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

Well here's another proof that $\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }= \frac{{\pi}^{2}}{6}$

We will start from the integral $\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln(cos(x))dx }$

now $cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$

$\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /2 }{ ln(\frac { { e }^{ ix }+{ e }^{ -ix } }{ 2 } )dx }$

$I = \displaystyle \int _{ 0 }^{ \pi /2 }{ ln(1+{ e }^{ 2ix })dx } - \int _{ 0 }^{ \pi /2 }{ (ln(2)+ix)dx }$

Considering taylor expansion of $ln(1+x)$ we have :

$\displaystyle I = \int _{ 0 }^{ \pi /2 }{ \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } { e }^{ 2irx } } dx } -(\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})$

Changing the order of integration and summation we have :

$\displaystyle I = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ r } \int _{ 0 }^{ \pi /2 }{ { e }^{ 2irx }dx } } -(\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})$

Integrating and putting limits we get :

$\displaystyle I= \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r-1 } }{ 2i{ r }^{ 2 } } ( } { e }^{ \pi ir }-1) - (\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})$

Now ${e}^{\pi ir} = {(-1)}^{r}$

Hence $\displaystyle I = (\sum _{ r=1 }^{ \infty }{ \frac { 1-{ (-1) }^{ r } }{ 2{ r }^{ 2 } } } )i - (\dfrac{\pi ln(2)}{2}+\frac{{\pi}^{2}i}{8})$

We will be calculating the value of our summation.

$S=(1+\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 5 }^{ 2 } } +......)$

$S=(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)-(\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +....)$

$S=(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)-\dfrac { 1 }{ 4 } (\dfrac { 1 }{ { 1 }^{ 2 } } +\dfrac { 1 }{ { 2 }^{ 2 } } +....)$

$S=(\dfrac { 3 }{ 4 } )(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)$

$S=(\dfrac { 3 }{ 4 } )\zeta (2)$

Putting the value of $S$ in our integral we have :

$I=\dfrac { -\pi ln(2) }{ 2 } +(\dfrac { 3\zeta (2) }{ 4 } -\dfrac { { \pi }^{ 2 } }{ 8 } )i$

Now since the integral is real hence imaginary part of our integral is $0$ hence :

$\dfrac { 3\zeta (2) }{ 4 } =\dfrac { { \pi }^{ 2 } }{ 8 }$

Finally :

$\zeta(2) = \dfrac{{\pi}^{2}}{6}$ Note by Ronak Agarwal
4 years, 9 months ago

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Sort by:

Wow wonderful !

- 4 years, 4 months ago

Its a superb proof.

- 4 years, 6 months ago

wow Ingenious .. and much impressive ! You have really genius brain . I'am much impress with your skills ! Hat's off

- 4 years, 7 months ago

wow ! ..... imaginary part must be zero ...... i would have been foolish enough to leave it seeing that we got a unreal value .... its great tht u carried it over ........ i learned from this :) ...... thanks for a beautiful derivation

- 4 years, 7 months ago

Hi Ronak , is your B'day on the 16th of March ?

- 4 years, 9 months ago

Nope on 16th March Result of INCHO will be announced. I am very much excited for that.

- 4 years, 9 months ago

Ok , so when is your B'day then ?

- 4 years, 9 months ago

15th July.

- 4 years, 9 months ago

My birthday is on 14th July!!! @Ronak Agarwal

- 4 years, 9 months ago

Ok , thanks :)

- 4 years, 9 months ago

Can this be done? -

$\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^2(cos(x))dx }$ if yes, can this also - $\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^n(cos(x))dx }$ $n \to N$

- 4 years, 9 months ago

Yes this can be done I will be posting this as a problem soon.

- 4 years, 9 months ago

Is this an original proof? If not, where did you get the proof from? It is a remarkable and elegant one.

- 4 years, 9 months ago

Hi dude , can't help but smile when I see your Profile pic xd .

Just curious to know why you changed your profile pic ?

- 4 years, 9 months ago

It is an original proof.

- 4 years, 9 months ago

That's wonderful! If you find analogues for $2n$ please do share with us.

- 4 years, 9 months ago

@Ronak Agarwal Could you tell me when can we change the order of summation and integration.

- 4 years, 9 months ago

Very slick! :D

Just curious though: what motivated you to start with that particular integral?

- 4 years, 9 months ago

This is amazing!!!

- 4 years, 9 months ago

Nicely Done $\ddot\smile$

- 4 years, 9 months ago