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

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

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 :

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

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 :

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

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Very slick! :D

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

Shashwat Shukla - 6 years, 3 months ago

This is amazing!!!

Kishlaya Jaiswal - 6 years, 3 months ago

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

Nishu sharma - 6 years, 1 month ago

Nicely Done $\ddot\smile$

A Former Brilliant Member - 6 years, 3 months ago

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

Jake Lai - 6 years, 3 months ago

It is an original proof.

Ronak Agarwal - 6 years, 3 months ago

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

@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 ?

Can this be done? -

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

U Z - 6 years, 3 months ago

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

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

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

Ok , so when is your B'day then ?

@A Former Brilliant Member – 15th July.

@Ronak Agarwal – Ok , thanks :)

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

Abdur Rehman Zahid - 6 years, 3 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

Abhinav Raichur - 6 years, 2 months ago

Its a superb proof.

Please tell the motivation :D

Jyothiraditya Datta Poduri - 6 years ago

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

Rajdeep Dhingra - 6 years, 3 months ago

Wow wonderful !

Tejas Suresh - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

Comments