# Evaluating int_0^((pi)/(2))((1)/(log(tan x))+(1)/(1-tan(x)))^3dx

Evaluating ${\int }_{0}^{\frac{\pi }{2}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}\left(x\right)}\right)}^{3}dx$
Using the method shown here, I have found the following closed form.
${\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}x}\right)}^{2}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x=3\mathrm{ln}2-\frac{4}{\pi }G-\frac{1}{2},$
where $G$ is Catalan's constant.
I can see that replicating the techniques for the following integral could be rather lenghty.
${\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}x}\right)}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x$
My question: Could someone have, ideally, a different idea to evaluate the latter integral?
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kregde84
I've found it.
We have
$\begin{array}{}\text{(1)}& {\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}x}\right)}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x=\frac{9}{2}\mathrm{ln}2-\frac{6}{\pi }G-\frac{3}{4}-\frac{\pi }{8}\end{array}$
where $G$ is Catalan's constant.
Proof. Set
$\begin{array}{}\text{(2)}& {I}_{n}:={\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}x}\right)}^{n}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x,\phantom{\rule{thinmathspace}{0ex}}n=0,1,2....\end{array}$
Clearly
${I}_{0}=\frac{\pi }{2}.$
Recall that
$\mathrm{tan}\phantom{\rule{negativethinmathspace}{0ex}}\left(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}-x\phantom{\rule{negativethinmathspace}{0ex}}\right)=\frac{1}{\mathrm{tan}x}$
giving
$\begin{array}{rl}\frac{1}{\mathrm{log}\mathrm{tan}\phantom{\rule{negativethinmathspace}{0ex}}\left(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}-x\phantom{\rule{negativethinmathspace}{0ex}}\right)}+\frac{1}{1-\mathrm{tan}\phantom{\rule{negativethinmathspace}{0ex}}\left(\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}-x\phantom{\rule{negativethinmathspace}{0ex}}\right)}& =\frac{1}{\mathrm{log}\left(\frac{1}{\mathrm{tan}x}\right)}+\frac{1}{1-\frac{1}{\mathrm{tan}x}}\\ \\ \text{(3)}& & =-\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}-\frac{1}{1-\mathrm{tan}x}+1\end{array}$
Hence, by the change of variable $x\to \frac{\pi }{2}-x$, we readily have
$\begin{array}{}\text{(4)}& {I}_{1}& ={\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\left(-\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}-\frac{1}{1-\mathrm{tan}x}+1\right)\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x=-{I}_{1}+\frac{\pi }{2}\end{array}$
${I}_{1}=\frac{\pi }{4}.$
Similarly,
$\begin{array}{}\text{(5)}& \begin{array}{rl}{\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\frac{1}{\mathrm{log}\left(\mathrm{tan}x\right)}+\frac{1}{1-\mathrm{tan}x}\right)}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x& ={\int }_{0}^{\phantom{\rule{negativethinmathspace}{0ex}}\frac{\pi }{2}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\left(\phantom{\rule{negativethinmathspace}{0ex}}-\frac{1}{\mathrm{log}\left(\mathrm{tan}x\phantom{\rule{negativethinmathspace}{0ex}}\right)}-\frac{1}{1-\mathrm{tan}x}+1\phantom{\rule{negativethinmathspace}{0ex}}\right)}^{3}\phantom{\rule{negativethinmathspace}{0ex}}\mathrm{d}x\end{array}\end{array}$
and, by the binomial expansion,
$\begin{array}{rl}{I}_{3}& =-{I}_{3}+3{I}_{2}-3{I}_{1}+{I}_{0}\end{array}$
${I}_{3}=-{I}_{3}+3{I}_{2}-3\frac{\pi }{4}+\frac{\pi }{2}$
$\begin{array}{}\text{(6)}& {I}_{3}=\frac{3}{2}{I}_{2}-\frac{\pi }{8}\end{array}$
we may conclude with this value obtained for ${I}_{2}$
A numerical approximation is
${I}_{3}=0.22709780663611705673940738484148718263....$