Integral $ \int_0^{\pi/2} x \sqrt{\cot x} \, dx $

Question

I cannot evaluate the following integral

$$ \int_0^{\pi/2} x \sqrt{\cot x} \, dx $$

which means

$$ \int_0^{\pi/2} \frac{x}{\sqrt{\tan x}} \, dx = 0.97482\ldots. $$

This integral seems to be equal to the value of

$$ \frac\pi{2\sqrt2} \left( \frac\pi2 - \ln2 \right) = 0.97482\ldots. $$

If anyone knows how to solve this integral, I would be very interested to know. (Or do you know what would be a generalization of this integral? And do you know how to prove it?)

P.S. This intgral is equal to

$$ \frac12 \int_0^\infty \frac{\sqrt{\frac1x} \tan^{-1} x + \sqrt{x} \tan^{-1} \frac1x}{1+x^2} \, dx. $$

Answer 1

Your integral is equivalent to $$ \int_0^{\infty} \frac{\arctan t}{\sqrt{t} \,(1+t^2)} \, dt $$ by substituting $x=\arctan t$. To evaluate this, we recall a well known integral from complex analysis: $$ \int_0^{\infty} \frac{t^{s-1}}{1+t} \, dt = \frac{\pi}{\sin(\pi s)}, \hspace{0.5cm} 0<\text{re}(s)<1 $$

Substituting $x^2t^2$ for $t$, where $x$ is some positive real number, and replacing $s$ with $\frac{s}{2}$ gives $$ \frac{\pi x^{-s}}{2\sin(\frac{\pi s}{2})} = \int_0^{\infty} \frac{t^{s-1}}{1+(xt)^2} \, dt, \hspace{0.5cm} 0<s<2, \; x>0 $$ Subtracting two such expressions gives $$ \frac{\pi (1-x^{2-s})}{2\sin(\frac{\pi s}{2})} = \int_0^{\infty} t^{s-1} \left(\frac{1}{1+t^2} - \frac{x^2}{1+(xt)^2} \right) dt = (1-x^2) \int_0^{\infty} \frac{t^{s-1}}{(1+t^2)(1+(xt)^2)} \, dt $$ and therefore $$ \frac{\pi}{2\sin(\frac{\pi s}{2})} \frac{1-x^{2-s}}{1-x^2} = \int_0^{\infty} \frac{t^{s-1}}{(1+t^2)(1+(xt)^2)} \, dt, \hspace{0.5cm} 0<s<2, \; x>0 $$ We now integrate both sides with respect to $x$ from $0$ to $1$ and apply Fubini's theorem on the right, which is justified by absolute convergence. This gives $$ \frac{\pi}{2\sin(\frac{\pi s}{2})} \int_0^1 \frac{1-x^{2-s}}{1-x^2} \, dx = \int_0^{\infty} \frac{t^{s-1}}{1+t^2} \int_0^1 \frac{dx}{1+(xt)^2} \, dt = \int_0^{\infty} \frac{t^{s-2} \arctan t}{1+t^2} \, dt $$ Finally setting $s=\frac{3}{2}$ we obtain the integral you're looking for on the right and $$ \frac{\pi}{\sqrt{2}} \int_0^1 \frac{1-\sqrt{x}}{1-x^2} \, dx = \frac{\pi}{\sqrt{2}} \int_0^1 \frac{1-t}{1-t^4} \, 2tdt = \frac{\pi}{\sqrt{2}} \int_0^1 \frac{2t}{(1+t)(1+t^2)} \, dt \\ = \frac{\pi}{\sqrt{2}} \int_0^1 \left(\frac{1+t}{1+t^2}-\frac{1}{1+t} \right) dt $$ on the left. Can you finish things off?

Answer 2

Substitute $t=\sqrt{\cot x}$

\begin{align} &\int_0^{\pi/2} x \sqrt{\cot x} \, dx \\ =& \int_0^\infty \frac{2t^2\cot^{-1}t^2}{1+t^4}dt = \int_0^\infty \int_0^1 \frac{2t^4}{(1+t^4)(y^2+t^4)}dy\ dt\\ =& \int_0^1 \int_0^\infty \frac2{1-y^2}\left(\frac1{1+t^4}-\frac{y^2}{y^2+t^4}\right)dt\ dy\\ =&\ \frac\pi{\sqrt2}\int_0^1 \frac{1-\sqrt y}{1-y^2}dy =\frac{\pi}{4\sqrt2}(\pi -2\ln2) \end{align}