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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.

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    $\begingroup$ Have you tried the obvious substitution $x+y=t$ and $x-y=u$? $\endgroup$ – Zacky Jun 5 '19 at 20:40
  • $\begingroup$ @Zacky Could you provide me with a reference regarding this type of substitution? I just have never done them before and would like to know how. $\endgroup$ – clathratus Jun 6 '19 at 16:58
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    $\begingroup$ I overthought it as I don't know how to get the new bounds. From there $x=\frac{t+u}{2}$ and $y=\frac{t-u}{2}$, which gives the Jacobian to be $dxdy=\frac12 dtdu$. $\endgroup$ – Zacky Jun 6 '19 at 17:56
  • $\begingroup$ Yeah I tried it as well... In general, how does one compute substitutions of this form? $\endgroup$ – clathratus Jun 6 '19 at 18:33
  • $\begingroup$ @clathratus There is a general method (unfortunately only avaiable in German) which you can find here. $\endgroup$ – mrtaurho Jun 6 '19 at 20:43
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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 $-\mathrm G$.

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

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    $\begingroup$ Note that most of the messy algebra can be avoided by exploiting the symmetry of the integrand. The integral with $x$ running only from $0$ to $y$ can be doubled to obtain the $\tan$-integral without any further substitutions. Otherwise, this seems to be the most direct approach. Very nice! $\endgroup$ – ComplexYetTrivial Jun 5 '19 at 22:02
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    $\begingroup$ @ComplexYetTrivial Frankly speaking I prefer stepwise integration over actually working with double integrals as I always confuse myself about borders of integration and substitutions while trying to do so. However, thank you! I appreciate your compliment. $\endgroup$ – mrtaurho Jun 5 '19 at 22:05
  • $\begingroup$ I did it the long way :) $\endgroup$ – clathratus Jun 6 '19 at 21:51
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$$I=\int_0^{\pi/4}\int_0^{\pi/4}\sec(x+y)+\sec(x-y)dxdy$$

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

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    $\begingroup$ Since I'm more familiar with the Clausen Functions it felt like ages ago the last time I really used those Fourier series in order to evaluate an integral like that ^^. Nevertheless it adds an interesting alternative to this post (+1) $\endgroup$ – mrtaurho Jun 7 '19 at 15:59
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    $\begingroup$ @mrtaurho Thank you so much! :) As I'm sure you know, the Fourier series for $\ln\sin x$ is really just $\mathrm{Cl}_1$ in disguise. $\endgroup$ – clathratus Jun 7 '19 at 16:02
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    $\begingroup$ Exactly. That's why I prefer using the machinery of the Clausen Functions as it is not only an elegant way of evaluating such integrals (not to forget it's close relation to the Polylogarithms) but also reduces what one has to write down quite tremendously :) $\endgroup$ – mrtaurho Jun 7 '19 at 16:10
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    $\begingroup$ @mrtaurho What are your thoughts on the identity with $\mathrm G/\pi^2$? Do you think it's correct? I have never seen this identity before, and I would like (your) professional opinion. It just seems a little too beautiful to be true... $\endgroup$ – clathratus Jun 7 '19 at 16:20
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    $\begingroup$ Took me a while to examine this series. However, using the relation between the Euler Numbers and the Dirichlet Beta Function this can be simplified to $$\sum_{n\geqslant0}\frac{\beta(2n+1)}{(2n+1)(n+1)}=\frac{4\mathrm G}\pi=2\beta'(-1)$$ Frankly speaking this one astonishines me; but at a second glance it is clear: the same sum may be obtained by integrating the secant function twice and evaluating at a suitable point, which on the other hand brings us back into the realm of the Clausen Functions where it's almost natural for Catalan's Constant to show up. $\endgroup$ – mrtaurho Jun 7 '19 at 20:59

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