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 :
\(\zeta(2) = \dfrac{{\pi}^{2}}{6}\)
Comments
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Top NewestVery slick! :D
Just curious though: what motivated you to start with that particular integral? – Shashwat Shukla · 2 years, 5 months ago
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This is amazing!!! – Kishlaya Jaiswal · 2 years, 5 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 · 2 years, 2 months ago
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Nicely Done \(\ddot\smile\) – Azhaghu Roopesh M · 2 years, 5 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 · 2 years, 5 months ago
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It is an original proof. – Ronak Agarwal · 2 years, 5 months ago
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That's wonderful! If you find analogues for \(2n\) please do share with us. – Jake Lai · 2 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 ? – Azhaghu Roopesh M · 2 years, 5 months ago
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Wow wonderful ! – Tejas Suresh · 2 years ago
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Its a superb proof.
Please tell the motivation :D – Pradyumna Datta Poduri · 2 years, 2 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 · 2 years, 3 months ago
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Hi Ronak , is your B'day on the 16th of March ? – Azhaghu Roopesh M · 2 years, 4 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 · 2 years, 4 months ago
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Ok , so when is your B'day then ? – Azhaghu Roopesh M · 2 years, 4 months ago
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15th July. – Ronak Agarwal · 2 years, 4 months ago
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My birthday is on 14th July!!! @Ronak Agarwal – Abdur Rehman Zahid · 2 years, 4 months ago
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Ok , thanks :) – Azhaghu Roopesh M · 2 years, 4 months ago
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Can this be done? -
\(\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^2(cos(x))dx } \) if yes, can this also - \(\displaystyle I=\int _{ 0 }^{ \pi /2 }{ ln^n(cos(x))dx } \) \( n \to N\) – Megh Choksi · 2 years, 4 months ago
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Yes this can be done I will be posting this as a problem soon. – Ronak Agarwal · 2 years, 4 months ago
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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. – Trevor B. · 2 years, 5 months ago
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I believe it must be \(\dfrac{3}{4}\) , I think you have quoted the wrong value as you can see \(\zeta{(2)}-1=\dfrac{{\pi}^{2}}{6} -1 > 0.5 \) – Ronak Agarwal · 2 years, 5 months ago
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Oops, I misremembered the value. Thanks for the catch. – Trevor B. · 2 years, 5 months ago
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Ha Ha Ha!!! you know very well that till date no such closed form of \(\zeta{(3)}\) has been found – Ronak Agarwal · 2 years, 5 months ago
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@Ronak Agarwal Could you tell me when can we change the order of summation and integration. – Rajdeep Dhingra · 2 years, 5 months ago
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