Question

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

Answer 1

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:

`1/(1-r)=1+r+r^2+r^3+r^4...=sum_(n=0)^oor^n` for `rlt1`

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

`int_0^x1/(1-r)dr =int_0^x 1+r+r^2+r^3+...dr`

Integrating the left hand side:

`int_0^x1/(1-r)dr=[-ln(1-r)]_0^x=-ln(1-x)`

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

`int_0^x 1+r+r^2+r^3+...dr=[r+r^2/2+r^3/3+r^4/4...]_0^x`

`=x+x^2/2+x^3/3+x^4/4+...`

So it follows that:

`-ln(1-x) = x+x^2/2+x^3/3+x^4/4+...`

`impliesln(1-x) =-x-x^2/2-x^3/3-x^4/4+...`

Now divide by `x`:

`ln(1-x)/x =(-x-x^2/2-x^3/3-x^4/4+...)/x`

`=-1-x/2-x^2/3-x^3/4-...`

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

`int_0^1ln(1-x)/x=int_0^1-1-x/2-x^2/3-x^3/4-...dx`

Integrating the right hand term by term side gives us:

`int_0^1ln(1-x)/x=-[x-x^2/4-x^3/9-x^4/16-...]_0^1`

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

`int_0^1ln(1-x)/x~~{-1-1^2/4-1^3/9-1^4/16}-{0}`

`=-(1+1/4+1/6+1/16+...)=-205/144~~-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_0^1ln(1-x)/x~~-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_0^1ln(1-x)/xdx=-sum_(n=0)^oo1/n^2`

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

`int_0^1ln(1-x)/xdx=-sum_(n=0)^oo1/n^2=-zeta(2)`

We actually already know the value of this to be: `zeta(2)=pi^2/6`.

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

`int_0^1ln(1-x)/xdx=-pi^2/6`