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How does one evaluate the following improper integral using Leibniz's integration rule?

$$\int^{\frac{\pi}{2}}_{0}x\cot (x)dx$$

I tried to add a new parameter $\ln(\sec(tx))$

$$f(t) = \int^{\frac{\pi}{2}}_{0}x\cot (x)\ln(\sec(tx))dx$$

$$\dfrac{\partial}{\partial t}f(t)= \int^{\frac{\pi}{2}}_{0}x\cot (x) \dfrac{\partial}{\partial t}\biggr (\ln(\sec(tx))\biggr)$$

$$\dfrac{\partial}{\partial t}f(t)= \int^{\frac{\pi}{2}}_{0}x\cot (x) x\tan(tx)dx$$

$$\dfrac{\partial}{\partial t}f(t)= \int^{\frac{\pi}{2}}_{0}x^2\cot (x) \tan(tx)dx$$

When $t = 1$,

$$\dfrac{\partial}{\partial t}f(1)= \int^{\frac{\pi}{2}}_{0}x^2\cot (x) \tan(x)dx = \int^{\frac{\pi}{2}}_{0}x^2dx$$

I could find $f(1)$ from there but I have to find $f(0)$.

Community
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Melz
  • 792

2 Answers2

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By integration by parts, $$ \int_{0}^{\pi/2}x\cot(x)\,dx = -\int_{0}^{\pi/2}\log\sin(x)\,dx = -\int_{0}^{1}\frac{\log(t)}{\sqrt{1-t^2}}\,dt=-\left.\frac{d}{d\alpha}\int_{0}^{1}\frac{t^\alpha\,dt}{\sqrt{1-t^2}}\right|_{\alpha=0} $$ and by Euler's Beta function $$ \int_{0}^{1}\frac{t^\alpha\,dt}{\sqrt{1-t^2}}=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{1+\alpha}{2}\right)}{\Gamma\left(1+\frac{\alpha}{2}\right)} $$ so $$\int_{0}^{\pi/2}x\cot(x)\,dx =\frac{\pi}{4}\left[\psi(1)-\psi\left(\tfrac{1}{2}\right)\right]=\frac{\pi}{2}\log(2).$$ There are plenty of other ways: exploiting symmetry, Fourier series, Weierstrass products etc.

Jack D'Aurizio
  • 353,855
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A more complete (and slightly different) version of Jack's answer.

The Gamma function is defined as $$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt.$$ Setting $t=x^2$ gives $$\Gamma(s)=2\int_0^\infty x^{2s-1}e^{-x^2}dx.$$ Thus, $$\Gamma(a)\Gamma(b)=4\int_0^\infty \int_0^\infty x^{2a-1}y^{2b-1}e^{-(x^2+y^2)}dxdy.$$ Then we convert the integrals to polar coordinates to get $$\begin{align} \Gamma(a)\Gamma(b)&=4\int_0^{\pi/2}\int_0^{\infty} r(r\cos\theta)^{2a-1}(r\sin\theta)^{2b-1}e^{-r^2}drd\theta\\ &=4\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}\int_0^{\infty} r^{2a+2b-1}e^{-r^2}drd\theta\\ &=2\left(2\int_0^{\infty} r^{2a+2b-1}e^{-r^2}dr\right)\left(\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta\right)\\ &=2\Gamma(a+b)\int_0^{\pi/2}\cos(\theta)^{2a-1}\sin(\theta)^{2b-1}d\theta . \end{align}$$ So we have the integral $$\int_0^{\pi/2}\sin(t)^{a}\cos(t)^{b}dt=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ which is equivalent to $$\int_0^1 t^a(1-t)^bdt=\mathrm B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ Anyway, as Jack noticed, $$\int_0^{\pi/2}x\cot x\ dx=-\int_0^{\pi/2}\ln\sin x\ dx.$$ But from the Leibniz integral rule, we know that $$\int_0^{\pi/2}\sin(t)^a\cos(t)^b\ln^{n}(\sin t)\ln^{m}(\cos t)dt=\partial_a^n\partial_b^m \frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$ so $$-\int_0^{\pi/2}\ln\sin x\ dx=-\partial_{a}\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{1}2)}{2\Gamma(\frac{a}2+1)}\Bigg|_{a=0}=-\left(\frac{\sqrt\pi}2\right)^2\left[\psi_0\left(\tfrac12\right)-\psi_0(1)\right]=\frac\pi4\ln2\ .$$

clathratus
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