Logarithmic Integral

[jacks]

Evaluation of $\displaystyle \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx$

[Tolaso J Kos]

Evaluation of $\displaystyle \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx$

Hello Jacks,

here is a solution. Consider the function:

$$f(s)= \int_{0}^{1}\frac{x^{s-1}}{\sqrt{1-x^2}}\, {\rm d}x = \frac{\sqrt{\pi}}{2}\frac{\Gamma \left ( \frac{s}{2} \right )}{\Gamma \left ( \frac{s+1}{2} \right )}, \; \; \mathfrak{Re}(s)>0$$

Differentiate with respect to $s$ once , hence:

$$\frac{\mathrm{d} }{\mathrm{d} s}f(s)= \int_{0}^{1}\frac{\partial }{\partial s}\frac{x^{s-1}}{\sqrt{1-x^2}}\, {\rm d}x= \int_{0}^{1}\frac{x^{s-1}\ln x}{\sqrt{1-x^2}}\, {\rm d}x = \frac{\sqrt{\pi}}{4}\cdot \frac{\Gamma \left ( \frac{s}{2} \right )}{\Gamma \left ( \frac{s+1}{2} \right )}\cdot \left [ \psi \left ( \frac{s}{2} \right )- \psi \left ( \frac{s+1}{2} \right ) \right] \tag{1} \label{*}$$

Plugging in \eqref{*} $s=2$ we have that:

$$\int_{0}^{1}\frac{x \ln x}{\sqrt{1-x^2}}\, {\rm d}x = \frac{\sqrt{\pi}}{4}\frac{\Gamma (1)}{\Gamma \left ( \frac{3}{2} \right )} \left [ \psi(1)- \psi \left ( \frac{3}{2} \right ) \right ]= \frac{\cancel{\sqrt{\pi}}}{4}\cdot \frac{2}{\cancel{\sqrt{\pi}}}\cdot \left [ \cancel{-\gamma} - 2+2\ln 2 +\cancel{\gamma} \right ] =\ln 2 -1$$



We used the known formulae:

• $\Gamma(1)=1$
• $\displaystyle \Gamma \left ( \frac{3}{2} \right )= \frac{\sqrt{\pi}}{2}$
• $\displaystyle \psi \left ( \frac{3}{2} \right )= 2 - 2\ln 2 - \gamma$
  
  while in more general it holds that:
  
  $$\psi \left ( n+\frac{1}{2} \right )= -2\ln 2 -\gamma + \sum_{k=1}^{n}\frac{2}{2k-1}$$

[jacks]

Thanks Admin. for Nice Solution.

Can we solve it Using Elementary Method.

[r9m]

\begin{align}\int_0^1 \frac{x\log x}{\sqrt{1-x^2}} &= \frac{1}{4}\int_0^1 \frac{\log y}{\sqrt{1-y}}\,dy\\&= \frac{1}{2}\int_0^1 \log (1-z^2)\,dz\\&= \frac{1}{2}[z\log (z) - z + (z+1)\log (z+1) - (z+1)]_0^1 \\&= \log 2 - 1\end{align}

Made the change of variable, $y = x^2$ in step $(1)$ and $z = \sqrt{1-y}$ in step $(2)$.

[jacks]

Thanks r9m.