Find $ \int_{0}^{2} \frac{\ln x}{x^2 - x + 1} \,dx $

Question

So I was looking at some previous year calculus exams of a math tournament.

$$\int_{0}^{2} \frac{\ln x}{x^2 - x + 1} \,dx $$

I could not solve this question after spending an hour on it.

So my original thought was to factorize the denominator in terms of imaginary numbers and convert the question into a form of partial fraction.

So I got $$x=\frac{1\pm \sqrt{-3}}{2}$$

Now $$ \frac{A}{x-(\frac{1 + i \sqrt{3}}{2})} + \frac{B}{x-(\frac{1 - i \sqrt{3}}{2})} = \ln x $$

However, this leads to failure.

I also tried trig sub as $x=\tan u$. But that too failed.

Now using methods like the Feynman integration technique also is not as reasonable as partially differentiating $2$ (assuming it to be t and then) with respect to $x$.

Also, methods like by parts also do not work because of the denominator.

Hence I am unable to proceed.

Would like if someone could share a hint or give a solution.

Answer 1

Let $$ I(t) = \int_0^2 \frac{\ln (t x )}{x^2 - x + 1} \, dx $$

Taking the derivative with respect to $t$ (by the dominated convergence theorem we may justify that):

$$ I'(t) = \int_0^2 \frac{1}{t} \cdot \frac{1}{x^2 - x + 1} \, dx $$

$$ = \frac{1}{t} \int_0^2 \frac{dx}{x^2 - x + 1} $$

It is known that:

$$ \int_0^2 \frac{dx}{x^2 - x + 1} = \frac{\pi}{\sqrt{3}} $$

Thus,

$$ I'(t) = \frac{1}{t} \cdot \frac{\pi}{\sqrt{3}} $$

Now, integrating both sides from $t = 1$ to $t = a $:

$$ \int_1^a I'(t) \, dt = I(a) - I(1) = \frac{\pi}{\sqrt{3}} \ln (a) $$

which gives

$$ I(1) = I(a) - \frac{\pi}{\sqrt{3}} \ln (a). $$

The thing is, mostly, it is easy to find an $a$ that makes $I(a)$ equal to zero - here, it is not so easy. Trying around with GeoGebra gives me a vague approximation for $a=1.71$, but this is not sufficient. Since you asked for hints, I hope you can continue from here. The value of the integral with a plugged into the formula above gives roughly $$I(1)=-0.97$$.

Answer 2

Don't be afraid by complex numbers. Write $$\frac 1{x^2-x+1}=\frac 1{(x-a)(x-b)}=\frac 1{a-b}\left(\frac{1}{x-a}-\frac{1}{x-b}\right)$$ and use twice $$I(c)=\int_0^t\frac {\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{t}{c}\right)+\log (t) \log \left(1-\frac{t}{c}\right)$$ Now, use $t=2$, $a=\frac{1+i \sqrt{3}}{2}$, $b=\frac{1-i \sqrt{3}}{2}$ to obtain the result $$-\frac{\pi \log (2)}{3 \sqrt{3}}+\frac{i}{\sqrt{3}}\left(\text{Li}_2\left(\frac{1+i \sqrt{3}}{4} \right)-\text{Li}_2\left(\frac{1-i \sqrt{3} }{4}\right)\right)$$ which is a real number.

Answer 3

$$ I=\int_0^2 \frac{\ln x}{x^2 - x + 1} \,dx $$

\int_0^\infty \frac{\ln x}{ax^2+bx+c} ,dx

There exists a really elaborate proof for the below generalization;

$$\color{red}{2b\int_0^z\frac{\ln(x+c)}{(x+a)^2+b^2}\,dx=\left[\arctan\left(\frac{z+a}{b}\right)-\arctan\left(\frac{a}{b}\right)\right]\ln\left((a-c)^2+b^2\right)+\operatorname{Cl}_2(2\alpha+2\gamma)-\operatorname{Cl}_2(2\beta+2\gamma)+\operatorname{Cl}_2(\pi-2\alpha)-\operatorname{Cl}_2(\pi-2\beta)}$$

