Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$.

Question

Evaluate $$\int_0^1 f(x) dx$$ where

$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$

I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to

$$\sin^2y\cdot \ln (\sin y) dy$$

Then I used the property of definite integrals that

$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$

Then too it wasn't getting simplified. I tried $e^z=\sin x$, but this gave no headway because after a while I reached a complete full-stop. How should I go about this?

Answer 1

An answer that does not use hypergeometric functions:

First integrate by parts using the functions $u(x)=x\ln x$ and $v(x)=-\sqrt{1-x^2}$, we have $u'(x)=\ln x +1$ and $v'(x)=\frac{x}{\sqrt{1-x^2}}$ and we get $$I=\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}\mathrm dx= \underbrace{\left[-x\ln x\sqrt{1-x^2}\right]_0^1}_{=0}+\int_0^1(\ln x+1)\sqrt{1-x^2}\,\mathrm dx.$$ The integral $\int_0^1\sqrt{1-x^2}\mathrm dx=\frac\pi4$ is easy. Let us concentrate on $$J=\int_0^1\ln x \,\sqrt{1-x^2}\mathrm dx$$ for which it seems a trigonometric change of variable will work. Let us set $x=\cos\theta$, $\mathrm dx=-\sin\theta\mathrm d\theta$, then $$J=\int_0^{\pi/2}\ln(\cos\theta)\sin^2\theta\mathrm d\theta =\int_0^{\pi/2}\ln(\cos\theta)(1-\cos^2\theta)\mathrm d\theta.\tag1$$

We can use the result $$\int_0^1\ln(\sin\theta)\,\mathrm d\theta=\int_0^{\pi/2}\ln(\cos\theta)\,\mathrm d\theta=-\frac\pi2\ln2$$ (see for instance this post for a derivation). Therefore we have $$J=-\frac\pi2\ln2-\int_0^{\pi/2}\ln(\cos\theta)\cos^2\theta\,\mathrm d\theta.\tag2$$

Adding up (1) and (2) we obtain $$2J=-\frac\pi2\ln2+\int_0^{\pi/2}\ln(\cos\theta)\left(\sin^2\theta-\cos^2\theta\right)\,\mathrm d\theta=-\frac\pi2\ln2-\int_0^{\pi/2}\ln(\cos\theta)\cos(2\theta)\,\mathrm d\theta.$$ Finally let us integrate by parts (with $u(x)=\ln(\cos\theta)$ and $v(x)=\frac12\sin(2\theta)$, $u'(x)=-\tan\theta$ and $v'(x)=\cos(2\theta)$) $$\begin{split}\int_0^{\pi/2}\ln(\cos\theta)\cos(2\theta)\,\mathrm d\theta&= \underbrace{\left[\ln(\cos\theta)\frac12\sin(2\theta)\right]_0^{\pi/2}}_{=0} +\int_0^{\pi/2}\tan\theta\,\frac12\sin2\theta\mathrm d\theta\\ &=\int_0^{\pi/2}\sin^2\theta\,\mathrm d\theta=\frac\pi4 \end{split}$$ We get $2J=-\frac\pi2\ln2-\frac\pi4$ and therefore $$I=-\frac\pi4\ln2-\frac\pi8+\frac\pi4=\boxed{\frac\pi8-\frac\pi4\ln2}.$$

