maduregimc

## Answered question

2021-12-31

Prove that:
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{ln}}^{2}\left(\mathrm{cos}x\right)dx=\frac{\pi }{2}{\mathrm{ln}}^{2}2+\frac{{\pi }^{3}}{24}$

### Answer & Explanation

soanooooo40

Beginner2022-01-01Added 35 answers

Let's get all powers of $\mathrm{ln}\left(\mathrm{cos}\left(x\right)\right)$ at once, using an exponential generating function:
$G\left(z\right)={\int }_{0}^{\frac{\pi }{2}}\sum _{j=0}^{\mathrm{\infty }}\frac{{\mathrm{ln}}^{j}\left(\mathrm{cos}\left(x\right)\right){z}^{j}}{j!}dx={\int }_{0}^{\frac{\pi }{2}}\mathrm{exp}\left(z\mathrm{ln}\left(\mathrm{cos}\left(x\right)\right)\right)dx$
Change variables: $\mathrm{cos}\left(x\right)={t}^{\frac{1}{2}}$, and using the Beta function:
$\frac{1}{2}{\int }_{0}^{1}\frac{{t}^{\frac{z-1}{2}}}{\sqrt{1-t}}dt=\frac{1}{2}B\left(\frac{1}{2},\frac{z+1}{2}\right)=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{z+1}{2}\right)}{2\mathrm{\Gamma }\left(1+\frac{z}{2}\right)}$
$=\frac{\pi }{2}-\frac{\pi \mathrm{ln}\left(2\right)}{2}z+\left(\frac{{\pi }^{3}}{48}+\frac{\pi {\left(\mathrm{ln}\left(2\right)\right)}^{2}}{4}\right){z}^{2}-\frac{\pi \left(4{\left(\mathrm{ln}\left(2\right)\right)}^{3}+{\pi }^{2}\mathrm{ln}\left(2\right)+6\zeta \left(3\right)\right)}{48}{z}^{3}+\dots$

lenkiklisg7

Beginner2022-01-02Added 29 answers

Have a look at this other question. We have that $\mathrm{log}\left(2\mathrm{cos}x\right)$ has a nice Fourier series:
$\mathrm{log}\left(2\mathrm{cos}x\right)=\sum _{n=1}^{+\mathrm{\infty }}\frac{{\left(-1\right)}^{n+1}}{n}\mathrm{cos}\left(2nx\right)$
and since:
${\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}\left(2nx\right)\mathrm{cos}\left(2mx\right)dx=\frac{\pi }{4}{\delta }_{m,n}$
in follows that:
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{log}}^{2}\left(2\mathrm{cos}x\right)dx=\frac{\pi }{4}\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{{n}^{2}}=\frac{\pi }{4}\zeta \left(2\right)=\frac{{\pi }^{3}}{24}$ (1)
while
${\int }_{0}^{\frac{\pi }{2}}\mathrm{log}\left(\mathrm{cos}x\right)dx=-\frac{\pi }{2}\mathrm{log}2$ (2)
is a well-known result. (1) and (2) proves your claim:
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{ln}}^{2}\left(\mathrm{cos}x\right)dx=\frac{\pi }{2}{\mathrm{ln}}^{2}2+\frac{{\pi }^{3}}{24}$
Since
$\mathrm{log}\left(2\mathrm{sin}x\right)=-\sum _{n=1}^{+\mathrm{\infty }}\frac{\mathrm{cos}\left(2nx\right)}{n}$
and ${\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}\left(2{n}_{1},x\right)\mathrm{cos}\left(2{n}_{2}x\right)\mathrm{cos}\left(2{n}_{3}x\right)dx$

it follows that
${\int }_{0}^{\frac{\pi }{2}}{\mathrm{log}}^{3}\left(2\mathrm{sin}x\right)dx=-\frac{3\pi }{4}\sum _{n=1}^{+\mathrm{\infty }}\frac{{H}_{n-1}}{{n}^{2}}=-\frac{3\pi }{4}\zeta \left(3\right)$

Vasquez

Expert2022-01-09Added 669 answers

Let $\mathrm{cos}x=\sqrt{t}$ and $dx=\frac{dt}{2\sqrt{t}\sqrt{1-t}}$, then
$\begin{array}{}{\int }_{0}^{\pi /2}{\mathrm{ln}}^{2}\left(\mathrm{cos}x\right)dx=\frac{1}{8}{\int }_{0}^{1}\frac{{\mathrm{ln}}^{2}t}{\sqrt{t}\sqrt{1-t}}dt\\ =\frac{1}{8}\underset{x\to \frac{1}{2}}{lim}\underset{y\to \frac{1}{2}}{lim}\frac{{d}^{2}}{d{x}^{2}}{\int }_{0}^{1}{t}^{x-1}\left(1-t{\right)}^{y-1}dt\\ =\frac{1}{8}\underset{x\to \frac{1}{2}}{lim}\underset{y\to \frac{1}{2}}{lim}B\left(x,y\right)\left[\left(\psi \left(x\right)-\psi \left(x+y\right){\right)}^{2}+{\psi }_{1}\left(x\right)-{\psi }_{1}\left(x+y\right)\right]\end{array}$
where $B\left(x,y\right)$ is beta function and ${\psi }_{k}\left(z\right)$ is polygamma function. The same approach also works for
$in{t}_{0}^{\pi /2}{\mathrm{ln}}^{3}\left(\mathrm{cos}x\right)dx$
but for this one, we use third derivative of beta function.

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