I have the following definite integral:
$$ \int_{0}^{\pi} cos(n\theta)[I_{1}(acos\theta) + cos\theta I_{0}(acos\theta)]d\theta $$
with $I_{1}$ and $I_{0}$ the first and zero-order modified bessel functions of the first kind respectively ; $n \in \mathbb{N}^{*}$ and $a \in \mathbb{R}^{+*}$.
with the hint that,
$$ I_{n}(z) = \frac{1}{\pi} \int_{0}^{\pi} e^{zcos\theta}cos(n\theta)d\theta $$
we can transform the previous integral,
$$ \int_{0}^{\pi} cos(\theta)I_{n}(acos\theta) + \frac{1}{2}(I_{n+1}(acos\theta) + I_{n-1}(acos\theta))d\theta $$ $$ \int_{0}^{\pi} cos(\theta)I_{n}(acos\theta) - \frac{1}{asin\theta}(I_{n}^{'}(acos\theta))d\theta $$
I couldn't go any further in the computation. But using the power series representation and considering $n$ odd and $n=2m+1$, we obtain:
$$ \pi \sum_{k=0}^{\infty} \frac{U^{2k+n-1}}{2^{2k+n}\Gamma (k+n+1) k!} \prod_{i=1}^{k+m} (\frac{2i-1}{2i}) (U\frac{2(k+m)+1}{2(k+m+1)} + 2k+n) $$
If someone has an idea ... Thanks!
Take the integral
$$ \int_{0}^{\pi} cos(\theta) I_{n}(acos(\theta)) + \frac{1}{2}(I_{n+1}(acos(\theta)) + I_{n-1}(acos(\theta))) d\theta $$
and apply the solution of the integral here, p724, 6.681-3:
$$ \int_{0}^{\frac{\pi}{2}} cos(2\mu x)I_{2\nu}(2acosx)dx = \frac{\pi}{2}I_{\nu-\mu}(a)I_{\nu+\mu}(a) $$
This is also valid for the sinus by symmetry in our case. Stating $n=2m+1$ we obtain,
$$ \pi I_{m}(\frac{a}{2})I_{m+1}(\frac{a}{2}) + \frac{\pi}{2}I_{m+1}(\frac{a}{2})^{2} + \frac{\pi}{2}I_{m}(\frac{a}{2})^{2} $$