TITLE: Show that $\int_{0}^{\pi/4}\int_{0}^{\pi/4}\left(\sec(x+y)+\sec(x-y)\right)\mathrm{d}x\mathrm{d}y=2G$
QUESTION [5 upvotes]: Show that
$$
\int_{0}^{\pi/4}\int_{0}^{\pi/4}
\left[\vphantom{\large A}\sec\left(x + y\right) +
\sec\left(x - y\right)\right]\,\mathrm{d}x\mathrm{d}y =
2G
$$
where $G$ is Catalan's constant.
I am not sure where to start or how to begin.

REPLY [3 votes]: $$I=\int_0^{\pi/4}\int_0^{\pi/4}\sec(x+y)+\sec(x-y)dxdy$$
We have that 
$$\int_0^{\pi/4}\sec(x+y)+\sec(x-y)dx=\ln c(y)+\ln c(-y)-\ln s(y)-\ln s(-y)$$
where 
$$
c(y)=\cos\left(\frac\pi8+\frac{y}2\right)\\
s(y)=\sin\left(\frac\pi8+\frac{y}2\right).
$$
Then we have 
$$I=\int_0^{\pi/4}\ln c(t)\ dt+\int_0^{\pi/4}\ln c(-t)\ dt-\int_0^{\pi/4}\ln s(t)\ dt-\int_0^{\pi/4}\ln s(-t)\ dt\\=C_++C_--S_+-S_-\ .$$
Then recall the Fourier series 
$$\ln\sin x=-\ln2-\sum_{k\geq1}\frac{\cos2kx}{k}$$
and $$\ln\cos x=-\ln2-\sum_{k\geq1}(-1)^k\frac{\cos2kx}{k}\ .$$
Thus
$$C_{\pm}=-\frac\pi4\ln2-\sum_{k\geq1}\frac{(-1)^k}{k}\int_0^{\pi/4}\cos\left(\frac{\pi k}{4}\pm kt\right)dt,$$
giving
$$C_++C_-=-\frac{\pi}{2}\ln2+\sum_{k\geq1}\frac{(-1)^{k+1}}{k^2}\sin\frac{\pi k}{2}\ .$$
Similarly
$$S_\pm=-\frac\pi4\ln2-\sum_{k\geq1}\frac1{k}\int_0^{\pi/4}\cos\left(\frac{\pi k}{4}\pm kt\right)dt,$$
and $$S_++S_-=-\frac\pi2\ln2-\sum_{k\geq1}\frac1{k^2}\sin\frac{\pi k}2\ .$$
Therefore 
$$\begin{align}
I&=-\frac{\pi}{2}\ln2+\sum_{k\geq1}\frac{(-1)^{k+1}}{k^2}\sin\frac{\pi k}{2}-\left(-\frac\pi2\ln2-\sum_{k\geq1}\frac1{k^2}\sin\frac{\pi k}2\right)\\
&=-\sum_{k\geq1}\frac{(-1)^{k}}{k^2}\sin\frac{\pi k}{2}+\sum_{k\geq1}\frac1{k^2}\sin\frac{\pi k}2\\
&=\sum_{k\geq1}\frac{1+(-1)^{k+1}}{k^2}\sin\frac{\pi k}{2}\\
&=2\sum_{k\geq0}\frac{1}{(2k+1)^2}\sin\frac{\pi (2k+1)}{2}\\
&=2\sum_{k\geq0}\frac{(-1)^k}{(2k+1)^2}\\
&=2\mathrm{G}.
\end{align}$$

Just an interesting side note, we recall that for $|x|<\pi/2$:
$$\sec(x)=\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}x^{2n}\ ,$$
where $E_n$ are the Euler numbers.
So 
$$\sec(x+y)+\sec(x-y)=2\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\sum_{k=0}^{n}{2n\choose 2k}x^{2n-2k}y^{2k}\ ,$$
from $$(x+y)^{2n}+(x-y)^{2n}=2\sum_{k=0}^{n}{2n\choose 2k}x^{2n-2k}y^{2k}.$$
So our integral is
$$\begin{align}
2\mathrm{G}&=\int_0^{\pi/4}\int_0^{\pi/4}2\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\sum_{k=0}^{n}{2n\choose 2k}x^{2n-2k}y^{2k}dxdy\\
&=2\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\sum_{k=0}^{n}{2n\choose 2k}\int_0^{\pi/4}x^{2n-2k}dx\int_0^{\pi/4}y^{2k}dy\\
&=2\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\sum_{k=0}^{n}{2n\choose 2k}\frac{\pi^{2n-2k+1}}{4^{2n-2k+1}(2n-2k+1)}\cdot\frac{\pi^{2k+1}}{4^{2k+1}(2k+1)}\\
&=2\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\left(\frac\pi4\right)^{2n+2}\sum_{k=0}^{n}\frac{{2n\choose 2k}}{(2n-2k+1)(2k+1)}.
\end{align}$$
So we finally have the beautiful identity
$$\frac1{16}\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\left(\frac\pi4\right)^{2n}\sum_{k=0}^{n}\frac{{2n\choose 2k}}{(2n-2k+1)(2k+1)}=\frac{\mathrm{G}}{\pi^2}$$