A Deviation from a Conventional Proof of the Basel Problem

Question

There's been many topics on the Riemann-Zeta function, specifically $\zeta(2)$.$$\zeta(2)=\sum_{n=1}^\infty\frac{1}{n^2}=\int_0^1\int_0^1\frac{1}{1-xy}dA$$This is the Basel Problem. Taking the multivariable calculus approach, one could make the change of variables $(x,y)=(\frac{u-v}{\sqrt2},\frac{u+v}{\sqrt2})$. Following this path, one would come across the following iterated integral: $$\int_0^\sqrt2\int_{|u-\frac{\sqrt2}2|-\frac{\sqrt2}2}^{\frac{\sqrt2}2-|u-\frac{\sqrt2}2|}\frac 2{v^2-u^2+2}dv\;du$$All proofs that I've found online that use multivariable calculus integrate with respect to $v$ then $u$, so I wanted to write a proof integrating with respect to $u$ then $v$. For the sake of maintaining this post's relative brevity, I won't write out the full proof here. Instead, my proof can be found here.

Integrating with respect to $u$ then $v$ would require this iterated integral:$$\int_{-\frac{\sqrt2}2}^{\frac{\sqrt2}2}\int_{|v|}^{\sqrt2-|v|}\frac2{v^2-u^2+2}du\;dv$$My question is this: is there any way to evaluate the following integral analytically?$$\int_0^{\arctan\frac 1 2}\sec\theta\ln{\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}}d\theta=\frac{\pi^2}{12}+\frac{1}2\ln^2{\frac{\sqrt5-1}2}+\frac{1}2\ln^2{\frac{\sqrt5+1}2}$$Rearranging the terms results in$$2\int_0^{\arctan\frac 1 2}\sec\theta\ln{\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}}d\theta-\ln^2{\frac{\sqrt5-1}2}-\ln^2{\frac{\sqrt5+1}2}=\frac{\pi^2}6$$Using Fubini's Theorem, one can see that this is in fact another way of stating the Basel Problem. Is there any other way to prove this result, though?

All of the steps in between are in my link above.
I've spent a week reviewing every number and symbol in my work, and I can guarantee there are no errors. Wolfram Alpha produces an approximation to within $0.000004$ of my result, although it doesn't list the exact value.

Answer 1

Using the tangent-half-angle substitution $t=\tan{\left(\frac{\theta}{2}\right)}$, partial fraction decomposition, and basic properties of logarithms, the integral in question may be reduced to a sum of six basic rational-log integrals. For $\alpha\in[0,\frac{\pi}{2}]$, we have

$$\begin{align} \mathcal{I}{(\alpha)} &=\int_{0}^{\alpha}\sec{\theta}\ln{\left(\frac{\cos{\theta}-\sin{\theta}+1}{\sin{\theta}-\cos{\theta}+1}\right)}\,\mathrm{d}\theta\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{1+t^2}{1-t^2}\ln{\left(\frac{1-t}{t(t+1)}\right)}\cdot\frac{2\,\mathrm{d}t}{1+t^2}\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{2}{1-t^2}\ln{\left(\frac{1-t}{t(t+1)}\right)}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(\frac{1-t}{t(t+1)}\right)}}{1+t}\,\mathrm{d}t+\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(\frac{1-t}{t(t+1)}\right)}}{1-t}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}-\ln{\left(t\right)}-\ln{\left(t+1\right)}}{1+t}\,\mathrm{d}t\\ &~~~~~ +\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}-\ln{\left(t\right)}-\ln{\left(t+1\right)}}{1-t}\,\mathrm{d}t\\ &=\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}}{1-t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t+1\right)}}{1+t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t\right)}}{1-t}\,\mathrm{d}t\\ &~~~~~ -\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t\right)}}{1+t}\,\mathrm{d}t+\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(1-t\right)}}{1+t}\,\mathrm{d}t-\int_{0}^{\tan{\frac{\alpha}{2}}}\frac{\ln{\left(t+1\right)}}{1-t}\,\mathrm{d}t.\\ \end{align}$$

The first two integrals are elementary, and the third is a well known integral representation of the dilogarithm function reflected about the unit interval. The remaining three integrals also have not-too-complicated anti-derivatives in terms of logs and dilogs. From there, obtaining a closed form is just a matter of turning the algebra cranks.

