Yet another proof for \(\zeta (2) = \frac { { \pi }^{ 2 } }{ 6 } \)

Let us consider the integral

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$

Now using

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta \\ \\ 1-{ e }^{ i\theta } = 1-cos\theta -isin\theta = 2sin(\frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \theta }{ 2 } ))\\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \theta }{ 2 } -\frac { \pi }{ 2 } ) }$

we have the integral as

$\displaystyle \int _{ 0 }^{ \pi }{ (i(ln(2) + ln(sin(\frac { \theta }{ 2 } )) - \left( \frac { \theta }{ 2 } -\frac { \pi }{ 2 } \right) )d\theta }$

Both its real and imaginary parts can be easily evaluated to get

answer as simply

$\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other)

Now, reconsider the integral

$\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$

Let us substitute

$\displaystyle z= { e }^{ i\theta }\\$

at $\displaystyle \theta =0 \quad z=1\\ \theta =\pi \quad z=-1\\$ also $dz\quad =\quad i{ e }^{ i\theta }$

We get the integral

$\displaystyle -\int _{ -1 }^{ 1 }{ ln(1-x)\frac { 1 }{ x } dx } \\$

Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past

using taylor expansion we get,

$\displaystyle \int _{ -1 }^{ 1 }{ (\frac { 1 }{ 1 } } +\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 3 } ...)\quad =\quad 2(\frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...)\quad =\frac { { \pi }^{ 2 } }{ 4 } (as\quad proved\quad before)\\$

$\displaystyle \zeta (2)\quad =\quad \frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...+\frac { 1 }{ 4 } \zeta (2)\quad \quad (which\quad you\quad yourself\quad can\quad check)$

$\displaystyle \frac { 3 }{ 4 } \zeta (2) = \frac { \pi ^{ 2 } }{ 8 } \quad\rightarrow \boxed{\zeta(2)= \frac { \pi ^{ 2 } }{ 6 }}$

Hence proved

Do point out any flaws i might have done

Entirely original, any resemblance is accidental

Inspritation - Ronaks proof (just inspiration to try to prove , not copy)