xn}=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^3}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}-\frac{i\pi}{4}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\\-\frac{1}{2}\zeta(3)-\frac{3}{8}\zeta(3)-i\frac{\pi^3}{24}
$$
And therefore
$$
I=\log(2)\frac{\pi^2}{8}+\frac{7}{16}\zeta(3)
$$
With the Riemann Zeta function $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$
A: This is rather a comment than an answer, but I don't have enough reputation to comment, sorry.
Mathematica evaluates this integral to $1/16 (\pi^2 Log[4] + 7 \zeta[3])$. Note the use of the Riemann-$\zeta$-function. 
This suggests to me that the integral cannot be solved elementary. Using the Taylor expansion it may be possible to write this as an expression than involves the Riemann-$\zeta$-function. Otherwise it may be possible to manipulate this until one gets a known integral identity involving the Riemann-$\zeta$-function.
