# Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$

Hi I am trying to show this astonishing result$$\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$$ Note we can to use $\ln(a/b)=\ln a-\ln b$ but that didn't help me much . After this I obtained integrals of the form $$\int_0^1 \log \big[\big(x\pm\sqrt{1-x^2}\big)^2\big] \frac{x\, dx}{1-x^2}$$ which I am not sure how to handle. Thanks.

• An approach I tried: let $x = \sin(t)$. Then the integral can be simplified to: $$\int_0^{\pi/2}\log^2(\tan(t+\pi/4))\tan(t)\;dt$$I don't know if that's helpful, but it's something different... May 6 '14 at 3:12
• I am confused at your question. Is the expression $\log\big(\frac{x+\sqrt{1-x^2}}{x-\sqrt{1-x^2}} \big)^2$ or $[\log\big(\frac{x+\sqrt{1-x^2}}{x-\sqrt{1-x^2}} \big)]^2$? May 6 '14 at 3:32
• @Rudstar Sorry I made that explicitly clear with my notation now., just edited. I realize that could be confusing thank you for pointing that out to me. May 6 '14 at 3:38
• OH! That makes a huge difference! :) Thanks. May 6 '14 at 3:39
• @anorton Sorry about that. And thank you! May 6 '14 at 3:39

Let



\begin{align} I(a)&=\int_0^1 2\, \log\frac{\big(x+a\, \sqrt{1-x^2}\big)}{\big(x-a\, \sqrt{1-x^2}\big)} \frac{x\, dx}{1-x^2} \tag 1\\ \therefore \frac{\partial}{\partial a}I(a) &= 2\, \int_0^1 \left(\frac{\sqrt{1-x^2}}{\big(x+a\, \sqrt{1-x^2}\big)}+\frac{\sqrt{1-x^2}}{\big(x-a\, \sqrt{1-x^2}\big)}\right) \frac{x}{1-x^2}\, dx\\ &= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{\tan{(t)}^2-a^2}\, dt \hspace{100pt}\text{(subst. $x=\sin{t}$)}\\ &= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{(\tan{(t)}^2-a^2)(1+\tan{(t)}^2)} \sec{(t)}^2 \, dt\\ &= \int_0^{\infty} \frac{4\, y^2}{(y^2-a^2)(1+y^2)} \, dy \hspace{82pt}\text{(subst. $\tan{t}=y$)}\\ &= \int_0^{\infty} \frac{1}{1+a^2}\left(\frac{2a}{y-a}-\frac{2a}{y+a}+\frac{4}{1+y^2}\right)\, dy\\ &= \frac{1}{1+a^2}\left(a\, \log\left(\frac{y-a}{y+a}\right)^2 +4\arctan{y}\right)\Bigg|_0^\infty\\ &= \frac{2\, \pi}{a^2+1}\\ \implies I(a)&=2\,\pi\arctan{a}+C \tag 2 \end{align}

Putting $a=0$ in $(1)$ and $(2)$, we find that $C=0$

Hence, $$I(a)=2\,\pi\arctan{a}$$

and the required integral $$I(1)=\frac{\pi^2}{2}$$

• Do you mean $\partial_a$? May 6 '14 at 4:20
• How do you do the integral in the line between (1) and (2), proof? Thank you for the clever solution, but this is my only question. May 6 '14 at 4:20
• Yes, thanks, corrected it..
– gar
May 6 '14 at 4:20
• I think a simple trigonometric substitution will do, like x=sin(t). Let me see.
– gar
May 6 '14 at 4:22
• You're welcome. I enjoy your integrals, keep them coming!
– gar
May 6 '14 at 12:19

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$\ds{\int_{0}^{1} \ln\pars{\bracks{x + \root{1 - x^{2}}}^2 \over \bracks{x-\root{1 - x^2}}^{2}}\, {x\,\dd x \over 1 - x^{2}} = {\pi^{2} \over 2}:\ {\large ?}}$

With $\ds{x \equiv \sin\pars{\theta}}$: \begin{align}&\int_{0}^{1} \ln\pars{\bracks{x + \root{1 - x^{2}}}^2 \over \bracks{x-\root{1 - x^2}}^{2}}\, {x\,\dd x \over 1 - x^{2}} = -\int_{0}^{\pi/2} \ln\pars{\bracks{\tan\pars{\theta} - 1 \over \tan\pars{\theta} + 1 }^{2}}\, \tan\pars{\theta}\,\dd\theta \\[5mm]= &\ -\int_{0}^{\pi/2}\ln\pars{\tan^{2}\pars{\theta - {\pi \over 4}}}\, \tan\pars{\theta}\,\dd\theta =-\int_{-\pi/4}^{\pi/4}\ln\pars{\tan^{2}\pars{\theta}}\, \tan\pars{\theta + {\pi \over 4}}\,\dd\theta \\[5mm]= &\ -\int_{0}^{\pi/4}\ln\pars{\tan^{2}\pars{\theta}}\, \bracks{{1 + \tan\pars{\theta} \over 1 - \tan\pars{\theta}} + {1 - \tan\pars{\theta} \over 1 + \tan\pars{\theta}}}\,\dd\theta \\[3mm]&=-2\int_{0}^{\pi/4}\ln\pars{\tan\pars{\theta}}\, {2\sec^{2}\pars{\theta} \over 1 - \tan^{2}\pars{\theta}}\,\dd\theta =-4\ \overbrace{\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t} ^{\ds{-\,{\pi^{2} \over 8}}} = \bbox[15px,#ffe,border:1px dotted navy]{\ds{\pi^{2} \over 2}} \end{align}

