Question

`int_(1/a)^a(tan^-1x)/xdx=?`

Answer 1

`int (tan^{-1}(x)dx)/x = i/2(`PolyLog`(2,-i x)-`PolyLog`(2,ix))`

Explanation:

We know that the `arctan(x)` series representation is given by

`tan^{-1}(x) = sum_{k=0}^oo(-1)^k (x^{2k+1})/(2k+1)` for `abs x < 1`

http://math.stackexchange.com/questions/29649/why-is-arctanx-x-x3-3x5-5-x7-7-dots

then

`tan^{-1}(x)/x = sum_{k=0}^oo(-1)^k (x^{2k})/(2k+1)` and

`int (tan^{-1}(x)dx)/x =sum_{k=0}^oo(-1)^k (x^{2k+1})/(2k+1)^2`

We know also that

PolyLog`(2,x) = sum_{k=1}^oo x^k/k^2`
https://en.wikipedia.org/wiki/Polylogarithm

so finally

`int (tan^{-1}(x)dx)/x = i/2(`PolyLog`(2,-i x)-`PolyLog`(2,ix))`

Answer 2

Let
`z=lnx=>x=e^z`
`x=a->z=lna and x=1/a->z=-lna`

`and dz==(dx)/x`

`I=int_(1/a)^a(tan^-1x)/xdx`

`=int_(-lna)^(lna)tan^-1(e^z)dz`

`=int_(-lna)^0tan^-1e^zdz+int_0^(lna)tan^-1e^zdz`

`=int_0^(lna)tan^-1e^-zdz+int_0^(lna)tan^-1e^zdz`

`=int_0^(lna)(cot^-1e^z+tan^-1e^z)dz`

`=int_0^(lna)(pi/2)dz`

`=pi/2lna`