Answer 2

I've got it!!! Contrary to my most recent reply to David H's answer, I didn't end up using Lebesgue integration. Instead, I had to teach myself the properties of polylogarithms in the span of about two days (that was fun). Anyhow, this is going to be somewhat of a long proof, so please bear with me. Also, you may have to refresh the page after clicking on links; Wolfram Alpha occasionally doesn't interpret the query correctly.
To reduce the amount of scrolling required, here is the identity that is to be proven:$$\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta=\frac{\pi^2}{12}+\frac12\ln^2\left({\frac{\sqrt5-1}2}\right)+\frac12\ln^2\left({\frac{\sqrt5+1}2}\right)$$By applying David's change of variable of $t=\tan\frac\theta2$, the integral above becomes$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1-t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(t)}{1-t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(t)}{1+t}\mathbb{d}t+\int_0^{\sqrt5-2}\frac{\ln{(1-t)}}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1-t}\mathbb{d}t\;.$$Tackling the first integral,$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1-t}\mathbb{d}t=-\int_{\sqrt5-2}^0\frac{\ln(1-t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)\text{Li}_0(t)}t\mathbb{d}t$$make the change of variable $k=\text{Li}_1(t)$ such that $\mathbb{d}k=\frac{\text{Li}_0(t)}t\mathbb{d}t$.$$\begin{align}\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)\text{Li}_0(t)}t\mathbb{d}t=\int_{\text{Li}_1(\sqrt5-2)}^0k\,\mathbb{d}k=\frac12\left[k^2\right]_{\text{Li}_1(\sqrt5-2)}^0=&-\frac12\text{Li}_1^2(\sqrt5-2)\\=&-\frac12\ln^2(3-\sqrt5)\end{align}$$Now, focusing on the second integral,$$-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1+t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1+t}\mathbb{d}t$$make the change of variable $m=\ln(1+t)$ such that $\mathbb{d}m=\frac{\mathbb{d}t}{1+t}$.$$\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1+t}\mathbb{d}t=\int_{\ln(\sqrt5-1)}^0m\,\mathbb{d}m=\frac12\left[m^2\right]_{\ln(\sqrt5-1)}^0=-\frac12\ln^2(\sqrt5-1)$$Taking the third integral by the horns,$$-\int_0^{\sqrt5-2}\frac{\ln(t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(t)}{1-t}\mathbb{d}t$$make the change of variable $p=1-t$ such that $\mathbb{d}p=-\mathbb{d}t$.$$\int_{\sqrt5-2}^0\frac{\ln(t)}{1-t}\mathbb{d}t=-\int_{3-\sqrt5}^1\frac{\ln(1-p)}p\mathbb{d}p=\left[\text{Li}_2(p)\right]_{3-\sqrt5}^1=\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)$$Tackling the fourth integral,$$\begin{align}-\int_0^{\sqrt5-2}\frac{\ln(t)}{1+t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\ln(t)}{1+t}\mathbb{d}t=&\left[\ln(t)\ln(1+t)\right]_{\sqrt5-2}^0-\int_{\sqrt5-2}^0\frac{\ln(1+t)}t\mathbb{d}t\\=&\int_0^{\sqrt5-2}\frac{\ln(1+t)}t\mathbb{d}t-\ln(\sqrt5-2)\ln(\sqrt5-1)&\end{align}$$make the change of variable $h=1+t$ such that $\mathbb{d}h=\mathbb{d}t$.$$\begin{align}\int_0^{\sqrt5-2}\frac{\ln(1+t)}t\mathbb{d}t-\ln(\sqrt5-2)\ln(\sqrt5-1)=&\int_1^{\sqrt5-1}\frac{\ln(h)}{h-1}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&\int_{\sqrt5-1}^1\frac{\ln(h)}{1-h}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)&\end{align}$$Now make a second change of variable, $n=1-h$, such that $\mathbb{d}n=-\mathbb{d}h$.