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Sum of 1/n^2


Date: 07/24/2000 at 06:35:53
From: Mike
Subject: Sum of 1/n^2 without Fourier series

Sirs,

Euler proved that the sum of 1/n^2 is equal to pi^2/6. It is easily 
shown that this sum is equal to INT(0->1)INT(0->1) 1/(1-xy) dxdy or 
the limit of the double integrals over the rectangle [0,t]X[0,t] as 
t->1(from the left).

I made the change of variables x=(u-v)/(2)^1/2 and y=(u+v)/(2)^1/2, 
which rotates the original rectangle about the origin by pi/4 in the 
uv plane. In changing the variables, the integral changed from f(x,y) 
to 2/(2-u^2+v^2); the Jacobian is 1, but in changing the bounds, you 
go from [0,1]X[0,1] to four functions in uv: v = u, v = -u, v = 1-u 
and v = u-1. which I do not know how to interpret in the bounds of a 
double integral. 

Also, how should I start in evaluating the integral? In its first 
integration (say wrt v), it is in the form of 2/(c+v^2), where 
c = 2+u^2, which is close to arctan(v) as a solution. 

I hope I have explained this clearly.
Thank you for your time.


Date: 07/24/2000 at 11:26:25
From: Doctor Rob
Subject: Re: Sum of 1/n^2 without Fourier series

Thanks for writing to Ask Dr. Math, Mike.

The square over which you are integrating has sides with inclination
Pi/4 and -Pi/4. To integrate over this region, you should probably
split it into two regions as follows:

   0 <= u <= sqrt(2)/2,  -u <= v <= u,

and

   sqrt(2)/2 <= u <= sqrt(2),  u - sqrt(2) <= v <= sqrt(2) - u.

Now to do the inside integration with respect to v, write

   2/(2+u^2-v^2) = A/(sqrt[2+u^2]+v) + B/(sqrt[2+u^2]-v),
   A = -1/sqrt[2+u^2],
   B = 1/sqrt[2+u^2].

Then the integral is

   A*ln(sqrt[2+u^2]+v) + B*ln(sqrt[2+u^2]-v),

evaluated between the limits. This can be simplified to

   4*Arctanh(u/sqrt[2+u^2])/sqrt(2+u^2) +
      4*Arctanh([sqrt(2)-u]/sqrt[2+u^2])/sqrt(2+u^2).

Now the first term can be integrated, and you get
2*Arctanh[1/Sqrt(5)]^2, but the second term seems intractable.

A variation of the method you suggest is actually successful in
computing SUM 1/n^2.  The trick is to compute the sum of the
reciprocals of the squares of the odd numbers only:

       infinity
   S =   SUM   1/(2*n-1)^2
         n=1

This can be shown to be identical to the double integral

               1         1
   S = INTEGRAL  INTEGRAL  1/(1-x^2*y^2) dy dx,
               0         0

using the same method. Now there is a very clever way to evaluate
this double integral: use the substitution

   x = sin(u)/cos(v),
   y = sin(v)/cos(u).

The Jacobian of this transformation is 1 - tan^2(u)*tan^2(v), and the 
integrand becomes 1/(1-tan^2[u]*tan^2[v]), so the new integrand is 
magically reduced to 1. The only problem is to figure out the new 
limits of integration. It turns out that the region over which to 
integrate is the triangle bounded by u = 0, v = 0, and u + v = Pi/2.  
Integrating 1 over this region gives the area of the triangle, which 
is obviously Pi^2/8 (half the base Pi/2 times the height Pi/2). Thus

   S = Pi^2/8.

Now to find the actual sum you want, call it T, observe that

   T - T/4 = S,
   T = 4*S/3 = Pi^2/6.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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