#int_0^1 lnx/(1+x)dx# ?
1 Answer
# 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) = - 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(0) = 0 #
# Li_2(-1) = -pi^2/12 #
Leading to the result:
# I = -pi^2/12 #