# Evaluate : int_0^(pi/2) ln^2(cos^2 x)dx

Evaluate :
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{ln}}^{2}\left({\mathrm{cos}}^{2}x\right)\text{d}x$
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Steinherrjm
I find a way to get the number using gamma functions, nothing is rigorous.
Consider the integral $I\left(\beta \right)={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\mathrm{cos}x{\right)}^{\beta }dx$. We know:
$2\frac{{d}^{2}}{d{\beta }^{2}}I\left(\beta \right){|}_{\beta =0}=4{\int }_{0}^{\frac{\pi }{2}}{\mathrm{ln}}^{2}\left(\mathrm{cos}x\right)dx$
is the integral we want. Introduce $u=\frac{1+\mathrm{sin}x}{2}$, we have:
$\begin{array}{rl}I\left(\beta \right)& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\mathrm{cos}x{\right)}^{\beta -1}d\mathrm{sin}x\\ & ={\int }_{0}^{1}\left(4u\left(1-u\right){\right)}^{\frac{\beta -1}{2}}d\left(2u\right)\\ & ={2}^{\beta }{\int }_{0}^{1}{u}^{\frac{\beta +1}{2}-1}\left(1-u{\right)}^{\frac{\beta +1}{2}-1}du\\ & ={2}^{\beta }\frac{\mathrm{\Gamma }\left(\frac{\beta +1}{2}{\right)}^{2}}{\mathrm{\Gamma }\left(\beta +1\right)}\end{array}$
Using the taylor expansion of various terms at $\beta =0$
$\begin{array}{rl}{2}^{\beta }& =1+\mathrm{ln}\left(2\right)\beta +\frac{{\mathrm{ln}}^{2}2}{2}{\beta }^{2}+\phantom{\rule{thinmathspace}{0ex}}...\\ \mathrm{\Gamma }\left(\frac{\beta +1}{2}\right)& =\sqrt{\pi }\left(1-\frac{\gamma +2\mathrm{ln}2}{2}\beta +\frac{{\pi }^{2}+2\left(\gamma +2\mathrm{ln}2{\right)}^{2}}{16}{\beta }^{2}+\phantom{\rule{thinmathspace}{0ex}}...\right)\\ \mathrm{\Gamma }\left(\beta +1\right)& =1-\gamma \beta +\frac{6{\gamma }^{2}+{\pi }^{2}}{12}{\beta }^{2}+\phantom{\rule{thinmathspace}{0ex}}...\end{array}$
We get:
$\begin{array}{rl}& I\left(\beta \right)=\pi \left(1-\mathrm{ln}\left(2\right)\beta +\frac{{\pi }^{2}+12{\mathrm{ln}}^{2}2}{24}{\beta }^{2}+\phantom{\rule{thinmathspace}{0ex}}...\right)\\ \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}& 2\frac{{d}^{2}}{d{\beta }^{2}}I\left(\beta \right){|}_{\beta =0}=2\pi \left(\frac{{\pi }^{2}}{12}+{\mathrm{ln}}^{2}2\right)\end{array}$
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Jaylyn Horne
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{2m-1}\left(x\right){\mathrm{cos}}^{2n-1}\left(x\right)dx=B\left(m,n\right)\phantom{\rule{0ex}{0ex}}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{2m-1}\left(x\right){\left({\mathrm{cos}}^{2}x\right)}^{\frac{2n-1}{2}}dx=B\left(m,n\right)$
On differentiating it twice w.r.t. to n and taking $m=\frac{1}{2}$ and $n=\frac{1}{2}$, we get
${\int }_{0}^{\frac{\pi }{2}}\left(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{\right)}^{2}=\frac{1}{2}\frac{{\left(\mathrm{\Gamma }\left(\frac{1}{2}\right)\right)}^{2}}{1}\left\{{\left(\psi \left(\frac{1}{2}\right)-\psi \left(1\right)\right)}^{2}+{\psi }^{\prime }\left(\frac{1}{2}\right)-{\psi }^{\prime }\left(1\right)\right\}$
$\therefore {\int }_{0}^{\frac{\pi }{2}}\left(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{\right)}^{2}=\frac{{\pi }^{3}}{6}+2\pi {\left(\mathrm{ln}2\right)}^{2}$