# How to calculate this? int_0^1 log(1-x)/xdx

Feb 11, 2018

See below.

#### Explanation:

Unfortunately the function inside the integral will not integrate to something that cannot be expressed in terms of elementary functions. You will have to use numerical methods to do this.

I can show you how to use a series expansion to get an approximate value.

Begin with the geometric series:

$\frac{1}{1 - r} = 1 + r + {r}^{2} + {r}^{3} + {r}^{4.} . . = {\sum}_{n = 0}^{\infty} {r}^{n}$ for $r < 1$

Now integrate with respect to $r$ and using the limits $0$ and $x$ to get this:

${\int}_{0}^{x} \frac{1}{1 - r} \mathrm{dr} = {\int}_{0}^{x} 1 + r + {r}^{2} + {r}^{3} + \ldots \mathrm{dr}$

Integrating the left hand side:

${\int}_{0}^{x} \frac{1}{1 - r} \mathrm{dr} = {\left[- \ln \left(1 - r\right)\right]}_{0}^{x} = - \ln \left(1 - x\right)$

Now integrate the right hand side by integrating term by term:

${\int}_{0}^{x} 1 + r + {r}^{2} + {r}^{3} + \ldots \mathrm{dr} = {\left[r + {r}^{2} / 2 + {r}^{3} / 3 + {r}^{4} / 4. . .\right]}_{0}^{x}$

$= x + {x}^{2} / 2 + {x}^{3} / 3 + {x}^{4} / 4 + \ldots$

So it follows that:

$- \ln \left(1 - x\right) = x + {x}^{2} / 2 + {x}^{3} / 3 + {x}^{4} / 4 + \ldots$

$\implies \ln \left(1 - x\right) = - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 + \ldots$

Now divide by $x$:

$\ln \frac{1 - x}{x} = \frac{- x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 + \ldots}{x}$

$= - 1 - \frac{x}{2} - {x}^{2} / 3 - {x}^{3} / 4 - \ldots$

So we now have power series expression for the function we originally started with. Finally, we can integrate again to get:

${\int}_{0}^{1} \ln \frac{1 - x}{x} = {\int}_{0}^{1} - 1 - \frac{x}{2} - {x}^{2} / 3 - {x}^{3} / 4 - \ldots \mathrm{dx}$

Integrating the right hand term by term side gives us:

${\int}_{0}^{1} \ln \frac{1 - x}{x} = - {\left[x - {x}^{2} / 4 - {x}^{3} / 9 - {x}^{4} / 16 - \ldots\right]}_{0}^{1}$

Evaluating the limits to four terms will give us an approximate value:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \approx \left\{- 1 - {1}^{2} / 4 - {1}^{3} / 9 - {1}^{4} / 16\right\} - \left\{0\right\}$

$= - \left(1 + \frac{1}{4} + \frac{1}{6} + \frac{1}{16} + \ldots\right) = - \frac{205}{144} \approx - 1.42361$

Now, this is only to four terms. If you would like a more accurate number simply use more terms in the series. For example, going to the 100th term:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \approx - 1.63498$

As an aside, if you work through the exact same process but use summation notation (i.e. with big sigma rather than writing out the terms of the series) you will find that:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2$

which is just the Riemann-Zeta function of 2, i.e:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\sum}_{n = 0}^{\infty} \frac{1}{n} ^ 2 = - \zeta \left(2\right)$

We actually already know the value of this to be: $\zeta \left(2\right) = {\pi}^{2} / 6$.

Hence the exact value of the integral can be deduced to be:

${\int}_{0}^{1} \ln \frac{1 - x}{x} \mathrm{dx} = - {\pi}^{2} / 6$