Show that $e^{\frac{1}{2}\lambda(z+\frac{1}{z})}= a_0 + \sum_{n=1}^\infty a_n(z^n+\frac{1}{z^n})$

Question

Here $\lambda \in \mathbb{C}$ and $$a_n = \frac{1}{\pi}\int_0^\pi e^{\lambda \cos(t)} \cos(nt)\,dt$$

I'm clueless about how to reach that expression for $a_n$. The things that came to my mind were to develop the Laurent series for some of the expressions, but didn't find myself nearer. I know the following:

$$e^{\frac{1}{2}\lambda(z+\frac{1}{z})}=\sum_{n=0}^\infty \frac{(\frac{1}{2} \lambda(z+\frac{1}{z}))^n}{n!}$$

$$e^{\frac{1}{2}\lambda(z+\frac{1}{z})}=e^{\frac{1}{2}\lambda z}e^{\frac{1}{2}\lambda\frac{1}{z}}=\sum_{n=0}^\infty\frac{(\frac{1}{2}\lambda z)^n}{n!}\sum_{n=0}^\infty\frac{(\frac{1}{2}\lambda \frac{1}{z})^n}{n!}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{\lambda^nz^{2k-n}}{2^nk!(n-k)!}$$

where the last equality is because of the Cauchy Product.

Answer 1

I think it's easier to use the fact that if $f$ is holomorphic in a punctured neighbourhood $U$ around $z = 0$ and $f(z) = \sum_{n = -\infty}^\infty a_n z^n$ is the Laurent series at $z = 0$, then $$ a_n = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(z)}{z^{n+1}} dz$$ for any closed contour $\gamma$ in $U$ that winds around $z = 0$ once anticlockwise.

For $f(z) = e^{\tfrac 1 2 \lambda (z + 1/z)}$, you probably want to take $\gamma$ to the unit circle, parameterised as $\gamma(t) = e^{it}$ for $ t \in [0, 2\pi].$ I hope this works out!

Answer 2

I'm gonna give my answer to this question for those who may need it.

Laurent's expansion of $f(z) =e^{\frac{\lambda}{2}(z + z^{-1})}$ is $$ f(z) = \sum_{n\in\mathbb{Z}}A_nz^n, $$ where $$ A_n = \frac{1}{2\pi i}\int_{|w|=r}\frac{e^{\frac{\lambda}{2}(w + w^{-1})}}{w^{n+1}}\,dw = \frac{1}{2\pi i}\int_{0}^{2\pi} \frac{e^{\frac{\lambda}{2}(re^{it} + r^{-1}e^{-it})}}{r^{n+1}e^{(n+1)it}}ire^{it}\,dt. $$ Lets separate this expression and solve $$ \frac{e^{\frac{\lambda}{2}(re^{it} + r^{-1}e^{-it})}}{r^{n+1}e^{(n+1)it}}re^{it} = \frac{e^{\frac{\lambda r}{2}e^{it} \frac{\lambda}{2r}e^{-it}}}{r^ne^{nit}} = \frac{e^{\frac{\lambda r}{2}\left(\cos t + i\sin t\right)}e^{\frac{\lambda}{2r}\left(\cos t - i\sin t\right)}}{r^ne^{nit}}, $$ taking $r=1$ this turns out to be $$ \frac{e^{\frac{\lambda r}{2}\left(\cos t + i\sin t\right)}e^{\frac{\lambda}{2r}\left(\cos t - i\sin t\right)}}{r^ne^{nit}} = e^{\lambda\cos t}e^{-n i t}= e^{\lambda\cos t}(\cos(nt) - i\sin(nt)). $$ Then $$ A_n = \frac{1}{2\pi}\int_{0}^{2\pi} e^{\lambda\cos t}(\cos(nt) - i\sin(nt))\,dt. $$ Because the function $g_n(t) = e^{\lambda\cos t}\sin t$ is odd and $h_n(t) = e^{\lambda\cos t}\cos(nt)$ es even then \begin{align*} &\int_0^{2\pi} e^{\lambda\cos t}\sin(nt)\,dt = 0,\\ &A_n = \frac{1}{2\pi}\int_{0}^{2\pi}e^{\lambda\cos t}\cos(nt)\,dt = \frac{1}{\pi}\int_{0}^{\pi}e^{\lambda\cos t}\cos(nt)\,dt. \end{align*} In addition to this, $\cos t$ is an even function, so$A_n = A_{-n}$, therefore Laurent's expansion of $f$ turns out to be $$ f(z) = \sum_{n=0}^{\infty} A_n(z + z^{-1}) = A_0 + \sum_{n=1}^{\infty} A_n(z + z^{-1}). $$ So far we've worked with $0<|z|<1$, but note that $f(z) = f\left(\frac{1}{z}\right)$, then we obtain what was desired for $0<|z|<\infty$.