# Using residue theorem to calculate following integral

I'm trying to evaluate the following integral using the residue theorem

\begin{align}\label{eq:int_1} S(z) = \dfrac{1}{2\pi}\int_{0}^{2\pi} \dfrac{e^{i\phi}+z}{e^{i\phi}-z} e^{-\lambda\sin^{2}(\phi/2)} \mathrm d\phi \end{align} where $$\lambda$$ is a real positive parameter. Now here's my attempt which I'm pretty sure is incorrect. We can transform the above real integral into a contour integral over the unit circle using the substitution $$w = e^{i\phi}$$. First note that - \begin{align} \exp(-\lambda\sin^{2}(\phi/2)) &= \exp(-\lambda/2)\exp((\lambda/4)(e^{i\phi} + e^{-i\phi})) \\ &= \exp(-\lambda/2)\exp((\lambda/4)(w + 1/w)) \end{align}

Therefore the above integral now becomes -

$$S(z) = \dfrac{\exp(-\lambda/2)}{2\pi i}\oint_{\mathcal{C}} \dfrac{w+z}{w-z} \exp\bigg(\dfrac{\lambda}{4}\bigg(w + \dfrac{1}{w}\bigg)\bigg)\dfrac{\mathrm dw}{w}$$

The integrand has singularities at $$w = z$$ and $$w = 0$$ (I'm interested in the case when $$|z|<1$$) so we can evaluate the residues at both of these singularities and then employ the residue theorem. The residue at $$w = z$$ is found to be

$$2 \exp\bigg(\dfrac{\lambda}{4}\bigg(z + \dfrac{1}{z}\bigg)\bigg)$$

whereas to evaluate the residues at $$w = 0$$, we need to expand the exponential as a power series and keep only all those terms which possess a simple pole at $$w=0$$. Since

$$\exp\bigg(\dfrac{\lambda}{4}\bigg(w + \dfrac{1}{w}\bigg)\bigg)\dfrac{1}{w} = \sum_{n=0}^{\infty} \dfrac{1}{n!}\bigg(\dfrac{\lambda}{4}\bigg)^{n}\dfrac{(w^2+1)^{n}}{w^{n+1}}$$ and from the binomial expansion we know that

$$(w^2+1)^{n} = \sum_{k=0}^{n} {n\choose k } w^{2k}$$

It's clear that only the terms with even n yield a simple pole since we must have $$n=2k$$. And I think finally the residue evaluates to a modified Bessel function of the form $$- I_{0}(\lambda/2)$$ where the negative sign comes about from the $$(w+z)/(w-z)$$ part. And so we obtain

$$S(z) = \exp(-\lambda/2)\bigg( 2 \exp\bigg(\dfrac{\lambda}{4}\bigg(z + \dfrac{1}{z}\bigg)\bigg) - I_{0}(\lambda/2) \bigg)$$

But something seems to have gone wrong here since the above expression has a singularity at $$z = 0$$ whereas the original expression for the function is clearly finite at $$z=0$$. I'm really unsure where I've made the mistake. Any help is hugely appreciated. Thanks !!

• Let $\lambda' = \lambda/2$. You're multiplying $$e^{\lambda' (w + 1/w)/2} = \sum_{k \in \mathbb Z} I_k(\lambda') w^k$$ by $$\frac {w + z} {w (w - z)} = -\frac 1 w - \frac 2 z \sum_{k \geq 0} \left( \frac w z \right)^k.$$ All terms from the second series contribute to the value of the residue. $S(z)$ is equal to $$2 e^{\lambda' (1 - z)^2/(2 z)} - e^{-\lambda'} I_0(\lambda') - 2 e^{-\lambda'} \sum_{k \geq 1} I_k(\lambda') z^{-k}.$$ – Maxim Jun 11 '19 at 14:53