I have a problem with the following integral:

$$ \int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\, {{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,} $$

The first idea was to use the integration by parts because

$$ \int{{\rm d}x \over \,\sqrt{x\left(\,1 - x\,\right)\,}\,} =\arcsin\left(\,2x - 1\,\right) + C $$

but what would be the next step is not clear. Another idea would be expand $\ln\left(\,\cdot\right)$ into Taylor series but it seems to be even worse option.

So, what are the other options?


Let $x = \sin(t)^2$ and $s = 2t$, we have

$$\int_0^1 \log\left(\frac{3+x}{3-x}\right)\frac{dx}{\sqrt{x(1-x)}} = \int_0^{\pi/2} \log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right)\frac{2\sin t\cos tdt}{ \sqrt{\sin(t)^2(1-\sin(t)^2)}}\\ = 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin(t)^2}{3-\sin(t)^2}\right) dt = \int_0^{\pi}\log\left(\frac{3 + \frac{1-\cos s}{2}}{3-\frac{1-\cos s}{2}}\right) ds\\ = \int_0^{\pi}\left(\log(7-\cos s) - \log(5+\cos s)\right) ds $$

Notice for any $a > 1$, we have

$$\frac{1}{\pi}\int_0^\pi \log(a \pm \cos s)ds = \cosh^{-1}(a) = \log\left(\frac{a + \sqrt{a^2-1}}{2}\right)\tag{*1}$$ The integral we desired is simply $$\pi\left( \cosh^{-1}(7) - \cosh^{-1}(5)\right) = \pi \log\left(\frac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\approx 1.072804016182156$$

I'm sure the identity in $(*1)$ has a name but I can't remember what it is. Let us prove it!

Notice for any $b > 1$, the function $\log(b+z)$ is analytic over and inside the unit circle $S^1$ in $\mathbb{C}$.
By Residue theorem, we have

$$\frac{1}{2\pi i}\int_{S^1} \log(b + z) \frac{dz}{z} = \log(b)$$

If one parametrize the unit circle by $z = e^{i\theta}$, we get

$$\frac{1}{2\pi}\int_0^{2\pi} \log(b + e^{i\theta}) d\theta = \log(b)$$

Take the real part on both sides, this leads to

$$\begin{align} &\frac{1}{2\pi}\int_0^{2\pi} \log(b^2 + 1 + 2b\cos\theta) d\theta = \log(b^2)\\ \iff & \frac{1}{2\pi}\int_0^{2\pi} \log\left(\frac{b+b^{-1}}{2} + \cos\theta\right)d\theta = \log\left(\frac{b}{2}\right)\end{align}$$ Substitute $\displaystyle\;\frac{b+b^{-1}}{2}\;$ by $a$, we have $\displaystyle\;\frac{b-b^{-1}}{2} = \sqrt{a^2-1}$ and it is clear $(*1)$ follows.

  • $\begingroup$ (+1). Nice. Actually we performed the same calculations by two different approaches: the Residue theorem and the Riemann sums. $\endgroup$ – Jack D'Aurizio Sep 11 '14 at 17:45
  • $\begingroup$ Thanks for the answer! Clear! $\endgroup$ – Martin Gales Sep 15 '14 at 14:06

We have:

$$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx = 2\pi\log\frac{2+\sqrt{3}}{2}.\tag{1}$$

This happens because: $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=2\int_{0}^{1}\frac{\log(3+x^2)}{\sqrt{1-x^2}}\,dx =\int_{-\pi/2}^{\pi/2}\log(3+\cos^2\theta)\,d\theta,$$ $$\int_{0}^{1}\frac{\log(3+x)}{\sqrt{x(1-x)}}\,dx=\frac{1}{2}\int_{-\pi}^{\pi}\log\left(\frac{7+\cos\theta}{2}\right)d\theta=-\pi\log 2+\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta.$$ Now comes an interesting technique - we have: $$\begin{eqnarray*}\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta &=& \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n}\log\left(7+\cos\frac{k\pi}{n}\right)\\&=&\lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n}\left(7+\cos\frac{k\pi}{n}\right)\end{eqnarray*}$$ but since: $$z^{2n}-1 = \prod_{k=1}^{2n}\left(z-e^{\frac{\pi i k }{n}}\right)=(z^2-1)\prod_{k=1}^{n-1}\left(z^2+1-2z\cos\frac{k\pi}{n}\right)$$ and the solutions of $$\frac{z^2+1}{2z}=-7$$ are $z=\pm 4\sqrt{3}-7$, it follows that: $$\int_{0}^{\pi}\log(7+\cos\theta)\,d\theta=\pi\log\left(\frac{7}{2}+2\sqrt{3}\right).$$ With the same technique we can prove:

$$\int_{0}^{1}\frac{\log(3-x)}{\sqrt{x(1-x)}}\,dx = \pi\log\left(\frac{5}{4}+\sqrt{\frac{3}{2}}\right),\tag{2}$$

hence we have:

$$\int_{0}^{1}\frac{\log\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\,dx = \pi\log\left((7+4\sqrt{3})(5-2\sqrt{6})\right).\tag{3}$$

