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

1 Answer
Jun 22, 2018

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)# 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 #