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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  
Newbie Joined: Jul 2017 From: London, England Posts: 2 Thanks: 1  Quote:
$\ln(x+1) = x  \frac{x^2}{2} + \frac{x^3}{3}  \frac{x^4}{4}+ \dots$ 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 $$S=\zeta(2)2\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\zeta(2)2\cdot\frac14\sum_{n=1}^{\infty}\frac{1}{n^2}= \zeta(2)\frac12\zeta(2)=\frac12\zeta(2)=\frac{\pi^2}{12}$$ . 
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.

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.

August 12th, 2017, 09:14 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,327 Thanks: 2451 Math Focus: Mainly analysis and algebra  Quote:
Beautiful!  

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