  • $\begingroup$ This just proves that, sometimes, the Residue theorem can be replaced by a Riemann sums argument. $\endgroup$ – Jack D'Aurizio Sep 11 '14 at 17:42
  • $\begingroup$ (+1) Actually, I first derive the result by a third method (differentiate under integral sign) but the derivation along that direction is pretty clumsy.... $\endgroup$ – achille hui Sep 11 '14 at 18:16
  • $\begingroup$ @achillehui I've just wanted to use that and I'm typing the answer. $\endgroup$ – Tunk-Fey Sep 11 '14 at 18:19
  • $\begingroup$ Mr. @achillehui, Tunk-Fey & Jack D'Aurizio: Could you please help me? I'm stuck ( つ﹏╰) See my answer below $\endgroup$ – Anastasiya-Romanova 秀 Sep 12 '14 at 10:13
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    $\begingroup$ Thank you! The product formula is especially nice! $\endgroup$ – Martin Gales Sep 15 '14 at 14:09

Split the integral into two forms by expanding the logarithm function $$ \int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx $$ Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have $$ 2\int_{0}^1\frac{\ln(3+t^2)}{\sqrt{1-t^2}}\ dt-2\int_{0}^1\frac{\ln(3-t^2)}{\sqrt{1-t^2}}\ dt $$ Let $t=\sin\theta\ \rightarrow\ dt=\cos\theta\ d\theta$, we have $$ 2\int_{0}^{\pi/2}\ln(3+\sin^2\theta)\ d\theta-2\int_{0}^{\pi/2}\ln(3-\sin^2\theta)\ d\theta $$ Using identity $\sin^2\theta=\dfrac12(1-\cos2\theta)$ and setting $y=2\theta$, we have $$ \int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^{\pi}\ln\left(7-\cos y\right)\ dy-\int_{0}^{\pi}\ln\left(5+\cos y\right)\ dy $$ We will use Feynman's way to evaluate integral above. Consider $$ I(k)=\int_{0}^{\large\pi}\ln\left(k\pm\cos y\right)\ dy $$ then \begin{align} I'(k)&=\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\ dy \end{align} Using formula $$ \int_0^\pi\frac{1}{a^2+b^2-2ab\cos x}dx=\frac{\pi}{a^2-b^2} $$ Setting $b=\pm\dfrac{1}{2a}$, we have $$ \int_0^\pi\frac{1}{a^2+\frac{1}{4a^2}\pm\cos x}dx=\frac{4\pi a^2}{4a^4-1} $$ Clearly $k=a^2+\dfrac{1}{4a^2}\ \rightarrow\ a^2=\dfrac{k+\sqrt{k^2-1}}{2}$ and $dk=\dfrac{4a^4-1}{2a^3}da$, then \begin{align} I(k)&=\int\int_{0}^{\large\pi}\frac{1}{k\pm\cos y}\,dy\,dk\\ &=\int\frac{4\pi a^2}{4a^4-1}\cdot\dfrac{4a^4-1}{2a^3}da\\ &=2\pi\int\frac{1}{a}\,da\\ &=2\pi\ln(a)+C\\ &=\pi\ln(a^2)+C\\ &=\pi\ln\left(\dfrac{k+\sqrt{k^2-1}}{2}\right)+C\\ \end{align} Finally \begin{align} \int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx&=I(7)-I(5)\\ &=\pi\ln\left(\dfrac{7+\sqrt{7^2-1}}{5+\sqrt{5^2-1}}\right)\\ &=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{align} Yeayyy, I'm done! (>‿◠)✌

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    $\begingroup$ Stuck? $\displaystyle\;\frac{4\pi a^2}{4a^4-1} = 2\pi \frac{k+\sqrt{k^2-1}}{2(k^2-1)+2k\sqrt{k^2-1}} = \frac{\pi}{\sqrt{k^2-1}}$ $\endgroup$ – achille hui Sep 12 '14 at 10:35
  • $\begingroup$ Mr. @achillehui Sorry for bothering you, I've just gotten another idea. Thanks anyway for your help (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 12 '14 at 10:36
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    $\begingroup$ Many thanks! Differentiation of the integrand is a cool thing! $\endgroup$ – Martin Gales Sep 15 '14 at 14:13
  • $\begingroup$ @MartinGales You're welcome (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Sep 15 '14 at 14:32

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