# 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

$\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (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 }\$/extract_itex] 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) Note by Mvs Saketh 6 years, 3 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote  # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or \[ ... $ to ensure proper formatting.
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FanTastic Proof. Hats off

- 6 years, 3 months ago

Thanks :)

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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 .

- 6 years, 3 months ago

Yes that too did inspire,

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

- 6 years, 3 months ago

Cheers :)

- 6 years, 3 months ago

Nice proof @Mvs Saketh :)

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

I was just kidding Saketh !!

- 6 years, 3 months ago

i know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P

- 6 years, 3 months ago

Sorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :)

- 6 years, 3 months ago

Cheers :)

- 6 years, 3 months ago

Me Too.

- 6 years, 3 months ago

It seems that your 100 follower question got a level 5 rating , cheers:)

- 6 years, 3 months ago

Yes but it should be level 3

- 6 years, 3 months ago

No worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of !

- 6 years, 3 months ago