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

Question

1891 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-20T13:38:42

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

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 $abs x < 1$

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

so finally

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

Answer 2 · 2016-07-21T02:35:16

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

$and dz==(dx)/x$

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

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

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

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

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

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

$=pi/2lna$

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