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Let $\lambda\in\mathbb{C}$. I want to get the laurent series of $f(z)=\exp({\frac{\lambda}{2}(z+z^{-1}))}$ on $C(0;0,\infty):=\mathbb{C}\setminus\{0\}$ as $$a_0+\sum_{n=1}^{\infty}a_n(z^n+z^{-n})$$ with $$a_n=1/\pi\int_0^\pi e^{\lambda \cos(t)}\cos(nt)dt.\qquad (1)$$

I know that the $a_n$ are computable with $$a_n=\frac{1}{2\pi i}\int_{|w|=\delta}\frac{f(w)}{w^{n+1}}dw$$ for every $0<\delta<\infty$. How can I get to the form $(1)$?

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  • $\begingroup$ As a Laurent series in $z$, this function generates the modified Bessel functions of the first kind $I_k(\lambda)$. See for instance equation (1) here. $\endgroup$ Jun 6 '16 at 16:28
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$$\frac{1}{2\pi i}\int_{|w|=\delta}\frac{f(w)}{w^{n+1}}dw=\frac{1}{2\pi i}\int_{|w|=\delta}\frac{\exp({\frac{\lambda}{2}(w+w^{-1}))}}{w^{n+1}}dw=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{\exp({\frac{\lambda}{2}(\delta^{-1} e^{-i\theta}+\delta e^{i\theta}))}}{e^{i\theta(n+1)}}\delta ie^{i\theta}d\theta$$

We put $\delta = 1$ : $$\frac{1}{2\pi }\int_{0}^{2\pi}e^{\lambda \cos(\theta)}e^{-i\theta n}d\theta=\frac{1}{2\pi }\int_{0}^{2\pi}e^{\lambda \cos(\theta)}\cos(-n\theta) d\theta+\frac{1}{2\pi }\int_{0}^{2\pi}ie^{\lambda \cos(\theta)}\sin(-n\theta) d\theta$$.

You can show that the second integral is zero (hint : use the symetrie of the fuction you want to integrate). Finally : $$\frac{1}{2\pi }\int_{0}^{2\pi}e^{\lambda \cos(\theta)}e^{-i\theta n}d\theta=\frac{1}{2\pi }\int_{0}^{2\pi}e^{\lambda \cos(\theta)}\cos(-n\theta) d\theta=\frac{1}{\pi}\int_0^\pi e^{\lambda \cos(t)}\cos(nt)dt$$

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  • $\begingroup$ Thank you! Just a question: Why haven't you the factor $(e^{i\theta})'=\theta e^{i\theta}$ after the second '='? $\endgroup$
    – user337060
    Jun 6 '16 at 16:26
  • $\begingroup$ oh my bad you are right I will fix it :) $\endgroup$
    – Bérénice
    Jun 6 '16 at 16:30
  • $\begingroup$ :) But now you have only added $e^{i\theta}$ instead of $te^{i\theta}$.. Is that right? $\endgroup$
    – user337060
    Jun 6 '16 at 16:40
  • $\begingroup$ iam really sorry for the mistakes... it is fixed $\endgroup$
    – Bérénice
    Jun 6 '16 at 16:45
  • $\begingroup$ No problem! So you are sure I have read your answer :P $\endgroup$
    – user337060
    Jun 6 '16 at 16:59

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