We can evaluate the summation as follows. When summing a meromorphic function f(z) over the zeroes of an analytic function g(z) we can consider the contour integral of f(z) g'(z)/g(z) dz over a counterclockwise circle of radius R. Then the zeroes of g(z) are the poles of the integrand (unless they are canceled by f(z)) and the residue there is just the value of f there. Then in the limit of R to infinity, the integral equals a pi i times the desired summation plus the sum of the residues of f(z) g'(z)/g(z) at the poles of f(z). If the integral tends to zero for R to infinity, then the desired summation is thus minus the sum of the residues of f(z) g'(z)/g(z) at the poles of f(z).

Since the function g(z) = sin(pi z) has its zeroes at the integers, we can sum ...

Solution Summary

A detailed calculation is given for the summation from minus infinity to infinity of 1/(n^4 + a^4).

... 3z^2 Sum over 1/n^4 from minus to plus infinity, while skipping n = 0 +. 5z^4 Sum over 1/n^6 from minus to plus infinity, while skipping n = 0. ...

... up a factor exp(i pi k) relative to the lower part and there is a minus sign because we ...Sum from = 0 to infinity of Integral from zero to infinity of x ...

... The sum of the coefficients |a_{j}|^2 is equal to 1... numerator of (4) is the integral of this from minus infinity to plus ... x^n exp(-x/a) from zero to infinity is a ...

... write this integral as the principal part obtained by removing the contribution to the integral from minus to plus ... f(t) = sum from n = 0 to infinity of (-1...

...1/(1-y)^k = sum from r = 0 to infinity of Binomial(r+k-1... the best approximation (note that by (2) negative p map x to minus infinity for z = 1). Let's first ...

... This means that P is the average of minus dE_r/dV ...Sum over n from zero to infinity of exp(-beta hbar omega n) = 1/[1 - exp ... So, for T ---> infinity Cv ---> 3 N k. ...

... Taking together the contributions from n and minus n in (1) gives for an even function f ...1/R Integral from 0 to R of f(t) dt + Sum over n = 1 to infinity of 2 ...

... Problem c). The probability of one or more mutations is the survival probability minus the probability of ... P_sur = Sum over n from zero to infinity of P(n...

... is the integral from zero to infinity, which yields: ... to an integration over x from minus to plus ... Let's consider numerically evaluating the summation: ∞ 1 (22 ...