Evaluate :

$${\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}({\mathrm{cos}}^{2}x)\text{d}x$$

$${\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}({\mathrm{cos}}^{2}x)\text{d}x$$

Rene Nicholson
2022-10-20
Answered

Evaluate :

$${\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}({\mathrm{cos}}^{2}x)\text{d}x$$

$${\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}({\mathrm{cos}}^{2}x)\text{d}x$$

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Steinherrjm

Answered 2022-10-21
Author has **12** answers

I find a way to get the number using gamma functions, nothing is rigorous.

Consider the integral $I(\beta )={\int}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{cos}x{)}^{\beta}dx$. We know:

$$2\frac{{d}^{2}}{d{\beta}^{2}}I(\beta ){{\textstyle |}}_{\beta =0}=4{\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}(\mathrm{cos}x)dx$$

is the integral we want. Introduce $u=\frac{1+\mathrm{sin}x}{2}$, we have:

$$\begin{array}{rl}I(\beta )& ={\int}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{cos}x{)}^{\beta -1}d\mathrm{sin}x\\ & ={\int}_{0}^{1}(4u(1-u){)}^{\frac{\beta -1}{2}}d(2u)\\ & ={2}^{\beta}{\int}_{0}^{1}{u}^{\frac{\beta +1}{2}-1}(1-u{)}^{\frac{\beta +1}{2}-1}du\\ & ={2}^{\beta}\frac{\mathrm{\Gamma}(\frac{\beta +1}{2}{)}^{2}}{\mathrm{\Gamma}(\beta +1)}\end{array}$$

Using the taylor expansion of various terms at $\beta =0$

$$\begin{array}{rl}{2}^{\beta}& =1+\mathrm{ln}(2)\beta +\frac{{\mathrm{ln}}^{2}2}{2}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...\\ \mathrm{\Gamma}(\frac{\beta +1}{2})& =\sqrt{\pi}(1-\frac{\gamma +2\mathrm{ln}2}{2}\beta +\frac{{\pi}^{2}+2(\gamma +2\mathrm{ln}2{)}^{2}}{16}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...)\\ \mathrm{\Gamma}(\beta +1)& =1-\gamma \beta +\frac{6{\gamma}^{2}+{\pi}^{2}}{12}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...\end{array}$$

We get:

$$\begin{array}{rl}& I(\beta )=\pi (1-\mathrm{ln}(2)\beta +\frac{{\pi}^{2}+12{\mathrm{ln}}^{2}2}{24}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...)\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& 2\frac{{d}^{2}}{d{\beta}^{2}}I(\beta ){{\textstyle |}}_{\beta =0}=2\pi (\frac{{\pi}^{2}}{12}+{\mathrm{ln}}^{2}2)\end{array}$$

Consider the integral $I(\beta )={\int}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{cos}x{)}^{\beta}dx$. We know:

$$2\frac{{d}^{2}}{d{\beta}^{2}}I(\beta ){{\textstyle |}}_{\beta =0}=4{\int}_{0}^{\frac{\pi}{2}}{\mathrm{ln}}^{2}(\mathrm{cos}x)dx$$

is the integral we want. Introduce $u=\frac{1+\mathrm{sin}x}{2}$, we have:

$$\begin{array}{rl}I(\beta )& ={\int}_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\mathrm{cos}x{)}^{\beta -1}d\mathrm{sin}x\\ & ={\int}_{0}^{1}(4u(1-u){)}^{\frac{\beta -1}{2}}d(2u)\\ & ={2}^{\beta}{\int}_{0}^{1}{u}^{\frac{\beta +1}{2}-1}(1-u{)}^{\frac{\beta +1}{2}-1}du\\ & ={2}^{\beta}\frac{\mathrm{\Gamma}(\frac{\beta +1}{2}{)}^{2}}{\mathrm{\Gamma}(\beta +1)}\end{array}$$

Using the taylor expansion of various terms at $\beta =0$

$$\begin{array}{rl}{2}^{\beta}& =1+\mathrm{ln}(2)\beta +\frac{{\mathrm{ln}}^{2}2}{2}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...\\ \mathrm{\Gamma}(\frac{\beta +1}{2})& =\sqrt{\pi}(1-\frac{\gamma +2\mathrm{ln}2}{2}\beta +\frac{{\pi}^{2}+2(\gamma +2\mathrm{ln}2{)}^{2}}{16}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...)\\ \mathrm{\Gamma}(\beta +1)& =1-\gamma \beta +\frac{6{\gamma}^{2}+{\pi}^{2}}{12}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...\end{array}$$

We get:

$$\begin{array}{rl}& I(\beta )=\pi (1-\mathrm{ln}(2)\beta +\frac{{\pi}^{2}+12{\mathrm{ln}}^{2}2}{24}{\beta}^{2}+\phantom{\rule{thinmathspace}{0ex}}...)\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}& 2\frac{{d}^{2}}{d{\beta}^{2}}I(\beta ){{\textstyle |}}_{\beta =0}=2\pi (\frac{{\pi}^{2}}{12}+{\mathrm{ln}}^{2}2)\end{array}$$

Jaylyn Horne

Answered 2022-10-22
Author has **5** answers

$${\int}_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^{2m-1}\left(x\right){\mathrm{cos}}^{2n-1}\left(x\right)dx=B(m,n)\phantom{\rule{0ex}{0ex}}{\int}_{0}^{\frac{\pi}{2}}{\mathrm{sin}}^{2m-1}\left(x\right){\left({\mathrm{cos}}^{2}x\right)}^{\frac{2n-1}{2}}dx=B(m,n)$$

On differentiating it twice w.r.t. to n and taking $m=\frac{1}{2}$ and $n=\frac{1}{2}$, we get

$${\int}_{0}^{\frac{\pi}{2}}(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{)}^{2}=\frac{1}{2}\frac{{\left(\mathrm{\Gamma}\left(\frac{1}{2}\right)\right)}^{2}}{1}\{{(\psi \left(\frac{1}{2}\right)-\psi \left(1\right))}^{2}+{\psi}^{\prime}\left(\frac{1}{2}\right)-{\psi}^{\prime}\left(1\right)\}$$

$$\therefore {\int}_{0}^{\frac{\pi}{2}}(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{)}^{2}=\frac{{\pi}^{3}}{6}+2\pi {(\mathrm{ln}2)}^{2}$$

On differentiating it twice w.r.t. to n and taking $m=\frac{1}{2}$ and $n=\frac{1}{2}$, we get

$${\int}_{0}^{\frac{\pi}{2}}(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{)}^{2}=\frac{1}{2}\frac{{\left(\mathrm{\Gamma}\left(\frac{1}{2}\right)\right)}^{2}}{1}\{{(\psi \left(\frac{1}{2}\right)-\psi \left(1\right))}^{2}+{\psi}^{\prime}\left(\frac{1}{2}\right)-{\psi}^{\prime}\left(1\right)\}$$

$$\therefore {\int}_{0}^{\frac{\pi}{2}}(\mathrm{ln}\left({\mathrm{cos}}^{2}x\right)dx{)}^{2}=\frac{{\pi}^{3}}{6}+2\pi {(\mathrm{ln}2)}^{2}$$

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