TeachingCalculus – Integration – Another remarkable integral
We will prove that \(\displaystyle\int_0^1\sqrt{\frac{x}{1-x}}\ln\left(\frac{x}{1-x}\right)\,dx=\pi\).
Start with the substitution \(\displaystyle\sqrt{\frac{x}{1-x}}=t\) which implies that \(x=\displaystyle\frac{t^2}{1+t^2}\). Then we have
\[I:=\int_0^1\sqrt{\frac{x}{1-x}}\ln\left(\frac{x}{1-x}\right)\,dx=\int_0^{\infty}t\ln(t^2)\cdot\frac{2t}{(1+t^2)^2}\,dt =4\int_0^{\infty}\frac{t^2}{(1+t^2)^2}\ln(t)\,dt.\]Now we use the fact that \(\displaystyle\frac{d}{dt}\left(\frac{1}{1+t^2}\right)=-\frac{2t}{(1+t^2)^2}\) to obtain that
\[I=-2\int_0^{\infty}t\ln(t)\,d\left(\frac{1}{1+t^2}\right)=-\frac{2t\ln(t)}{1+t^2}\bigg|_0^{\infty}+2\int_0^{\infty}\frac{1}{1+t^2}\,d\left(t\ln(t)\right) =2\int_0^{\infty}\frac{1+\ln(t)}{1+t^2}\,dt,\]since by l'Hospital's rule we have
\[\lim\limits_{t\downarrow 0}t\ln(t)=\lim\limits_{t\downarrow 0}\frac{\ln(t)}{1/t}=\lim\limits_{t\downarrow 0}\frac{1/t}{-1/t^2}=\lim\limits_{t\downarrow 0}(-t)=0\]and
\[\lim\limits_{t\to\infty}\frac{t\ln(t)}{1+t^2}=\lim\limits_{t\to\infty}\frac{\ln(t)+1}{2t}=\lim\limits_{t\to\infty}\frac{1/t+0}{2}=0.\]Now we have
\[I=2\int_0^{\infty}\frac{dt}{1+t^2}+2\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt=2\arctan(t)\bigg|_0^{\infty}+2\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt =\pi+2\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt.\]Finally we prove that \(\displaystyle\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt=0\).
Start with
\[\int_0^{\infty}\frac{\ln(t)}{1+t^2}\,dt=\int_0^1\frac{\ln(t)}{1+t^2}\,dt+\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt.\]Now we use the substitution \(t=1/u\) to obtain that
\[\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt=\int_1^0\frac{\ln(1/u)}{1+1/u^2}\cdot\left(-\frac{1}{u^2}\right)\,du =-\int_0^1\frac{\ln(u)}{1+u^2}\,du.\]Note that we used the fact that \(\displaystyle\int_1^{\infty}\frac{\ln(t)}{1+t^2}\,dt=-\int_0^1\frac{\ln(t)}{1+t^2}\,dt\).
It turns out that the value of this integral equals Catalan's constant, which will be considered later.
Last modified on March 8, 2021
