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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. $\endgroup$ – SinTan1729 Dec 4 '15 at 6:33
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    $\begingroup$ It doesn't seem simple to me. $\endgroup$ – Yves Daoust Dec 4 '15 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 '15 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 '15 at 9:31
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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 '15 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 '15 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 '15 at 15:01
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    $\begingroup$ @BCLC I like your attitude. $\endgroup$ – Skeleton Bow Oct 16 '16 at 14:13
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    $\begingroup$ @SkeletonBow Thanks XD $\endgroup$ – BCLC Oct 16 '16 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 '15 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. $\endgroup$ – Marco Cantarini Dec 4 '15 at 19:24
  • $\begingroup$ Marco Cantarini, must he/she? The tag is 'calculus' not 'analysis' :| $\endgroup$ – BCLC Dec 4 '15 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. $\endgroup$ – Marco Cantarini Dec 4 '15 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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