# 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
3 years, 2 months ago

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Very slick! :D

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

- 3 years, 2 months ago

This is amazing!!!

- 3 years, 2 months ago

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

- 2 years, 11 months ago

Nicely Done $$\ddot\smile$$

- 3 years, 2 months ago

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

- 3 years, 1 month ago

It is an original proof.

- 3 years, 1 month ago

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

- 3 years, 1 month 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 ?

- 3 years, 1 month ago

Its a superb proof.

- 2 years, 11 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

- 3 years ago

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Ok , so when is your B'day then ?

- 3 years, 1 month ago

15th July.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Ok , thanks :)

- 3 years, 1 month ago

Can this be done? -

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Comment deleted Feb 23, 2015

As @Ronak Agarwal said, there is no closed form for $$\zeta(3),$$ and there isn't a closed form for any $$\zeta(2k+1)$$ given positive integer $$k$$.

An interesting result I found a while back is that all of the $$\zeta(2k)$$ values with all of their $$\pi^{2k}$$ numerators and strange denominators can be manipulated to sum to a nice result: $\sum_{k=1}^\infty(\zeta(2k)-1)=\dfrac{3}{4}$ It just goes to show how looking at a problem from a new perspective can lead you to a surprisingly simple result.

- 3 years, 2 months ago

I believe it must be $$\dfrac{3}{4}$$ , I think you have quoted the wrong value as you can see $$\zeta{(2)}-1=\dfrac{{\pi}^{2}}{6} -1 > 0.5$$

- 3 years, 2 months ago

Oops, I misremembered the value. Thanks for the catch.

- 3 years, 2 months ago

Ha Ha Ha!!! you know very well that till date no such closed form of $$\zeta{(3)}$$ has been found

- 3 years, 2 months ago

Wow wonderful !

- 2 years, 9 months ago

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

- 3 years, 2 months ago