Proof that ζ(2)=π26\zeta(2)=\frac{{\pi}^{2}}{6}

Well here's another proof that n=11n2=π26\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }= \frac{{\pi}^{2}}{6}

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

now cos(x)=eix+eix2cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}

I=0π/2ln(eix+eix2)dx\Rightarrow I=\displaystyle \int _{ 0 }^{ \pi /2 }{ ln(\frac { { e }^{ ix }+{ e }^{ -ix } }{ 2 } )dx }

I=0π/2ln(1+e2ix)dx0π/2(ln(2)+ix)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)ln(1+x) we have :

I=0π/2r=1(1)r1re2irxdx(πln(2)2+π2i8)\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 :

I=r=1(1)r1r0π/2e2irxdx(πln(2)2+π2i8)\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 :

I=r=1(1)r12ir2(eπir1)(πln(2)2+π2i8)\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 eπir=(1)r{e}^{\pi ir} = {(-1)}^{r}

Hence I=(r=11(1)r2r2)i(πln(2)2+π2i8)\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+132+152+......)S=(1+\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 5 }^{ 2 } } +......)

S=(1+122+132+142+......)(122+142+....)S=(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)-(\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +....)

S=(1+122+132+142+......)14(112+122+....)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=(34)(1+122+132+142+......)S=(\dfrac { 3 }{ 4 } )(1+\dfrac { 1 }{ { 2 }^{ 2 } } +\dfrac { 1 }{ { 3 }^{ 2 } } +\dfrac { 1 }{ { 4 }^{ 2 } } +......)

S=(34)ζ(2)S=(\dfrac { 3 }{ 4 } )\zeta (2)

Putting the value of SS in our integral we have :

I=πln(2)2+(3ζ(2)4π28)iI=\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 00 hence :

3ζ(2)4=π28\dfrac { 3\zeta (2) }{ 4 } =\dfrac { { \pi }^{ 2 } }{ 8 }

Finally :

ζ(2)=π26\zeta(2) = \dfrac{{\pi}^{2}}{6}

Note by Ronak Agarwal
4 years, 6 months ago

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Very slick! :D

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

Shashwat Shukla - 4 years, 6 months ago

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This is amazing!!!

Kishlaya Jaiswal - 4 years, 6 months ago

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Nicely Done ¨\ddot\smile

A Brilliant Member - 4 years, 6 months ago

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

Nishu sharma - 4 years, 3 months ago

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

Jake Lai - 4 years, 5 months ago

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It is an original proof.

Ronak Agarwal - 4 years, 5 months ago

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That's wonderful! If you find analogues for 2n2n please do share with us.

Jake Lai - 4 years, 5 months ago

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@Jake Lai

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

Just curious to know why you changed your profile pic ?

A Brilliant Member - 4 years, 5 months ago

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Can this be done? -

I=0π/2ln2(cos(x))dx\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^2(cos(x))dx } if yes, can this also - I=0π/2lnn(cos(x))dx\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^n(cos(x))dx } nN n \to N

U Z - 4 years, 5 months ago

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

Ronak Agarwal - 4 years, 5 months ago

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

A Brilliant Member - 4 years, 5 months ago

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

Ronak Agarwal - 4 years, 5 months ago

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Ok , so when is your B'day then ?

A Brilliant Member - 4 years, 5 months ago

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@A Brilliant Member 15th July.

Ronak Agarwal - 4 years, 5 months ago

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@Ronak Agarwal Ok , thanks :)

A Brilliant Member - 4 years, 5 months ago

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

Abdur Rehman Zahid - 4 years, 5 months ago

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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

Abhinav Raichur - 4 years, 4 months ago

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Its a superb proof.

Please tell the motivation :D

Jyothiraditya Datta Poduri - 4 years, 2 months ago

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

Rajdeep Dhingra - 4 years, 6 months ago

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Wow wonderful !

Tejas Suresh - 4 years ago

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