Where,

$$\color{red}{\alpha = \arctan\left(\frac{a}{b}\right) ,\quad \beta = \arctan\left(\frac{z+a}{b}\right) ,\quad \gamma = \arctan\left(\frac{c-a}{b}\right)} $$

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Re-write the integral,

$$I = \int_0^2 \frac{\ln x}{\left(x - \tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2} \, dx$$

Set $z=2,c=0, b=\frac{\sqrt3}{2}, a=-\frac12$

$$\color{blue}{\sqrt3\int_0^2 \frac{\ln x}{\left(x - \tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2} \, dx=\operatorname{Cl}_2(2\alpha+2\gamma)-\operatorname{Cl}_2(2\beta+2\gamma)+\operatorname{Cl}_2(\pi-2\alpha)-\operatorname{Cl}_2(\pi-2\beta)}$$

where,

$$\color{blue}{\alpha = -\frac{\pi}{6} ,\quad \beta = \frac{\pi}{3} ,\quad \gamma = \frac{\pi}{6}} $$

Therefore,

• $\operatorname{Cl}_2(2\alpha+2\gamma)=\operatorname{Cl}_2(0)=0$

• $\operatorname{Cl}_2(2\beta+2\gamma)=\operatorname{Cl}_2(\pi)=0$

• $\operatorname{Cl}_2(\pi-2\alpha)=\operatorname{Cl}_2\left(\frac{4\pi}{3}\right)=\Im \left[ \mathrm{Li}_2(e^{i\frac{4\pi}{3}}) \right]=-\frac{\psi^{(1)}\left(\tfrac{1}{3}\right)}{6\sqrt{3}} + \frac{\psi^{(1)}\left(\tfrac{2}{3}\right)}{6\sqrt{3}} $

• $\operatorname{Cl}_2(\pi-2\beta)=\operatorname{Cl}_2\left(\frac{\pi}{3}\right)=\Im \left[ \mathrm{Li}_2(e^{i\frac{\pi}{3}}) \right]=\frac{\psi^{(1)}\left(\tfrac{1}{6}\right)}{24\sqrt{3}} + \frac{\psi^{(1)}\left(\tfrac{1}{3}\right)}{24\sqrt{3}} - \frac{\psi^{(1)}\left(\tfrac{2}{3}\right)}{24\sqrt{3}} - \frac{\psi^{(1)}\left(\tfrac{5}{6}\right)}{24\sqrt{3}} $

All of the above derived from $\boxed{\mathrm{Cl}_2(\theta) = \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^2} = \Im \left[ \mathrm{Li}_2(e^{i\theta}) \right]}$

To wrap it up, we get,

$$\therefore I=\frac{1}{\sqrt3}\left(\Im \left[ \mathrm{Li}_2(e^{i\frac{4\pi}{3}}) \right]-\Im \left[ \mathrm{Li}_2(e^{i\frac{\pi}{3}}) \right]\right)=\frac{1}{72} \left( 5\psi^{(1)}\left(\tfrac{2}{3}\right) - 5\psi^{(1)}\left(\tfrac{1}{3}\right) - \psi^{(1)}\left(\tfrac{1}{6}\right) + \psi^{(1)}\left(\tfrac{5}{6}\right) \right) $$

Here are the numerical checks - 1 and 2

Must say, a pretty tacky problem for math tournament.

Edit:

Had the question been with the limits $0->1$, you can see this post, which gives a good insight.

Had the question been with the limits $0->\infty$, this post

Answer 4

\begin{align} &\int_{0}^{2} \frac{\ln x}{x^2 - x + 1} \,dx =\frac1{\sin\frac\pi3}\ \Im \int_{0}^{2} \frac{\ln x}{x -e^{i\frac\pi3}} \overset{x\to 2x} {dx }\\ =&\ \frac1{\sin\frac\pi3}\ \Im \int_{0}^{1} \frac{\ln 2+\ln x}{x -\frac12e^{i\frac\pi3}} dx =\frac1{\sin\frac\pi3}\left[\frac\pi2\ln2 +\Im \text{Li}_2(2e^{-i\frac\pi3} )\right] \end{align}