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#### sbhatnagar

##### Active member

- Jan 27, 2012

- 95

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

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

- Thread starter sbhatnagar
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- Thread starter
- #1

- Jan 27, 2012

- 95

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

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

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- Jan 27, 2012

- 95

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.

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- Jan 27, 2012

- 95

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

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

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