How Do You Evaluate the Residues in the Cauchy Integral Formula for G_0(u)?

[marcusl]

I have a function
$$G_0=\frac{1-\alpha/u^2}{1-\alpha u^2} .$$

Since $$0<\alpha<1$$, $$G_0$$ has zeroes but no poles inside the unit circle.

I need to evaluate
$$\Gamma(\tau)=\frac{1}{2\pi i}\oint{\frac{\ln{G_0 (u)}}{u-\tau}du}$$
where the integral is around the unit circle. How do I evaluate the poles of the integrand so I can evaluate this using residues?

EDIT: Oops, G0 has a second order singularity at u=0, too.

 

[mathwonk]

maybe power series wouldwork, first the geometric series for G0 then the series for log(1+w)?

 

[tacman]

To evaluate the poles of the integrand, we can use the fact that the integrand has a logarithmic singularity at u=0. This means that we can expand the integrand as a Laurent series around u=0:

\ln{G_0(u)} = \ln{\left(\frac{1-\alpha/u^2}{1-\alpha u^2}\right)} = -\ln{(1-\alpha u^2)} - \ln{(1-\alpha/u^2)} = -\sum_{n=1}^{\infty} \frac{\alpha^n}{n} u^{2n} - \sum_{n=1}^{\infty} \frac{\alpha^n}{n} \frac{1}{u^{2n}}

We can see that the first term in the series has a pole of order 2 at u=0, while the second term has a pole of order 2 at u=\infty. This means that the integrand has poles at u=0 and u=\infty, both of order 2.

To evaluate the residue at u=0, we can use the formula for the coefficient of the Laurent series:

Res(f,u_0) = \frac{1}{2\pi i} \oint{\frac{f(u)}{(u-u_0)^{n+1}}du}

In this case, we have n=2 and u_0=0. Plugging in the Laurent series for \ln{G_0(u)}, we get:

Res(\ln{G_0(u)}, 0) = \frac{1}{2\pi i} \oint{\frac{-\sum_{n=1}^{\infty} \frac{\alpha^n}{n} u^{2n}}{u^3}du} = \frac{-\alpha}{2\pi i} \oint{\frac{1}{u}du} = -\frac{\alpha}{2}

Similarly, we can evaluate the residue at u=\infty by using the series expansion for \ln{G_0(u)} around u=\infty:

Res(\ln{G_0(u)}, \infty) = \frac{1}{2\pi i} \oint{\frac{\sum_{n=1}^{\infty} \frac{\alpha^n}{n} \frac{1}{u^{2n}}