Calculating $\int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx$

Question

I'm stuck on this problem and would appreciate any help.

I'm trying to compute the integral $$ \int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx. $$ To approach it, I define a parametric function $$ F(a) = \int_0^1 \frac{\ln(ax)}{\sqrt{x - x^2}}\, dx, $$ and then I differentiate under the integral sign: $$ F'(a) = \int_0^1 \frac{1}{a\sqrt{x - x^2}}\, dx. $$ This simplifies to $$ F'(a) = \frac{1}{a} \int_0^1 x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\, dx = \frac{1}{a} B\left(\frac{1}{2}, \frac{1}{2}\right), $$ where $ B $ denotes the Beta function. Since $ B\left(\frac{1}{2}, \frac{1}{2}\right) = \pi $, we get $$ F'(a) = \frac{\pi}{a}. $$ Integrating with respect to $ a $ , we obtain $$ F(a) = \pi \ln(a) + C. $$ Now I want to compute the original integral, which is $ F(1) $ , but I don't know how to determine the constant $ C $ . Any ideas on how to proceed?

Answer 1

I´m afraid Feynman's trick won't work here, like that.

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Let the integral be $I$. Let $x = \sin^2 \theta$, then $$I = \int_0^{\pi/2} \frac{\ln(\sin^2 \theta)}{\sin \theta \cos \theta} (2 \sin \theta \cos \theta)\, d\theta=\int_0^{\pi/2} 2 \ln(\sin^2 \theta)\, d\theta =4 \int_0^{\pi/2} \ln(\sin \theta)\, d\theta$$ where we used $$ \sqrt{x - x^2} = \sqrt{\sin^2 \theta - \sin^4 \theta} = \sqrt{\sin^2 \theta (1 - \sin^2 \theta)} = \sqrt{\sin^2 \theta \cos^2 \theta} =\sin \theta \cos \theta, $$ since for $\theta \in [0, \pi/2]$, $\sin \theta \ge 0$ and $\cos \theta \ge 0.$

Let $J = \int_0^{\pi/2} \ln(\sin \theta)\, d\theta$. This is a well-known definite integral (cf. George's answer here), and we have $$J = -\frac{\pi}{2} \ln 2.$$

Finally, substitute this value of $J$ back into the expression for $I$: $$\boxed{I = 4J = -2\pi \ln 2.}$$

Answer 2

Suppose for $a\in \mathbb{R}$, we have $$\Omega(a)=\int_0^1\frac{x^a}{\sqrt{x-x^2}}\,dx$$ it's clear that $$\Omega'(a)=\int_0^1\frac{x^a\ln(x)}{\sqrt{x-x^2}}\,dx \qquad \text{and} \qquad \Omega'(0)=\int_0^1\frac{\ln (x)}{\sqrt{x-x^2}}\,dx$$ To this end, $$\Omega(a)=\int_0^1\frac{x^a}{\sqrt{x-x^2}}\,dx=\int_0^1x^{a-1/2}(1-x)^{-1/2}\,dx=\beta(a+1/2,1/2)=\frac{\Gamma(a+1/2)\Gamma(1/2)}{\Gamma(a+1)}$$ Upon differentiating, $$\Omega'(a)=\frac{\sqrt{\pi}\Gamma(a+1/2)}{\Gamma(a+1)}\left(\psi^{(0)}(a+1/2)-\psi^{(0)}(a+1)\right)$$ Hence at $a=0$, $$\Omega'(0)=\int_0^1\frac{\ln (x)}{\sqrt{x-x^2}}\,dx=\pi\left(\psi^{(0)}(1/2)-\psi^{(0)}(1)\right)=\pi(-2\ln(2)-\gamma-(-\gamma))$$ Finally, $$\int_0^1\frac{\ln (x)}{\sqrt{x-x^2}}\,dx=-2\pi\ln(2)$$

Some side notes:

1. $\psi^{(0)}(\cdot)$ is the digamma function
2. $\beta(m,n)$ is the beta function
3. $\gamma$ is the Euler-Mascheroni constant

Answer 3

If we did continue your work, $$ F(a) = \pi \ln(a) + C. $$

Put $a=1$,

$$C=F(1)=\int_0^1 \frac{\ln(x)}{\sqrt{x - x^2}}\, dx=\int_0^1 \frac{\ln(x)}{\sqrt{x}\sqrt{1 - x}}\, dx$$

$$\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,dx\implies F(1)\implies\lim_{(m,n)\to(\frac12,\frac12)}\frac{\partial}{\partial m}\beta(m,n)$$

$$F(1)=\pi \left( \gamma + \psi\left(0, \tfrac{1}{2} \right) \right)=-2\pi\ln(2)\implies C=-2\pi\ln(2)$$

$$\therefore \int_0^1 \frac{\ln(ax)}{\sqrt{x - x^2}}\, dx = \pi \ln\left(\frac{a}{4}\right) $$

Answer 4

Let $2x-1=\cos (2 \theta)$, then $$ \begin{aligned} I&= 2 \int_0^1 \frac{\ln x}{\sqrt{1-(2 x-1)^2}} d x \\&=2 \int_0^{\frac{\pi}{2}} \ln \left(\frac{\cos (2 \theta)+1}{2}\right) d \theta \\ & =2 \int_0^{\frac{\pi}{2}} \ln \left(\cos ^2 \theta\right) d \theta \\ & =4\left(-\frac{\pi}{2} \ln 2\right) \\ & =-2 \pi \ln 2 \end{aligned} $$

Answer 5

$$I=\int\frac{\log(x)}{\sqrt{x - x^2}}\, dx$$

$$x=\sin ^2(t) \quad \implies \quad I=4\int \log (\sin (t))\,dt$$ $$\int \log (\sin (t))\,dt=\frac{1}{2} i \left(t^2+ \text{Li}_2\left(e^{2 it}\right)\right)-t \log \left(1-e^{2 i t}\right)+t \log (\sin (t))$$ $$\int_0^{\frac \pi 2} \log (\sin (t))\,dt=\left(-\frac{1}{2} \pi \log (2)+\frac{i \pi ^2}{12}\right)-\frac{i \pi ^2}{12}=-\frac{1}{2} \pi \log (2)$$

$$J=\int_0^{1}\frac{\log(x)}{\sqrt{x - x^2}}\, dx=-2\pi \log (2)$$

You could even go much further $$K=\int_0^{1-\epsilon}\frac{\log(x)}{\sqrt{x - x^2}}\, dx$$ $$K=-2\pi \log (2)+\frac{2}{3} \epsilon ^{3/2}\Bigg(1+\frac{3 \epsilon }{5}+\frac{23 \epsilon ^2}{56}+\frac{11 \epsilon^3}{36}+ \frac{1689 \epsilon^4}{7040}+O\left(\epsilon ^5\right) \Bigg)$$

Answer 6

$$ \int_0^1 \frac{\ln x}{\sqrt{x - x^2}}\, dx = -\int_0^1 \int_0^1 \frac{2y\sqrt{\frac1x-1}}{(1-x)y^2+x}dy \ dx\\ =-\int_0^1\frac{2\pi}{1+y}dy=-2\pi\ln2 $$