A Reading Mistake :)

I won't be providing justification of my steps :

Put $x=tan(y)$ to get our integral as :

$\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ yln({ sec }^{ 2 }y)dy }$

$\displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ yln(cos(y))dy }$

Now we would be calculating : $\displaystyle J = \int _{ 0 }^{ \frac { \pi }{ 2 } }{ yln(cos(y))dy }$

Now $\displaystyle cos(x) = \frac{e^{ix}+e^{-ix}}{2}$ .

Hence we have :

$\displaystyle J=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ yln(\dfrac { { e }^{ iy }+{ e }^{ -iy } }{ 2 } )dy }$

$\displaystyle J=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ yln(1+{ e }^{ 2iy })dy } - \int _{ 0 }^{ \frac { \pi }{ 2 } }{ y(ln(2)+iy)dy }$

I will be using the taylor expansion of $ln(1+x)$ we would be getting :

$\displaystyle J=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ y\sum _{ r=1 }^{ \infty }{ \dfrac { { (-1) }^{ r-1 } }{ r } { e }^{ (2iry) } } dy } - (\dfrac { { \pi }^{ 2 } }{ 8 } ln(2)+\dfrac { { \pi }^{ 3 }i }{ 24 })$

Let :

$K= \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ y{ e }^{ 2iry }dy } }$

Put $2iry =x$ , Note in this step my path of this complex integration is a straight line between $0 , r\pi i$

$\displaystyle K=\sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r } }{ { 4r }^{ 3 } } \int _{ 0 }^{ r\pi i }{ { xe }^{ x }dx } }$

$\Rightarrow K=\sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r } }{ { 4r }^{ 3 } } ((r\pi i-1){ (-1) }^{ r }+1) }$

$\displaystyle K = (\pi i)\sum _{ r=1 }^{ \infty }{ \frac { 1 }{ { 4r }^{ 2 } } } +\sum _{ r=1 }^{ \infty }{ \frac { -1+{ (-1) }^{ r } }{ { 4r }^{ 3 } } }$

$K=(\frac { \pi i }{ 4 } )\zeta (2)+(\frac { -1 }{ 2 } )(1+\frac { 1 }{ { 3 }^{ 3 } } +\frac { 1 }{ { 5 }^{ 3 } } +.......)$

$(\frac { \pi i }{ 4 } )\zeta (2)+(\frac { -1 }{ 2 } )((1+\frac { 1 }{ 2^{ 3 } } +\frac { 1 }{ { 3 }^{ 3 } } +.......)-\frac { 1 }{ 8 } (1+\frac { 1 }{ 2^{ 3 } } +\frac { 1 }{ { 3 }^{ 3 } } +.......))$

$K= \frac { 4i\pi \zeta (2)-7\zeta (3) }{ 16 }$

Putting the value of $K$ in $J$ we get :

$\frac { -7\zeta (3) }{ 16 } -\frac { { \pi }^{ 2 }ln(2) }{ 8 } +i(\frac { 6\zeta (2)-{ \pi }^{ 2 } }{ 4 } )$

Interesting thing to observe that since $J$ is purely real hence we have :

$\zeta(2)=\frac{{\pi}^{2}}{6}$

And we have proved an very interesting result.

Also we get :

$J=\frac { -7\zeta (3) }{ 16 } -\frac { { \pi }^{ 2 }ln(2) }{ 8 }$

Finally we have :

$I=\frac { { \pi }^{ 2 }ln(2) }{ 4 } +\frac { 7\zeta (3) }{ 8 }$