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)

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## Comments

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TopNewestFanTastic Proof. Hats off

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Thanks :)

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Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series?

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Well, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :)

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That's nice :) ...I had thought that it might have in part been inspired by \[\int_{0}^{1}\frac{ln(1-x)}{x}dx\] (both of them have ln(1-z) )

Either ways, it's pretty cool.

Oh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D .

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Ok , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks

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Nice proof @Mvs Saketh :)

@Ronak Agarwal Beware !! Here's some competition for you !

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thankyou but that was never the intention, i just posted it because i liked it,

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I was just kidding Saketh !!

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