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Evaluate , if possible in a closed form, the integral:

$$\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$$

Basically, I have not done that much. I broke the integral

\begin{align*} \int_{0}^{\pi} \left ( \frac{\pi}{2}-x \right ) \frac{\tan x}{x} \, {\rm d}x &= \int_{0}^{\pi/2} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x + \int_{\pi/2}^{\pi} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x\\ &\!\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u + \int_{-\pi/2}^{0} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\ &= \int_{-\pi/2}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\ &\approx 2.13897 \end{align*}

I have no idea how to evaluate this. I was thinking of IBP and then some kind of Fourier , but I cannot get it to work. Any ideas?

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    $\begingroup$ What is the origin of this integral? What makes you believe that there is a closed-form solution? $\endgroup$ – Mark Viola Aug 23 '16 at 17:38
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    $\begingroup$ @Dr.MV Well I found it at an integration marathon. By closed form I mean whatever kind. It many contain special functions for example. $\endgroup$ – Tolaso Aug 23 '16 at 17:56
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    $\begingroup$ where was this marathon, on AOPS? $\endgroup$ – Dr. Sonnhard Graubner Aug 23 '16 at 18:45
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    $\begingroup$ \begin{align} u & = \frac \pi 2 - x \\ \\ x & = \frac \pi 2 - u\\ \\ \text{As $x$ goes from $0$ to $\frac \pi 2$, } & \text{ $u$ goes from $\frac\pi2$ to $0$.} \\ \\ du & = -dx \\ \\ \int_0^{\pi/2} \left( \frac \pi 2 - x \right) \frac{\tan x} x \, dx & = \int_{\pi/2}^0 u \, \frac{\tan\left(\frac \pi 2 - u \right)}{\frac\pi2 - u} \, (-du) = \int_0^{\pi/2} u\, \frac{\cot u}{\frac\pi2-u} \, du \\ \\ & = \int_0^{\pi/2} \frac 1 {\displaystyle \left( \frac \pi 2 -u \right) \frac{\tan u} u} \, du \end{align} $\endgroup$ – Michael Hardy Aug 31 '18 at 21:40
  • $\begingroup$ The integral of this function is equal to the integral of its reciprocal, and I wonder if something should be made of that. $\endgroup$ – Michael Hardy Aug 31 '18 at 21:40
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Too long for a comment. We have the following representations:

\begin{align*} \int_{0}^{\pi} \left(\frac{\pi}{2} - x\right) \frac{\tan x}{x} \, dx &= \frac{\pi^2}{4} + \frac{\pi}{2}\int_{0}^{1} \psi\left(\frac{1+x}{2}\right) \sin(2\pi x) \, dx \\ &= \frac{\pi^2}{4} - \pi \int_{0}^{\infty} \frac{dt}{(1+t^2)(e^{2\pi t} + 1)}. \end{align*}

At this moment I have no good idea of dealing with these, though I am posting this in hope of providing an alternative start for others as well as for a self record.

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  • $\begingroup$ It seems we can't use the digamma reflection formula to crack the first equality the one with the digamma integral. :( :( $\endgroup$ – Tolaso Jul 8 '18 at 14:49

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