$$\begin{align}\int_{\sqrt5-1}^1\frac{\ln(h)}{1-h}\mathbb{d}h-\ln(\sqrt5-2)\ln(\sqrt5-1)=&-\int_{2-\sqrt5}^0\frac{\ln(1-n)}{n}\mathbb{d}n-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&\left[\text{Li}_2(n)\right]_{2-\sqrt5}^0-\ln(\sqrt5-2)\ln(\sqrt5-1)\\=&{-\text{Li}_2(2-\sqrt5)}-\ln(\sqrt5-2)\ln(\sqrt5-1)\end{align}$$Now, one could evaluate the fifth and sixth integrals separately, but evaluating them together turns out to be much simpler.$$\int_0^{\sqrt5-2}\frac{\ln(1-t)}{1+t}\mathbb{d}t-\int_0^{\sqrt5-2}\frac{\ln(1+t)}{1-t}\mathbb{d}t=\int_{\sqrt5-2}^0\frac{\text{Li}_1(t)}{1+t}\mathbb{d}t+\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t\\\begin{align}&=\left[\text{Li}_1(t)\ln(1+t)\right]_{\sqrt5-2}^0-\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t+\int_{\sqrt5-2}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t\\&=\left[\text{Li}_1(t)\ln(1+t)\right]_{\sqrt5-2}^0=-\text{Li}_1(\sqrt5-2)\ln(1+(\sqrt5-2))=\ln(3-\sqrt5)\ln(\sqrt5-1)\end{align}$$Gathering these results yields$$\begin{align}&\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta\\=&\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)-\text{Li}_2(2-\sqrt5)-\frac12\ln^2(\sqrt5-1)-\frac12\ln^2(3-\sqrt5)+&\\&\ln(\sqrt5-1)\ln(3-\sqrt5)-\ln(\sqrt5-2)\ln(\sqrt5-1)\end{align}$$According to Wolfram Alpha, this is equivalent to the desired identity. In order to verify the alleged congruence, assume that the following is true:$$\begin{align}&\frac{\pi^2}{12}+\frac12\ln^2\left(\frac{\sqrt5-1}2\right)+\frac12\ln^2\left(\frac{\sqrt5+1}2\right)\\=&\frac{\pi^2}6-\text{Li}_2(3-\sqrt5)-\text{Li}_2(2-\sqrt5)-\frac12\ln^2(\sqrt5-1)-\frac12\ln^2(3-\sqrt5)+\\&\ln(\sqrt5-1)\ln(3-\sqrt5)-\ln(\sqrt5-2)\ln(\sqrt5-1)\;.\end{align}$$Using dilogarithmical identities, the right-hand side becomes $$\begin{align}&\frac{\pi^2}{12}+\text{Li}_2(\sqrt5-2)-\text{Li}_2(3-\sqrt5)+\frac12\text{Li}_2(4\sqrt5-8)-\frac12\ln^2(3-\sqrt5)-\frac12\ln^2(\sqrt5-1)+\\&\ln(\sqrt5-2)\ln(3-\sqrt5)+\ln(\sqrt5-1)\ln(3-\sqrt5)\;.\end{align}$$One could find other equivalent expressions, but this one is useful because it, like the original identity, contains the term $\frac{\pi^2}{12}$.
This means the assumed-to-be-true equation can be simplified to$$\frac12\ln^2\left(\frac{\sqrt5-1}2\right)+\frac12\ln^2\left(\frac{\sqrt5+1}2\right)=\text{Li}_2(\sqrt5-2)-\text{Li}_2(3-\sqrt5)+\frac12\text{Li}_2(4\sqrt5-8)-\frac12\ln^2(3-\sqrt5)-\frac12\ln^2(\sqrt5-1)+\ln(\sqrt5-2)\ln(3-\sqrt5)+\ln(\sqrt5-1)\ln(3-\sqrt5)\;.$$I've tried to simplify this further, yet I always seem to end up going in circles. However, there is another way to approach the equation: graphically. Replacing $\sqrt5-1$ with the variable $z$ results in the functional equation $$\frac12\ln^2\left(\frac{z}2\right)+\frac12\ln^2\left(\frac{z+2}2\right)=\text{Li}_2(z-1)-\text{Li}_2(2-z)+\frac12\text{Li}_2(2z-z^2)-\frac12\ln^2(2-z)-\frac12\ln^2(z)+\ln(z-1)\ln(2-z)+\ln(z)\ln(2-z)\;.$$The two graphs aren't very similar, but it's only necessary to check for an intersection at $z=\sqrt5-1$. It's not apparent looking at the two graphs side by side, but moving everything to one side of the equation appears to result in a zero at $z=\sqrt5-1$. Checking this gives a result of $0$, which means that the functions do intersect at $z=\sqrt5-1$ and, therefore,$$\int_0^{\arctan{\frac12}}\sec\theta\ln\left(\frac{\cos\theta-\sin\theta+1}{\sin\theta-\cos\theta+1}\right)\mathbb{d}\theta=\frac{\pi^2}{12}+\frac12\ln^2\left({\frac{\sqrt5-1}2}\right)+\frac12\ln^2\left({\frac{\sqrt5+1}2}\right)\;.$$