MHB Proving Definite Integral: \(\ln x/\sqrt{x(1-x^2)}\)

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The integral \(\int_0^1 \frac{\ln x}{\sqrt{x(1-x^2)}}dx\) is proven to equal \(-\frac{\sqrt{2\pi}}{8} \left(\Gamma\left(\frac{1}{4} \right)\right)^2\). The proof involves substituting \(t=x^c\) in the integral and applying properties of the Gamma function. By setting specific values for \(a\), \(b\), and \(c\), the integral is transformed and evaluated using known results. The reflection rule of the Gamma function and properties of the digamma function are utilized to simplify the expression. This leads to the final result confirming the original claim about the integral's value.
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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.
 
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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.
 
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

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