# Definite Integral

#### sbhatnagar

##### Active member
Prove that

$\int_0^1 \frac{\ln x}{\sqrt{x(1-x^2)}}dx=-\frac{\sqrt{2\pi}}{8} \left(\Gamma\left(\frac{1}{4} \right)\right)^2$

$$\Gamma (x)$$ is the Gamma Function.

#### sbhatnagar

##### Active member
Here's a Hint.

Differentiate the identity,

$\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$

with respect to the parameter $$a$$.

$\int_0^1 x^{a-1}(1-x)^{b-1}\ln x \ dx = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} \left( \psi(a)-\psi(a+b)\right)$

where $$\psi(x)$$ is the Digamma Function.

#### sbhatnagar

##### Active member
I will post the solution now. In the integral,

$I=\int_0^1 x^{a-1}(1-x^c)^{b-1}\ln x \ dx$

substitute $$t=x^c$$, and obtain

$I=\frac{1}{c^2}\int_0^1 t^{\frac{a}{c}-1}(1-t)^{b-1}\ln t \ dt$

This integral can be evaluated using the result obtained in my previous post.

$I=\frac{\Gamma(a/c) \Gamma(b)}{c^2\Gamma(a/c +b)}\left( \psi(a/c)-\psi(a/c+b)\right)$

If we put $$a=1/2,b=1/2,c=2$$, we will obtain

$\int_0^1\frac{\ln x}{\sqrt{x(1-x^2)}}=\frac{\Gamma(1/4)\Gamma(1/2)}{4\Gamma (3/4)}\left( \psi(1/4)-\psi(3/4)\right)$

From the reflection rule of gamma function, we have

$$\displaystyle \Gamma \left(\frac{3}{4} \right)=\frac{\pi}{\sin(\frac{\pi}{4})\Gamma(\frac{1}{4})}=\frac{\pi\sqrt{2}}{\Gamma(\frac{1}{4})}$$

Also, $$\psi(1-x)-\psi(x)=\pi\cot \pi x$$. Therefore

$$\psi(1/4)-\psi(3/4)=\pi \cot(3\pi/ 4)=-\pi$$

From all these we obtain

$\int_0^1 \frac{\ln x}{\sqrt{x(1-x^2)}}=-\frac{\sqrt{2\pi}}{8}\left( \Gamma \left ( \frac{1}{4}\right)\right)^2$