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

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.

Answer 1

Essentially we only need to find the rather easy inner integral (namely the one depending on $x$). Starting to do so we get

$$\small\begin{align*} \int\sec(x+y)+\sec(x-y)\mathrm dx=&-\log\left(\cos\left(\frac{x-y}2\right)-\sin\left(\frac{x-y}2\right)\right)+\log\left(\cos\left(\frac{x-y}2\right)+\sin\left(\frac{x-y}2\right)\right)\\&-\log\left(\cos\left(\frac{x+y}2\right)-\sin\left(\frac{x+y}2\right)\right)+\log\left(\cos\left(\frac{x+y}2\right)+\sin\left(\frac{x+y}2\right)\right) \end{align*}$$

Plugging in the values for $x$, $0$ and $\frac\pi4$, we obtain after some messy algebra (including the substitutions $\frac y2-\frac\pi8\mapsto y$ and $\frac y2+\frac\pi8\mapsto y$) that the integrals equals

$$\int_0^\frac\pi4\int_0^\frac\pi4\sec(x+y)+\sec(x-y)\mathrm dx\mathrm dy=2\int_0^\frac\pi4\log\left(\frac{1+\tan y}{1-\tan y}\right)\mathrm dy$$

Now enforcing $\tan y\mapsto y$ followed up by $\frac{1-y}{1+y}\mapsto y$ we obtain

$$2\int_0^\frac\pi4\log\left(\frac{1+\tan y}{1-\tan y}\right)\mathrm dy=-2\int_0^1\frac{\log\left(\frac{1-y}{1+y}\right)}{1+y^2}\mathrm dy=-2\int_0^1\frac{\log(y)}{1+y^2}\mathrm dx$$

The latter one is a standard integral well-known to equal the negative of Catalan's Constant $-\mathrm G$.

$$\therefore~\int_0^\frac\pi4\int_0^\frac\pi4\sec(x+y)+\sec(x-y)\mathrm dx\mathrm dy~=~2\mathrm G$$

Answer 2

$$I=\int_0^{\pi/4}\int_0^{\pi/4}\sec(x+y)+\sec(x-y)dxdy$$

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 $$\begin{align} I&=-\frac{\pi}{2}\ln2+\sum_{k\geq1}\frac{(-1)^{k+1}}{k^2}\sin\frac{\pi k}{2}-\left(-\frac\pi2\ln2-\sum_{k\geq1}\frac1{k^2}\sin\frac{\pi k}2\right)\\ &=-\sum_{k\geq1}\frac{(-1)^{k}}{k^2}\sin\frac{\pi k}{2}+\sum_{k\geq1}\frac1{k^2}\sin\frac{\pi k}2\\ &=\sum_{k\geq1}\frac{1+(-1)^{k+1}}{k^2}\sin\frac{\pi k}{2}\\ &=2\sum_{k\geq0}\frac{1}{(2k+1)^2}\sin\frac{\pi (2k+1)}{2}\\ &=2\sum_{k\geq0}\frac{(-1)^k}{(2k+1)^2}\\ &=2\mathrm{G}. \end{align}$$

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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 $$\frac1{16}\sum_{n\geq0}(-1)^n\frac{E_{2n}}{(2n)!}\left(\frac\pi4\right)^{2n}\sum_{k=0}^{n}\frac{{2n\choose 2k}}{(2n-2k+1)(2k+1)}=\frac{\mathrm{G}}{\pi^2}$$