Evaluate the integral:
$$\int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}\, {\rm d}x$$
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:
$$\int_{0}^{1}\frac{\partial }{\partial s}\frac{x^{s-1}}{\sqrt{1-x^2}}\, {\rm d}x= \int_{0}^{1}\frac{x^{s-1}\ln x}{\sqrt{1-x^2}}\, {\rm d}x = \frac{\sqrt{\pi}}{4}\cdot \frac{\Gamma \left ( \frac{s}{2} \right )}{\Gamma \left ( \frac{s+1}{2} \right )}\cdot \left [ \psi \left ( \frac{s}{2} \right )- \psi \left ( \frac{s+1}{2} \right ) \right] \tag{1} \label{*}$$
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:
The exercise can also be found in mathimatikoi.org
$$\int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}\, {\rm d}x$$
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:
$$\int_{0}^{1}\frac{\partial }{\partial s}\frac{x^{s-1}}{\sqrt{1-x^2}}\, {\rm d}x= \int_{0}^{1}\frac{x^{s-1}\ln x}{\sqrt{1-x^2}}\, {\rm d}x = \frac{\sqrt{\pi}}{4}\cdot \frac{\Gamma \left ( \frac{s}{2} \right )}{\Gamma \left ( \frac{s+1}{2} \right )}\cdot \left [ \psi \left ( \frac{s}{2} \right )- \psi \left ( \frac{s+1}{2} \right ) \right] \tag{1} \label{*}$$
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:
$$\psi \left ( n+\frac{1}{2} \right )= -2\ln 2 -\gamma + \sum_{k=1}^{n}\frac{2}{2k-1}$$
The exercise can also be found in mathimatikoi.org
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