Definite Integral $\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$ for Logarithm and Algebraic.

Question

Evaluate $$\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx$$

I have been thinking for this question quite some time, I've tried all the methods I have learnt but still getting nowhere. Hope that someone can explain it for me. Thanks in advance.

Answer 1

Note that $\frac{1}{x(1-x)}=\frac{1}{x}+\frac{1}{1-x}$, and the change of variable $u=1-x$ show that your integral is $I=2\int_0^1\frac{\log(1-x+x^2)}{x} dx=2J$. Now $(1-x+x^2)(1+x)=1+x^3$, this show that $$\int_0^1\frac{\log(1+x^3)}{x}dx=\int_0^1\frac{\log(1+x)}{x}dx+J$$ By the change of variable $x^3=u$, in the first integral, we have $$J=-\frac{2}{3}\int_0^1\frac{\log(1+x)}{x}=-\frac{2}{3}(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}\int_0^1 x^{n-1}dx)=-\frac{2}{3}(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2})$$

It suffice now to use the fact that $\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}$ to finish.

Answer 2

Using expansion of logarithm we have \begin{align} \int_0^1\dfrac{\ln(x^2-x+1)}{x-x^2}dx &= \int_0^1\dfrac{\ln(1-(x-x^2))}{x-x^2}dx \\ &= -\int_0^1\dfrac{1}{x-x^2}\sum_{n=1}^\infty\dfrac{1}{n}(x-x^2)^ndx \\ &= -\sum_{n=1}^\infty\dfrac1n\int_0^1(x-x^2)^{n-1}dx \\ &= -\sum_{n=1}^\infty\dfrac{\beta(n,n)}{n} \\ &= -\sum_{n=1}^\infty\dfrac{2}{n^2\binom{2n}{n}} \\ &= -\dfrac{\pi^2}{9} \end{align} where the last equality proved here.

Answer 3

If you want to avoid series altogether one could always try Feynman's trick of differentiating under the integral sign.

Consider $$I(a) = \int^1_0 \frac{\ln \left [a(x^2 - x) + 1 \right ]}{x - x^2} \, dx, \quad a \geqslant 0.$$ Note that $I(0) = 0$ and we require $I(1)$.

Differentiating under the integral sign with respect to the parameter $a$ we have $$I'(a) = - \int^1_0 \frac{dx}{a(x^2 - x) + 1}.$$ Completing the square in the denominator before integrating we have \begin{align*} I'(a) &= -\frac{1}{a} \int^1_0 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \left (\frac{1}{2} \sqrt{\frac{4 - a}{a}} \right )^2}\\ &= -\frac{2}{a} \sqrt{\frac{a}{4 - a}} \left \{\tan^{-1} \left [2 \sqrt{\frac{a}{4 - a}} \left (x - \frac{1}{2} \right ) \right ]\right \}^1_0\\ &= -\frac{4}{a} \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}}. \end{align*}

Thus $$I(1) = -4 \int^1_0 \sqrt{\frac{a}{4 - a}} \tan^{-1} \sqrt{\frac{a}{4 - a}} \cdot \frac{da}{a}.$$ Letting $$u^2 = \frac{a}{4 - a},$$ observing that $$a = \frac{4u^2}{1 + u^2},$$ we have $$da = \frac{8 u}{(1 + u^2)^2} \, du,$$ while for the limits of integration: $(0,1) \mapsto (0,1/\sqrt{3})$. Thus $$I(1) = - 8 \int^{1/\sqrt{3}}_0 \frac{\tan^{-1} u}{1 + u^2} \, du.$$ This last integral can be readily found by setting $t = \tan^{-1} u$. Doing so yields $$I(1) = -8 \int^{\pi/6}_0 t \, dt = -8 \left [\frac{t^2}{2} \right ]^{\pi/6}_0,$$ or $$\int^1_0 \frac{\ln \left (x^2 - x + 1 \right )}{x - x^2} \, dx = -\frac{\pi^2}{9}.$$

Answer 4

You can use, since you're in the range $[0, 1]$, the logarithm expansion

$$\ln(x^2 - x + 1) = \sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} (x^2-x)^k$$

together with the fact that $x - x^2 = -(x^2 - x)$, hence the integration becomes

$$\int_1^0$$

due to the minus sign, incorporated into the extrema.

After that, treat the fraction

$$\frac{(x^2-x)^{k}}{x^2 - x} = (x^2-x)^{k-1}$$

With the binomial theorem, which transforms it into

$$\sum_{j = 0}^{k-1} \binom{k-1}{j} (-1)^{k-1-j}$$

Giving you the integral

$$\sum_{k = 1}^{+\infty} \frac{(-1)^k}{k} \sum_{j = 0}^{k-1} \binom{k-1}{j} (-1)^{k-1+j} \int_1^0 x^{k-1+j} dx$$

Which is trivial.

The final result in terms of series:

$$-\sum_{k = 1}^{+\infty} \frac{(-1)^{2k-j}}{k} \sum_{j = 0}^{k-1} \binom{k-1}{j} \left(\frac{1}{k+j}\right)$$

The series can be summed and the final result is numerical

$$-\frac{\pi^2}{9}$$

Answer 5

Note that $t=x(1-x)$ is increasing in $(0,1/2)$ and decreasing in $(1/2,1)$. So $$ x_\pm(t)=\frac12(1\pm\sqrt{1-4t}) $$ from which $$x_-(t):[0,\frac14]\to[0,1\frac12], x_+(t):[0,\frac14]\to[0,\frac12].$$ Thus \begin{eqnarray} &&\int_{0}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\ &=&\int_{0}^{1/2}\frac{\ln(x^2-x+1)}{x-x^2}dx+\int_{1/2}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\ &=&\int_{0}^{1/2}\frac{\ln(x^2-x+1)}{x-x^2}dx+\int_{1/2}^{1}\frac{\ln(x^2-x+1)}{x-x^2}dx\\ &=&2\int_{0}^{1/4}\frac{\ln(1-t)}{t\sqrt{1-4t}}dt\\ &=&-2\int_{0}^{1/4}\sum_{n=1}^\infty\frac1n \frac{t^{n-1}}{\sqrt{1-4t}}dt\\ &=&-2\sum_{n=1}^\infty\frac{\sqrt{\pi}4^{-n}(n-1)!}{n\Gamma(n+\frac12)}\\ &=&-4\arcsin^2(\frac12)\\ &=&-\frac{\pi^2}{9}. \end{eqnarray} Here $$ \int_0^{1/4}\frac{t^{n-1}}{\sqrt{1-4t}}dt=\frac{\sqrt\pi4^{-n}(n-1)!}{\Gamma(n+\frac12)},\sum_{n=1}^\infty\frac{\sqrt{\pi}(n-1)!}{n\Gamma(n+\frac12)}x^{2n}=2\arcsin^2(x) $$ are used.