A Deviation from a Conventional Proof of the Basel Problem 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.
 A: 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.
