×

# 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

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

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
2 years ago

Sort by:

FanTastic Proof. Hats off · 2 years ago

Thanks :) · 2 years ago

Nice proof @Mvs Saketh :)

@Ronak Agarwal Beware !! Here's some competition for you ! · 2 years ago

thankyou but that was never the intention, i just posted it because i liked it, · 2 years ago

I was just kidding Saketh !! · 2 years ago

i know, please dont use more than one $$!$$ at a time, it becomes hard to know whether you are shouting or telling :P · 2 years 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 :) · 2 years ago

Cheers :) · 2 years ago

Me Too. · 2 years ago

It seems that your 100 follower question got a level 5 rating , cheers:) · 2 years ago

Yes but it should be level 3 · 2 years 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 ! · 2 years ago

Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series? · 2 years 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 :) · 2 years 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 . · 2 years 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 · 2 years ago