The last integral can be evaluated by expanding $\ds{\pars{1 - t^{2}}^{-1}}$ in powers of $\ds{t}$. Namely,

\begin{align}&\color{#c00000}{\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t} =\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}t^{2n}\,\dd t =\left.\sum_{n = 0}^{\infty}\partiald{}{\mu}\int_{0}^{1}t^{2n + \mu}\,\dd t\, \right\vert_{\,\mu = 0} =-\sum_{n = 0}^{\infty}{1 \over \pars{2n + 1}^{2}} \\[3mm]&=-\sum_{n = 1}^{\infty}{1 \over n^{2}} + \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}} =-\,{3 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}} = -\,{3 \over 4}\,{\pi^{2} \over 6}=\color{#c00000}{-\,{\pi^{2} \over 8}} \end{align}

I hope it is not too late. Let $$I(a)=\int_0^1 \log\frac{\big(x+a\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}, -1\le a\le 1.$$ Clearly $I(-1)=0$ and \begin{eqnarray} I'(a)&=&2\int_0^1 \frac{x}{(x+a\sqrt{1-x^2})\sqrt{1-x^2}}dx\\ &=&2\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt. \end{eqnarray} Define $$A=\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt, B=\int_0^{\frac{\pi}{2}}\frac{\cos t}{\sin t+a\cos t}dt$$ and then $$A+aB=\frac{\pi}{2}, B-aA=-\log |a|.$$ Thus $A=\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}$. However $\frac{a\log |a|}{1+a^2}$ is an odd function of $a$ and hence $$I(1)=2\int_{-1}^1\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}da=2\int_{-1}^1\frac{\frac{\pi}{2}}{1+a^2}da=\frac{\pi^2}{2}.$$

• I would like to point out that the integrand of your $I'(a)$ has a pole at $x = -a/\sqrt{1+a^2}$ if $a$ is negative. So this solution also shares the same issue with @gar's answer. Circumventin this problem, though, seems not hard. Oh, and +1 upvote, of course. Dec 10 '14 at 22:46
• @sos440, Thank for pointing out this. I will fix it. Dec 11 '14 at 0:24

So here is a solution which follows the logic of both @gar's and @xpaul's but avoids the issue of poles.

Let $I(a)$ be defined by

$$I(a) = \int_{0}^{1} \frac{2x}{1-x^{2}} \log \left| \frac{x + \sqrt{1-x^{2}}}{x + a\sqrt{1-x^{2}}} \right| \, dx.$$

The value we want to calculate is $I(-1)$. With the substitution $t = x/\sqrt{1-x^{2}}$ (this substitution is essentially the same as in @Felix's answer), it follows that

$$I(a) = \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left| \frac{1+t}{a+t} \right| \, dt = \Re \int_{0}^{\infty} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt,$$

where the branch of logarithm is chosen to avoid the upper half-plane. Then by DCT,

$$I(a) = \lim_{\epsilon \downarrow 0} \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{1+t^{2}} \log \left( \frac{1+t}{a+t} \right) \, dt =: \lim_{\epsilon \downarrow 0} I_{\epsilon}(a).$$

Now, for $0 < \epsilon \ll 1$, we have

\begin{align*} I_{\epsilon}'(a) &= - \Re \int_{i\epsilon}^{\infty+i\epsilon} \frac{2t}{(1+t^{2})(t+a)} \, dt \\ &= - \frac{2a}{1+a^{2}} \Re \int_{i\epsilon}^{\infty+i\epsilon} \left( \frac{1}{a} \frac{1}{1+t^{2}} + \frac{t}{1+t^{2}} - \frac{1}{t + a} \right) \, dt \\ &= - \frac{2a}{1+a^{2}} \Re \left( \frac{\pi}{2a} - \frac{1}{a}\arctan(i\epsilon) + \log(a+i\epsilon) - \log\sqrt{1-\epsilon^{2}} \right) \\ &= - \frac{2a}{1+a^{2}} \left( \frac{\pi}{2a} + \log\sqrt{a^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right). \end{align*}

This gives, together with the initial condition $I_{\epsilon}(1) = 0$, that

$$I_{\epsilon}(a) = \int_{a}^{1} \frac{2s}{1+s^{2}} \left( \frac{\pi}{2s} + \log\sqrt{s^{2}+\epsilon^{2}} - \log\sqrt{1-\epsilon^{2}} \right) \, ds.$$

Taking $\epsilon \downarrow$, again by DCT, we get

$$I(a) = \int_{a}^{1} \frac{\pi + 2s\log|s|}{1+s^{2}} \, ds.$$

Plugging $a = -1$ we finally obtain the desired answer.