I haven't real knowledge in Bessel's function and I'd like to know how to switch integral and sum in these two equations. I've already tried a lot of ideas but nothing really works. The first one is :
$$ h(r, \theta, t) = \int_{\lambda = 0}^{+ \infty}\sum_{n \geq 1} e^{- \lambda^2 t/2} \lambda \sin\big(\frac{n \pi \theta}{\beta}\big) \sin\big(\frac{n \pi \theta_0}{\beta}\big) \mathbf{J}_{\frac{n\pi}{\beta}}(\lambda r_0) \mathbf{J}_{\frac{n \pi}{\beta}}(\lambda r) \text{ d}\lambda $$ where $\mathbf{J}_{\frac{n \pi}{\beta}}$ is the Bessel's function of the first kind of order $\frac{n \pi}{\beta}$ and $\beta$, $\theta_0$ and $r_0$ are constants.
I've tried to apply the Fubini/Tonelli theorem's but I can't easily show up (and even know if it's true or not) that: $$ \sum_{n \geq 1} |\mathbf{J}_{\frac{n \pi}{\beta}} (\lambda r_0) \mathbf{J}_{\frac{n \pi}{\beta}} (\lambda r) | < \infty. $$
The other equation is, I think, harder :
$$F(t)= \int_{r=0}^{+\infty} \int_{\theta=0}^{\beta} \frac{2r}{\beta t} \sum_{n\geq 1} \sin\big( \frac{n \pi \theta_0}{\beta} \big) \sin\big(\frac{n \pi \theta}{\beta} \big) \exp\big( - \frac{r_0^2 + r^2}{2t} \big) \mathbf{I}_{\frac{n \pi}{\beta}} \big( \frac{r r_0}{t} \big) \text{ d}r \text{ d}\theta $$ where $\mathbf{I}_{\frac{n \pi}{\beta}}$ is the modified Bessel's function of the first kind.
I thank you in advance,
Hiz.
EDIT : Ok, I've found my answer. We've to use these two inegalities : $$ \mathbf{J}_{\nu}(x) \leq x^{\nu} \frac{1}{\Gamma(\nu + 1)} \\ \mathbf{I}_{\nu}(x) \leq \frac{x^{\nu}}{2^{\nu} \Gamma(\nu +1)} e^x $$ for $x \in \mathbb{R}$.With the Stirling formula and the Alembert rule we can conclude that the sums converge. To finish this proof,I just need to calculate explicitly these two sum :
$$ \sum_n \Big(\frac{\lambda r}{2}\Big)^{\frac{n\pi}{\beta}}\frac{1}{\Gamma\big(\frac{n \pi}{\beta} + 1\big)} \\ \sum_n \Big(\frac{rr_0}{2t}\Big)^{\frac{n \pi}{\beta}}\frac{1}{\Gamma\big( \frac{n \pi}{\beta} + 1\big)}. $$