I need to calculate the following integral. $$\int _{\frac{-2 \pi}{L}}^{\frac{2\pi}{L}} \left( \sum _{n=0}^{\infty} \frac{\left(j \chi \rho \cos\left(\xi - \theta\right) \right)^n}{n!}\right) d\xi$$
where $j$ is the imaginary unit, $\chi \geq 0, \chi \in \Bbb R$, $\rho \in \Bbb R$, $\theta \in \Bbb R$ and $L>0 \in \Bbb N$.
Is there a theorem as of how to move the infinite series out of the integral? Or a solution for this kind of integral?
For the Riemann integral, we know that if $\sum _{n=0}^{\infty} \frac{\left(j \chi \rho cos\left(\xi - \theta\right) \right)^n}{n!}$ converges uniformly for $\xi \in [\frac{-2 \pi}{L},\frac{2\pi}{L}]$ as $n \to \infty$, then we can move the sum outside the integral. Try the Weierstrass M-test to prove uniform convergence.
The Lebesgue theory of integration provides additional criteria that allow moving the sum outside the integral. See, for example, the dominated convergence theorem .
added
Note that $$ \left|\frac{\big(j\chi\rho\cos(\xi-\theta)\big)^n}{n!}\right| \le\frac{\big(\chi|\rho|\big)^n}{n!} , $$ and the series $$ \sum_{n=1}^\infty \frac{\big(\chi|\rho|\big)^n}{n!} $$ converges [to $e^{\chi|\rho|}$]. Therefore, by the Weierstrass $M$-test, $$ \sum_{n=0}^\infty\frac{\big(j\chi\rho\cos(\xi-\theta)\big)^n}{n!} $$ converges uniformly for $\xi \in \mathbb R$.