Joseph Krupa

2022-01-01

How to calculate the value of the integrals
${\int }_{0}^{1}{\left(\frac{\mathrm{arctan}x}{x}\right)}^{2}dx$,
${\int }_{0}^{1}{\left(\frac{\mathrm{arctan}x}{x}\right)}^{3}dx$
and ${\int }_{0}^{1}\frac{{\mathrm{arctan}}^{2}x\mathrm{ln}x}{x}dx$ ?

Bukvald5z

For the first one,
${\int }_{0}^{1}{\left(\frac{\mathrm{arctan}x}{x}\right)}^{2}dx={\int }_{0}^{\frac{\pi }{4}}{x}^{2}{\mathrm{csc}}^{2}xdx$
$=-{x}^{2}\mathrm{cot}x{\mid }_{0}^{\frac{\pi }{4}}+2{\int }_{0}^{\frac{\pi }{4}}x\mathrm{cot}xdx$
$=-\frac{{\pi }^{2}}{16}+4\sum _{n=1}^{\mathrm{\infty }}{\int }_{0}^{\frac{\pi }{4}}x\mathrm{sin}\left(2nx\right)dx$
$=-\frac{{\pi }^{2}}{16}+\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(n\frac{\pi }{2}\right)}{{n}^{2}}-\frac{\pi }{2}\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{cos}\left(n\frac{\pi }{2}\right)}{n}$
$=G+\frac{\pi }{4}\mathrm{ln}2-\frac{{\pi }^{2}}{16}$

godsrvnt0706

For the second one,
${\int }_{0}^{1}\frac{{\mathrm{arctan}}^{3}x}{{x}^{3}}dx={\int }_{0}^{\frac{\pi }{4}}{x}^{3}\mathrm{cot}x{\mathrm{csc}}^{2}xdx$
$=-\frac{1}{2}{x}^{3}{\mathrm{cot}}^{2}x{\mid }_{0}^{\frac{\pi }{4}}+\frac{3}{2}{\int }_{0}^{\frac{\pi }{4}}{x}^{2}{\mathrm{cot}}^{2}xdx$
$=-\frac{{\pi }^{3}}{128}-\frac{3}{2}{\int }_{0}^{\frac{\pi }{4}}{x}^{2}+\frac{3}{2}{\int }_{0}^{\frac{\pi }{4}}{x}^{2}{\mathrm{csc}}^{2}xdx$
$=-\frac{{\pi }^{3}}{64}+\frac{3}{2}\left(G+\frac{\pi }{4}\mathrm{ln}2-\frac{{\pi }^{2}}{16}\right)$
$=\frac{3}{2}G-\frac{{\pi }^{3}}{64}+\frac{3p}{8}\mathrm{ln}2-\frac{3{\pi }^{2}}{32}$

karton

For the third one,
$\begin{array}{}{\int }_{0}^{1}\frac{{\mathrm{arctan}}^{2}x\mathrm{ln}x}{x}dx\\ =-\int \frac{\mathrm{arctan}x{\mathrm{ln}}^{2}x}{1+{x}^{2}}dx\\ =-\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{n}\frac{\left(-1{\right)}^{n}}{2k+1}{\int }_{0}^{1}{x}^{2n+1}{\mathrm{ln}}^{2}xdx\\ =-\frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}\left({H}_{2n+1}-\frac{1}{2}{H}_{n}}{\left(n+1{\right)}^{3}}\\ =\frac{1}{4}\sum _{n=1}\frac{\left(-1{\right)}^{n}{H}_{2n}}{{n}^{3}}-\frac{1}{8}\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}{H}_{n}}{{n}^{3}}\end{array}$

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