Question

`int_0^1 lnx/(1+x)dx` ?

Answer 1

`int_0^1 \ lnx/(1+x) \ dx = -pi^2/12`

Explanation:

We seek:

`I = int_0^1 \ lnx/(1+x) \ dx`

We can apply Integration By Parts:

Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/(1+x), => v,=ln(1+x) ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int_0^1 \ (lnx)(1/(1+x)) \ dx = [(lnx)(ln(1+x))]_0^1 - int_0^1 \ (ln(1+x))(1/x) \ dx`

`:. I = - int_0^1 \ (ln(1+x))/x \ dx`

The resulting integral, is cannot be expressed in terms of elementary functions, in fact by defining the polylogarithm function, `Li_2(x)` we write:

`Li_2(-x) = - int \ (ln(1+x))/x \ dx`

So that

`:. I = -[Li_2(-x)]_0^1 = Li_2(-1) - Li_2(0)`

And (fortunately) special values of `Li_2(x)` are:

`Li_2(0) = 0`
`Li_2(-1) = -pi^2/12`

Leading to the result:

`I = -pi^2/12`