I'm trying to understand the derivation of:
$$ I_0(x) = \frac{1}{\pi}\int_{0}^{\pi} \exp(x\cos\theta) \, d\theta$$
I'm trying to use this generating function:
$$ \exp\left(\frac{x}{2}(z+z^{-1})\right) = \exp(x\cos\theta) = \sum_{n=-\infty}^\infty I_n(x) \exp (in\theta)$$
Is this correct? (left side real and right side complex). Then, i'm using:
$$ \int_0^\pi \exp(x\cos\theta) \,d\theta = \frac 1 \pi \int_0^\pi \sum_{n=-\infty}^\infty I_n(x) \exp (in\theta) \, d\theta$$
But this seems not quite right.
Thanks!
Let us put few things in their proper place. I believe we want to prove that $$ I_0(x)=\frac{1}{\pi}\int_{0}^{\pi}\exp(x\cos\theta)\,d\theta \tag{1}$$ where the modified Bessel functions of the first kind are defined as the coefficients of a Laurent series, $$ \exp\left[\frac{x}{2}\left(z\color{red}{+}\frac{1}{z}\right)\right] = \sum_{n\in\mathbb{Z}} I_n(x) z^n . \tag{2}$$ It does not make sense to ask if your generating function is correct, since no relation between $z,\theta$ and $\varphi$ is given.
By the residue theorem we have that $(2)$ implies $$ I_n(x)=\operatorname*{Res}_{z=0}\frac{1}{z^{n+1}}\exp\left[\frac{x}{2}\left(z+\frac{1}{z}\right)\right] = \frac{1}{2\pi i}\oint_{|z|=1}\frac{1}{z^{n+1}}\exp\left[\frac{x}{2}\left(z+\frac{1}{z}\right)\right]\,dz.\tag{3}$$ Now we may enforce the substitution $z=e^{i\theta}$, turning the RHS of $(3)$ into $$ \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ni\theta}\exp(x\cos\theta)\,d\theta.\tag{4} $$ If $n=0$ this leads to $$ I_0(x) = \frac{1}{2\pi}\int_{0}^{2\pi}\exp(x\cos\theta)\,d\theta\stackrel{\text{symmetry}}{=}\frac{1}{\pi}\int_{0}^{\pi}\exp(x\cos\theta)\,d\theta \tag{5} $$ as was to be shown. In equivalent terms, if we know that $I_n(x)$ (for $n\geq 0$) are the coefficients of the Fourier cosine series of $\exp(x\cos\theta)$, then $I_0(x)$ clearly has to be the average value of the function $\exp(x\cos\theta)$ over the interval $(0,2\pi)$.