Definite Integral (08)

[jacks]

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

[Grigorios Kostakos]

\begin{align*}
I(m)&=\int_{0}^{1}{\frac{x^{m-1}}{\sqrt{1-x^2}}\,dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,x^2}\\
{\frac{1}{2}\,t^{-\frac{1}{2}}dt\,=\,dx}\\
\end{subarray}}\,\frac{1}{2}\int_{0}^{1}{\frac{t^{\frac{m}{2}-1}}{\sqrt{1-t}}\,dt} \\
&=\frac{1}{2}\,{\rm{B}}\bigl({\tfrac{m}{2},\tfrac{1}{2}}\bigr)=\frac{1}{2}\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)\,\Gamma\bigl({\tfrac{1}{2}}\bigr)}{\Gamma\bigl({\tfrac{m}{2}+\tfrac{1}{2}}\bigr)}=\frac{\sqrt{\pi}}{2}\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)}{\Gamma\bigl({\tfrac{m+1}{2}}\bigr)}\quad\Longrightarrow\\
\frac{d}{dm}I(m)&=\int_{0}^{1}{\frac{x^{m-1}\log{x}}{\sqrt{1-x^2}}\,dx}\\
&=\frac{\sqrt{\pi}}{2}\,\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)}{2\,\Gamma\bigl({\tfrac{m+1}{2}}\bigr)}\bigl({\psi\bigl({\tfrac{m}{2}}\bigr)-\psi\bigl({\tfrac{m+1}{2}}\bigr)}\bigr)\quad\stackrel{m=3}{=\!=\!\Longrightarrow}\\
\int_{0}^{1}{\frac{x^2\log{x}}{\sqrt{1-x^2}}\,dx}&=\frac{\sqrt{\pi}}{4}\frac{\Gamma\bigl({\tfrac{3}{2}}\bigr)}{\Gamma(2)}\bigl({\psi\bigl({\tfrac{3}{2}}\bigr)-\psi(2)}\bigr)\\
&=\frac{\sqrt{\pi}}{4}\frac{\sqrt{\pi}}{2}\,({2-\gamma-2\log2-1+\gamma})\\
&=\frac{\pi}{8}\,(1-2\log{2})\,.
\end{align*}

[Papapetros Vaggelis]

Here is a 2nd solution assuming that \(\displaystyle{I=\int_{0}^{1}\dfrac{x^2\,\ln\,x}{\sqrt{1-x^2}}\,\mathrm{d}x<\infty}\) .

We apply the sub \(\displaystyle{t=\arcsin\,x\,,t\in\left[0,\dfrac{\pi}{2}\right]}\) and we get :

\(\displaystyle{\mathrm{d}t=\dfrac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\,\,,x=\sin\,t}\) , so :

\(\displaystyle{I=\int_{0}^{\dfrac{\pi}{2}}\sin^2\,t\,\ln\,\left(\sin\,t\right)\,\mathrm{d}t\,\,(1)}\) .

On the other hand,

\(\displaystyle{\begin{aligned} I&=\int_{0}^{1}\left(\dfrac{-x}{\sqrt{1-x^2}}\right)\,\left(-x\,\ln\,x\right)\,\mathrm{d}x\\&=\left[-x\,\ln\,x\,\sqrt{1-x^2}\right]_{0}^{1}+\int_{0}^{1}\sqrt{1-x^2}\,\left(\ln\,x+1\right)\,\mathrm{d}x\\&\stackrel{(\ast)}{=}\int_{0}^{1}\sqrt{1-x^2}\,\left(\ln\,x+1\right)\,\mathrm{d}x\end{aligned}}\)

\(\displaystyle{(\ast) : \lim_{x\to 0^{+}}x\,\ln\,x=0}\)

If \(\displaystyle{J=\int_{0}^{1}\sqrt{1-x^2}\,\left(\ln\,x+1\right)\,\mathrm{d}x}\) , then \(\displaystyle{I=J}\) .

By substituting \(\displaystyle{x=\sin\,y}\) we have that :

\(\displaystyle{\begin{aligned} J&=\int_{0}^{\frac{\pi}{2}}\left[\ln\,\left(\sin\,y\right)+1\right]\,\cos^2\,y\,\mathrm{d}y\\&=\int_{0}^{\frac{\pi}{2}}\left[\ln\,\left(\sin\,y\right)+1\right]\,\left(1-\sin^2\,y\right)\,\mathrm{d}y\\&=\int_{0}^{\frac{\pi}{2}}\left[\ln\,\left(\sin\,y\right)-\sin^2\,y\,\ln\,\left(\sin\,y\right)+\cos^2\,y\right]\,\mathrm{d}y\\&\stackrel{(I)}{=}\int_{0}^{\frac{\pi}{2}}\ln\,\left(\sin\,y\right)\,\mathrm{d}y-I+\int_{0}^{\frac{\pi}{2}}\cos^2\,y\,\mathrm{d}y\\&=-\dfrac{\pi\,\ln\,2}{2}-I+\int_{0}^{\frac{\pi}{2}}\dfrac{\cos\,2\,y+1}{2}\,\mathrm{d}y\\&=-\dfrac{\pi\,\ln\,2}{2}-I+\left[\dfrac{y}{2}+\dfrac{\sin\,2\,y}{4}\right]_{0}^{\frac{\pi}{2}}\\&=-\dfrac{\pi\,\ln\,2}{2}-I+\dfrac{\pi}{4}\\&=-I+\dfrac{\pi}{4}\,\left(1-2\,\ln\,2\right)\end{aligned}}\)

so,

\(\displaystyle{I=-I+\dfrac{\pi}{4}\,\left(1-2\,\ln\,2\right)\iff 2\,I=\dfrac{\pi}{4}\,\left(1-2\,\ln\,2\right)\iff I=\dfrac{\pi}{8}\,\left(1-2\,\ln\,2\right)}\) .