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$$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$

Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$

$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2})}{t} dt=\int_0^1 \frac{\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(1-\frac{t}{z_2})}{t} dt$$

where $\ln(z_1)+\ln(z_2)=0$ and $\frac{1}{z_1}=z_2, \frac{1}{z_2}=z_1$

$$I=-\sum_{n=1}^\infty \frac{1}{n} \int_0^1 \frac{1}{t}\left(\frac{t}{z_1}\right)^n+\frac{1}{t}\left(\frac{t}{z_2}\right)^n dt = -\sum_{n=1}^\infty \frac{1}{n}\left( \frac{z_1^n+z_2^n}{n} \right)$$

where $z_1^n+z_2^n=2\cos(\frac{n\pi}{6})$

$$I=-2\sum_{n=1}^\infty \frac{\cos(\frac{n\pi}{6})}{n^2}$$

How to proceed next?

MathFail
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2 Answers2

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Continue with the identity $\operatorname{Li}_2(z)+\operatorname{Li}_2(1/z)=-\frac{\pi^2}6-\frac12\ln^2(-z)$ \begin{align} I= & \int_0^1 \frac{\ln(1-e^{-i\frac{\pi}6}{t})}{t}+\frac{\ln(1-e^{i\frac{\pi}6}t)}tdt\\ = & -\operatorname{Li}_2\left(e^{-i\frac{\pi}6}\right)-\operatorname{Li}_2\left(e^{i\frac{\pi}6}\right)\\ = & \frac{\pi^2}6+ \frac12\ln^2\left(e^{i\frac{5\pi}6}\right) = \frac{\pi^2}6+\frac12\left(i\frac{5\pi}6\right)^2\\= & -\frac{13\pi^2}{72} \end{align}

Frank W
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Quanto
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As @Frank W pointed out in comment...(although I saw the comment after I came up with the proof). Actually, I found the integral $I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$ in the book "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik. Then I came up with the proof.

For $a \in (0, \pi/2]$, let $$I(a) := \int_0^1 \frac{\ln(1 - 2t \cos a + t^2)}{t}\,\mathrm{d} t.$$

We have \begin{align*} I'(a) &= \int_0^1 \frac{2\sin a}{1 - 2t\cos a + t^2}\, \mathrm{d} t\\ &= 2\arctan \frac{t - \cos a}{\sin a}\Big\vert_0^1 \\ &= 2\arctan \frac{1 - \cos a}{\sin a} + 2\arctan\frac{\cos a}{\sin a}\\ &= \pi - a \end{align*}

We have $I(a) = \pi a - \frac{a^2}{2} + C$ on $(0, \pi/2]$ for some constant $C$.

Since $I(a)$ is continuous on $[0, \pi/2]$, taking limit as $a \to 0^{+}$, we get $$C = \int_0^1 \frac{2\ln(1 - t)}{t}\,\mathrm{d} t = \int_0^1 (-2)\sum_{k=1}^\infty \frac{1}{k}t^{k-1}\,\mathrm{d} t = -2\sum_{k=1}^\infty \frac{1}{k^2} = - \pi^2/3.$$

Thus, we have $I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$.

Thus, $I(\pi/6) = -\frac{13}{72}\pi^2$.

We are done.

River Li
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