Well here's another proof that n=1∑∞n21=6π2
We will start from the integral I=∫0π/2ln(cos(x))dx
now cos(x)=2eix+e−ix
⇒I=∫0π/2ln(2eix+e−ix)dx
I=∫0π/2ln(1+e2ix)dx−∫0π/2(ln(2)+ix)dx
Considering taylor expansion of ln(1+x) we have :
I=∫0π/2r=1∑∞r(−1)r−1e2irxdx−(2πln(2)+8π2i)
Changing the order of integration and summation we have :
I=r=1∑∞r(−1)r−1∫0π/2e2irxdx−(2πln(2)+8π2i)
Integrating and putting limits we get :
I=r=1∑∞2ir2(−1)r−1(eπir−1)−(2πln(2)+8π2i)
Now eπir=(−1)r
Hence I=(r=1∑∞2r21−(−1)r)i−(2πln(2)+8π2i)
We will be calculating the value of our summation.
S=(1+321+521+......)
S=(1+221+321+421+......)−(221+421+....)
S=(1+221+321+421+......)−41(121+221+....)
S=(43)(1+221+321+421+......)
S=(43)ζ(2)
Putting the value of S in our integral we have :
I=2−πln(2)+(43ζ(2)−8π2)i
Now since the integral is real hence imaginary part of our integral is 0 hence :
43ζ(2)=8π2
Finally :
ζ(2)=6π2
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:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestVery slick! :D
Just curious though: what motivated you to start with that particular integral?
Log in to reply
This is amazing!!!
Log in to reply
wow Ingenious .. and much impressive ! You have really genius brain . I'am much impress with your skills ! Hat's off
Log in to reply
Nicely Done ⌣¨
Log in to reply
Is this an original proof? If not, where did you get the proof from? It is a remarkable and elegant one.
Log in to reply
It is an original proof.
Log in to reply
That's wonderful! If you find analogues for 2n please do share with us.
Log in to reply
@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 ?
Log in to reply
Can this be done? -
I=∫0π/2ln2(cos(x))dx if yes, can this also - I=∫0π/2lnn(cos(x))dx n→N
Log in to reply
Yes this can be done I will be posting this as a problem soon.
Log in to reply
Hi Ronak , is your B'day on the 16th of March ?
Log in to reply
Nope on 16th March Result of INCHO will be announced. I am very much excited for that.
Log in to reply
Ok , so when is your B'day then ?
Log in to reply
Log in to reply
Log in to reply
@Ronak Agarwal
My birthday is on 14th July!!!Log in to reply
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
Log in to reply
Its a superb proof.
Please tell the motivation :D
Log in to reply
@Ronak Agarwal Could you tell me when can we change the order of summation and integration.
Log in to reply
Wow wonderful !
Log in to reply