1. ## Integral/Power Series

Just a quick question that I'm stuck on:

Let $\Gamma$ be the positively oriented circle with centre 0 and radius R. Show that

$\int_{\Gamma} \ z^{-2} e^{t(z + z^{-1})} dz = \sum_{m=0}^{\infty} \ b_{m} t^{2m+1}$

where the $b_{m}$ are constants you should calculate.

Is it not just a case of observing singularities and using Cauchy's Residue Theorem? Maybe there's something I'm missing but I'm stumped.

Thanks

pomp.

2. OK, I think I might have it, but it involved Laurent expansions...and I'm not very good at these kind of things.

So the integrand has a singularity about 0, but is holomorphic on any punctured disc about zero. Therefore we can expand about z=0 and look for the coefficient of the $z^{-1}$ term.

$f(z) = z^{-2} e^{tz} e^{t z^{-1}} = \frac{1}{z^2} \left( \sum_{n=0}^{\infty} \ \frac{(tz)^n}{n!} \right) \left( \sum_{n=0}^{\infty} \ \frac{(t/z)^n}{n!} \right)$

$f(z) = \left( z^{-2} + tz^{-1} + {t^2}/2 + zt^3 /6 + \ldots \right) \left( 1 + tz^{-1} + t^2 z^{-2}/2 + t^3 z^{-3}/6 + \ldots \right)$

So reading of the $z^{-1}$ terms,

$\int_{\Gamma} \ f(z) dz = 2 \pi i \times (t + \frac{t^3}{2} + \ldots )$

I can see now why only odd powers of t contribute but I can't see the quick way of finding all the coefficients of these powers, can anyone see how? Or possibly even a better way of approaching the problem?

-pomp

3. $\sum_{n=0}^{\infty}\frac{t^n z^{n-2}}{n!}\sum_{n=0}^{\infty}\frac{t^n}{n! z^n}=\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{t^k z^{k-2}}{k!} \frac{t^{n-k}}{z^{n-k} (n-k)!}$

we need the z terms with $k-2-(n+k)=-1$ or $n=2k-1, k\geq 1$ then the sum of these coefficients would be

$\lim_{n\to\infty} \sum_{k=1}^{n}\frac{t^k t^{2k-1-k}}{k! (2k-1-k)!}$

$=\sum_{k=1}^{\infty} \frac{t^{2k-1}}{k! (k-1)!}=\sum_{k=0}^{\infty} \frac{t^{2k+1}}{k!(k+1)!}$

Interesting exercise:

Show solution of Bessel equation $z^2 y''+zy'-(z^2+1)y=0$ is $y(z)=\sum_{n=0}^{\infty} \frac{z^{2n+1}}{n! (n+1)!}$ and therefore:

$y(z)=\frac{1}{2\pi i}\oint e^{z(s+1/s)} s^{-2}ds$.

I think it's surprising a complex contour integral function is a solution to a differential equation.

4. Ah nice one, perfect. That's a nice link, I've seen that solution of the Bessel equation before. Maybe there were a couple of marks credited at the end of the question for spotting this (this was from an old exam question)

thanks.

pomp