Answer 2

Consider the integral $$I\left(t\right)=\frac{1}{2}\int_{0}^{1}\frac{x^{2t}}{\sqrt{1-x^{2}}}dx$$ we can observe that $$I'\left(1\right)=\int_{0}^{1}\frac{x^{2}\log\left(x\right)}{\sqrt{1-x^{2}}}dx.$$ So we have $$I\left(t\right)=\frac{1}{2}\int_{0}^{1}\frac{x^{2t}}{\sqrt{1-x^{2}}}dx\stackrel{x^{2}=v}{=}\frac{1}{4}\int_{0}^{1}\frac{v^{t-1/2}}{\sqrt{1-v}}dv$$ and using the identity involving the hypergeometric function $$_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{v^{b-1}\left(1-v\right)^{c-b-1}}{\left(1-vz\right)^{a}}$$ under the hypothesis $$\textrm{Re}\left(c\right)>\textrm{Re}\left(b\right)>0\wedge\left|\textrm{arg}(1-z)\right|<\pi$$ we get $$I\left(t\right)=\frac{\Gamma\left(t+1/2\right)}{4\Gamma\left(t+3/2\right)}{}_{2}F_{1}\left(\frac{1}{2},t+\frac{1}{2};t+\frac{3}{2};1\right)$$ and again using the closed form $$_{2}F_{1}\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)},\,\textrm{Re}\left(c-a-b\right)>0$$ we have $$I\left(t\right)=\frac{\Gamma\left(t+1/2\right)}{4\Gamma\left(t+3/2\right)}\frac{\Gamma\left(t+3/2\right)\Gamma\left(1/2\right)}{\Gamma\left(t+1\right)\Gamma\left(1\right)}=\frac{\Gamma\left(1/2\right)}{4}\frac{\Gamma\left(t+1/2\right)}{\Gamma\left(t+1\right)}$$ so now we can take the derivative and evalutate it at $t=1$ $$I'\left(1\right)=\frac{\Gamma\left(1/2\right)}{4}\frac{\Gamma\left(3/2\right)\left(\psi\left(3/2\right)-\psi\left(2\right)\right)}{\Gamma\left(2\right)}=-\frac{1}{8}\pi\left(\log\left(4\right)-1\right)\approx-0.151697$$ where $\psi(x)$ is the digamma function.

Answer 3

$$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$ Using Beta function we have $$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$$ then with Digamma function $\psi$ \begin{align} I(\alpha) = \dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx \\ = \dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)\left(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha+2}{2})\right) \end{align} now let $\alpha=2$, $$\psi(\frac{3}{2})-\psi(\frac{4}{2})=2\sum_{n=0}^\infty\left(\dfrac{1}{2n+4}-\dfrac{1}{2n+3}\right)=1-\ln4$$ therefore $$I(2)=\color{blue}{\dfrac{\pi}{8}(1-\ln4)}$$

Answer 4

$$ \begin{aligned} I &\stackrel{x=\sin \theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \theta \ln (\sin \theta)}{\cos \theta} \cdot \cos \theta d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \sin ^{2} \theta \ln (\sin \theta) d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)\\& \stackrel{IBP}{=} \left[\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right) \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{\theta}{2}-\frac{\sin 2 \theta}{4}\right)d(\ln(\sin \theta) )\\& =-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\theta}{2} d(\ln (\sin \theta))}_{J}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{4} d(\ln (\sin \theta))}_{K} \end{aligned} $$

Integration by parts yields $$\begin{aligned} J &=\left[\frac{\theta}{2} \ln (\sin \theta)\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta =\frac{\pi}{4} \ln 2 \end{aligned}$$ $$ \begin{aligned} K &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 \theta}{\sin \theta} \cos \theta d \theta =\frac{1}{4} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 \theta) d \theta =\frac{\pi}{8} \\ \end{aligned} $$

Now we can conclude that $$\boxed{I =\frac{\pi}{8}(1-\ln 4)}$$

Answer 5

Noting $$ d\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]=\frac{x^2}{\sqrt{1-x^2}}, \int_0^{\pi/2}\ln\sin xdx=-\frac{\pi}{2}\ln2$$ one has \begin{eqnarray} \int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx&=&\int_0^1\ln xd\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\\ &=&\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\ln x\bigg|_0^1-\int_0^1\bigg[-\frac12x\sqrt{1-x^2}+\frac12\arcsin(x)\bigg]\frac1xdx\\ &=&\frac12\int_0^1\sqrt{1-x^2}dx-\frac12\int_0^1\frac{\arcsin x}{x}dx\\ &=&\frac{\pi}{8}-\frac12\int_0^{\pi/2}\frac{x}{\sin x}\cos xdx\\ &=&\frac{\pi}{8}-\frac12\int_0^{\pi/2}xd\ln\sin x\\ &=&\frac{\pi}{8}-\frac12x\ln\sin x\bigg|_0^{\pi/2}+\frac12\int_0^{\pi/2}\ln\sin xdx\\ &=&\frac{\pi}{8}-\frac\pi4\ln 2. \end{eqnarray}

Answer 6

\begin{eqnarray} \int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx&=& \frac12\int_0^1\frac{\ln x}{\sqrt{1-x^2}}dx - \frac12\int_0^1\ln x\overset{ibp} d\left(x\sqrt{1-x^2}\right)\\ &=& \frac12\cdot (-\frac\pi2\ln 2)+\frac12\int_0^1 \sqrt{1-x^2}\ dx =-\frac\pi4\ln 2+\frac\pi8 \end{eqnarray}