TITLE: Finding $\int_{0}^{\pi/2} x \log(\sec{x}) \mathrm{d}x$
QUESTION [9 upvotes]: 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.

REPLY [9 votes]: 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:
$$ I = \frac{\pi^2}{8}\log 2 +\frac{7}{16}\zeta(3).\tag{4}$$