Sum of 1/n^2Date: 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/ |
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