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 :

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

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## Comments

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TopNewestVery slick! :D

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

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This is amazing!!!

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

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Nicely Done \(\ddot\smile\)

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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.

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It is an original proof.

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

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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 ?

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Its a superb proof.

Please tell the motivation :D

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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 derivationLog in to reply

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

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

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Ok , so when is your B'day then ?

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@Ronak Agarwal

My birthday is on 14th July!!!Log in to reply

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Can this be done? -

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

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

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Comment deleted Feb 23, 2015

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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.

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

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

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Wow wonderful !

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

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