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 July 30th, 2017, 10:31 AM #1 Member   Joined: Oct 2016 From: Arizona Posts: 63 Thanks: 15 Math Focus: Still deciding! Proving a definite integral Hi, I used Wolfram Alpha to get that $\int_0^1 \! \frac{\ln(x+1)}{x} \, \mathrm{d}x= \frac{\pi^2}{12}$ but is there any way to prove this? How would you come to this without using Wolfram Alpha? Any ideas are appreciated.
July 30th, 2017, 01:16 PM   #2
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Quote:
 Originally Posted by ProofOfALifetime Hi, I used Wolfram Alpha to get that $\int_0^1 \! \frac{\ln(x+1)}{x} \, \mathrm{d}x= \frac{\pi^2}{12}$ but is there any way to prove this? How would you come to this without using Wolfram Alpha? Any ideas are appreciated.
Using Maclaurin's Expansion.

$\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ \dots$

Hence,

$\frac{\ln(x+1)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \dots$

Therefore the integral is

$1 - \frac{1}{4} + \frac{1}{9} -\frac{1}{16} + \dots$

which is $\frac{\pi^2}{12}$

Last edited by skipjack; July 30th, 2017 at 08:05 PM.

 August 2nd, 2017, 09:29 AM #3 Member   Joined: Oct 2016 From: Arizona Posts: 63 Thanks: 15 Math Focus: Still deciding! Thank you MaxBenjamin. I seem to not have the background for the $1 - \frac{1}{4} + \frac{1}{9} -\frac{1}{16} + \dots$ part. Can you give me advice on any books/sources that go over this material? In my calculus book, even in the series sections it does not go over this. How would I learn more about recognizing that as $\frac{\pi^2}{12}$? I do understand the Maclaurin expansion and integration steps, just not the last step. How do I practice problems similar to this?
 August 2nd, 2017, 10:30 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Of interest is the Riemann Zeta function, in particular, $\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}= \frac{\pi^2}{6}$. From this $$S=\zeta(2)-2\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\zeta(2)-2\cdot\frac14\sum_{n=1}^{\infty}\frac{1}{n^2}= \zeta(2)-\frac12\zeta(2)=\frac12\zeta(2)=\frac{\pi^2}{12}$$ . Thanks from ProofOfALifetime
 August 2nd, 2017, 01:17 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra Also you can generate such results from Fourier series. Thanks from ProofOfALifetime
 August 2nd, 2017, 09:27 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 For a sketch of an elementary proof that $\displaystyle \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$, see my post here. Thanks from ProofOfALifetime
 August 3rd, 2017, 01:46 PM #7 Member   Joined: Oct 2016 From: Arizona Posts: 63 Thanks: 15 Math Focus: Still deciding! Thank you all very much! I look forward to studying all these suggestions. I appreciate it. Thanks from greg1313
August 12th, 2017, 09:14 PM   #8
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Quote:
 Originally Posted by skipjack For a sketch of an elementary proof that $\displaystyle \zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$, see my post here.
And here's Euler's method

Beautiful!

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