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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?

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  • $\begingroup$ Try integrating by parts. $\endgroup$ Dec 4, 2015 at 6:33
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    $\begingroup$ It doesn't seem simple to me. $\endgroup$
    – user65203
    Dec 4, 2015 at 8:30
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    $\begingroup$ Note that that's an improper integral on both sides, as you have divergence issues at both 0 and 1 $\endgroup$
    – Alan
    Dec 4, 2015 at 8:38
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    $\begingroup$ @YvesDaoust. No it is not simple. It somehow belongs to a family of integrals that contains logarithms and trigonometric functions that can be computed by linear combination of itself. See my answer below and the answer under the link in it. $\endgroup$
    – Tom-Tom
    Dec 4, 2015 at 9:31
  • $\begingroup$ Reduction formula (A22) $$\int_0^1\frac{x^{2n}\ln x}{\sqrt{1-x^2}}dx =I_n=\frac{2n-1}{2n}I_{n-1}+\frac1{2n}\int_0^1 x^{2(n-1)}\sqrt{1-x^2}\>dx $$ $\endgroup$
    – Quanto
    Apr 5, 2022 at 13:15

5 Answers 5

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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 $$I=-\frac\pi4\ln2-\frac\pi8+\frac\pi4=\boxed{\frac\pi8-\frac\pi4\ln2}.$$

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    $\begingroup$ I agree, I tried it later and used the same method. I'm still I'm high school so I have no idea what the hypergeometry guy above did :x $\endgroup$
    – Sat D
    Dec 4, 2015 at 11:41
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    $\begingroup$ @BCLC. No, the computation does not use any complex analysis. It only involves manipulations of the trigonometric functions. And actually, the only function that is integrated is a constant ! $\endgroup$
    – Tom-Tom
    Dec 4, 2015 at 14:51
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    $\begingroup$ @SatD So what if you're still in highschool? I have a bachelor's in applied math and am about to finish my master's in applied math, and I don't know what those are $\endgroup$
    – BCLC
    Dec 4, 2015 at 15:01
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    $\begingroup$ @BCLC I like your attitude. $\endgroup$ Oct 16, 2016 at 14:13
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    $\begingroup$ @SkeletonBow Thanks XD $\endgroup$
    – BCLC
    Oct 16, 2016 at 14:22
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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.

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    $\begingroup$ Do you really expect someone taking basic calculus to understand this? This seems kind of, what's the term, highfalutin? $\endgroup$
    – BCLC
    Dec 4, 2015 at 14:58
  • $\begingroup$ @BCLC I only answered the question, and the question is to solve the integral. If the OP wants some "particular ways" to solve it, then he/she must write it in the question. And he can also ask me the passages that he /she doesn't understand. $\endgroup$ Dec 4, 2015 at 19:24
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    $\begingroup$ Marco Cantarini, must he/she? The tag is 'calculus' not 'analysis' :| $\endgroup$
    – BCLC
    Dec 4, 2015 at 19:39
  • $\begingroup$ @BCLC There are a lot of question like this with the tag "calculus" and without the tag "analysis" with answers "similar" to mine. Normally if the OP has some specific request, it must write it in the question. Anyway, it's my last comment on this. $\endgroup$ Dec 4, 2015 at 20:48
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$$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$ 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 $$I(2)=\color{blue}{\dfrac{\pi}{8}(1-\ln4)}$$

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

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 $$\boxed{I =\frac{\pi}{8}(1-\ln 4)}$$

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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}

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