I'm trying to compute

$$\int_{0}^{\pi/2} (\frac{\pi}{2} - x)\tan(x)\ dx.$$

This is an improper integral and according to Mathematica, its value is $\frac{\pi \log(2)}{2}$. I can't expand the product because $\int_{0}^{\pi/2} \tan(x)$ and $\int_{0}^{\pi/2} x\tan(x)$ both diverge. I also don't see any obvious substitutions. If possible, I'd like to see a solution using elementary methods (i.e. no contour integration).

In case this is relevant, I came across this integral when trying to find an alternate solution for this question. I started with

$$ \int_1^{\infty}\frac{\ln x}{x\sqrt{x^2-1}}\ dx. $$

By integration by parts, this is equal to

$$ -\log(x) \arctan(\frac{1}{\sqrt{x^2-1}})\Big|_1^\infty + \int_1^\infty \frac{\arctan(\frac{1}{\sqrt{x^2-1}})}{x}\ dx $$

The left term is $0$. For the right term, I substituted $x = \sec(u)$ and got the integral $$\int_0^{\pi/2} \arctan(\frac{1}{\tan(u)}) \tan(u) \ du.$$ Then, since $\tan(u)\geq 0$ for $u \in [0,\pi/2]$, I used the identity $\arctan(y) + \arctan(1/y) = \pi/2$ for all $y > 0$ to get the integral I'm asking about.


2 Answers 2


Your are asking $$\int_0^{\frac{\pi}{2}} x\cot x dx$$

Note first the identity:

$$\sum_{n=1}^{N} \sin(2nx) = \frac{1}{2}\cot x - \frac{\cos(2N+1)x}{2\sin x}$$ Hence $$\int_0^{\frac{\pi }{2}} {x\cot xdx} = 2\sum\limits_{n = 1}^N {\int_0^{\frac{\pi }{2}} {x\sin (2nx)dx} } + \int_0^{\frac{\pi }{2}} {\frac{x}{{\sin x}}\cos (2N + 1)xdx}$$ Since $x/\sin x$ is continuous on $[0,\pi/2]$, Riemann-Lebesgue lemma says the last term tends to 0. Hence $$\int_0^{\frac{\pi }{2}} {x\cot xdx} = 2\sum\limits_{n = 1}^\infty {\int_0^{\frac{\pi }{2}} {x\sin (2nx)dx} } = 2\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^{n + 1}\pi}}}{{4 n}}} = \frac{\pi }{2}\ln 2$$

Yet another proof comes from integration by parts:$$\int_0^{\frac{\pi }{2}} {x\cot xdx} = - \int_0^{\frac{\pi }{2}} {\ln (\sin x)dx} = -I$$ We have $$ I = \frac{1}{2}\int_0^\pi {\ln (\sin x)dx} = \int_0^{\frac{\pi }{2}} {\ln (\sin 2x)dx} = \frac{\pi }{2}\ln 2 + \int_0^{\frac{\pi }{2}} {\ln (\sin x)dx} + \int_0^{\frac{\pi }{2}} {\ln (\cos x)dx} $$ Therefore $$I = \frac{\pi }{2}\ln 2 + 2I$$

I believe the second proof is the standard way to evaluate this integral (this is a famous integral actually).


A symmetry argument is enough. By integration by parts your integral equals:

$$ I = -\int_{0}^{\pi/2}\log\cos(t)\,dt = -\int_{0}^{\pi/2}\log\sin(t)\,dt = -\int_{0}^{\pi/2}\log\frac{\sin(2t)}{2\sin t}\,dt $$ and since $-\int_{0}^{\pi/2}\log\sin(2t)\,dt=-\frac{1}{2}\int_{0}^{\pi}\log\sin(t)\,dt =I$ the previous line leads to

$$ I = I - I + \frac{\pi}{2}\log 2 $$ from which $I=\frac{\pi}{2}\log 2$.

  • $\begingroup$ Thanks, this is also really cool. $\endgroup$
    – aras
    Jul 8, 2017 at 18:52

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