A hint to calculate the integral $\int^1_0 \frac{\log x}{1-x}dx$

Question

I'm struggling in trying to solve integral. Could you help me? Please, don't solve the problem, give some hints of what I should do.
$$\int^1_0 \frac{\log x}{1-x}dx$$

Answer 1

This is a non-elementary function integral. Specifically via a simple substitution of $u=1-x \to du = -dx$, we yield the known special function integral :

$$\int \frac{\ln x}{1-x}\mathrm{d}x=-\int\frac{\ln(1-u)}{u}\mathrm{d}u = \dots = \text{Li}_2(1-x) + C$$

thus we need to find another way to work around it.

For an analytical approach, continuing Jack's idea (credits) :

$$\int x^n\ln(x)\mathrm{d}x = \frac{x^{n+1}\ln(x)}{n+1}-\frac{1}{n+1}\int x^n\mathrm{d}x= \frac{x^{n+1}\big[(n+1)\ln(x)-1\big]}{(n+1)^2} + C$$

for which we used integration by parts as :

$$\int f\mathrm{d}g = fg - \int g\mathrm{d}f$$

$$\mathrm{d}f = \frac{1}{x}\mathrm{d}x$$ $$g=\frac{x^{n+1}}{n+1}$$

Now, exploiting the well known geometric series :

$$\frac{1}{1-x} = \sum_{k = 0}^{\infty} x^k$$

will lead to the expression for our calculation, of :

$$\sum_{n = 0}^{\infty} \int_0^1 x^n\ln(x)\ dx \to \int_0^1 x^n\ln(x)= -\frac{1}{(n+1)^2} \space \text{for} \space \Re\{n\} > -1$$

which finally leads to the calculation via the usage of the Zeta Function :

$$- \sum_{n = 0}^{\infty} \frac{1}{(n+1)^2} = -\zeta(2) = -\frac{\pi^2}{6}$$

Truly, our calculations are correct, as :

$$\int_0^1 \frac{\ln x}{1-x}\mathrm{d}x = \text{Li}_2(0) - \text{Li}_2(1) = - \frac{\pi^2}{6}$$

Alternatively, if you're not familiar with such a procedure, the best possible outcome will come via numerical analysis, using Simpson's rule for example :

$$\int_0^1 \frac{\ln(x)}{1-x}\mathrm{d}x \approx \frac{1-0}{6}\bigg(f(0) + + 4f\bigg(\frac{1}{2}\bigg) + f(1)\bigg), \space \text{where} \space f(x) = \frac{\ln(x)}{1-x}$$

Answer 2

Since the range of the integral is from $0$ to $1$, you may like to use Geometric Series

$$\frac{1}{1-x} = \sum_{k = 0}^{+\infty} x^k$$

Hence

$$ \sum_{k = 0}^{+\infty} \int_0^1 x^k\ln(x)\ dx$$

The integral is pretty standard, you can do it by parts many times or in other ways. Eventually

$$\int x^k\ln(x)\ dx = \frac{x^{k+1} ((k+1) \log (x)-1)}{(k+1)^2}$$

hence

$$\int_0^1 x^k\ln(x) = -\frac{1}{(k+1)^2}$$

This holds for $\Re(k) > -1$ which is true.

Then all together

$$- \sum_{k = 0}^{+\infty} \frac{1}{(k+1)^2} = -\zeta(2) = -\frac{\pi^2}{6}$$

Answer 3

Things to try:

• Make the change of variable $x \mapsto 1-x$. This simplifies the denominator, leaves the interval of integration unchanged (not important, but sometimes useful, especially if a linear combination of the original and new integrands simplifies), and makes the numerator something whose Taylor series you should know.
• Attempt integration by parts to allow you to differentiate the logarithm, turning it into a rational function. (I'm pretty sure this doesn't help for this problem, but is in the constellation of things to do to integrals that look like this.)
• Try solving the differential equation $f'(x) = \frac{\log x}{1-x}$, which can be turned into a relatively simple recurrence for the Taylor coefficients.
• Look it up in tables. For instance, in Gradshteyn & Ryzhik, 7th ed., this is 4.231.2.