I have some trouble computing $\int_0^{x}e^{-u}I_n(\alpha u)du$, where $I_n(u)$ is the modified Bessel function of the first kind, $n$ is an integer and $0<\alpha<1$ and $x>0$ are real numbers.
Here is how I start (I use the integral representation of this modified Bessel function for an integer $n$):
\begin{align} \int_0^{x}e^{-u}I_n(au)du&=\int_0^{x}\frac{1}{\pi}\int_0^\pi e^{-u}\cos(n\theta)e^{u\alpha \cos(\theta)} d\theta du\\ &=\frac{1}{\pi}\int_0^\pi\int_0^{x} e^{u(\alpha \cos(\theta)-1)} \cos(nu)d\theta du\\ &=\frac{1}{\pi}\int_0^\pi\frac{e^{x(\alpha \cos(\theta)-1)}-1}{\alpha \cos(\theta)-1} \cos(n\theta)d\theta \\ &=\frac{1}{2\pi}\int_0^\pi\frac{e^{x(\alpha \cos(\theta)-1)}-1}{\alpha \cos(\theta)-1} (e^{in\theta}+e^{-in\theta})d\theta \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi\frac{e^{x(\alpha \cos(\theta)-1)}-1}{\alpha \cos(\theta)-1}e^{in\theta}d\theta \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi\frac{e^{x(\alpha \cos(\theta)-1)}}{\alpha \cos(\theta)-1}e^{in\theta}d\theta - \frac{1}{2\pi}\int_{-\pi}^\pi\frac{1}{\alpha \cos(\theta)-1}e^{in\theta}d\theta \end{align}
Now we use a complex contour and we compute the residue. We will just compute the first integral since the second one is the same with $x=0$:
\begin{align} \frac{1}{2\pi}\int_{-\pi}^\pi\frac{e^{x(\alpha \cos(\theta)-1)}}{\alpha \cos(\theta)-1}e^{in\theta}d\theta &= \frac{1}{2\pi}\int_{|z|=1}\frac{e^{x(\frac{\alpha}{2} (z+z^{-1})-1)}}{\frac{\alpha}{2} (z+z^{-1})-1}z^n\frac{dz}{iz} \\ &= \frac{1}{i2\pi}\int_{|z|=1}\frac{e^{x(\frac{\alpha}{2} (z+z^{-1})-1)}}{\frac{\alpha}{2} (z²+1)-z}z^ndz \\ &= \frac{1}{i2\pi}\int_{|z|=1}\frac{e^{x(\frac{\alpha}{2} (z+z^{-1})-1)}}{\frac{\alpha}{2}(z-z_0^+)(z-z_0^⁻)}z^ndz \end{align}
where the poles are $z_0^{\pm}=\frac{1\pm\sqrt{1-\alpha²}}{\alpha}$.
$z_0^{-}$ is the only one which contributes to the integral since it is enclosed by the contour (the unit circle). Then:
\begin{align} \frac{1}{2\pi}\int_{-\pi}^\pi\frac{e^{x(\alpha \cos(\theta)-1)}}{\alpha \cos(\theta)-1}e^{in\theta}d\theta&= \lim_{z\rightarrow z_0^{-}}\frac{e^{x(\frac{\alpha}{2} (z+z^{-1})-1)}}{\frac{\alpha}{2}(z-z_0^+)}z^n\\ &= \frac{e^{x(\frac{\alpha}{2} (z_0^{-}+(z_0^{-})^{-1})-1)}}{\frac{\alpha}{2}(z_0^{-}-z_0^+)}(z_0^{-})^n \end{align}
Great,...but this is false. Indeed: $$\frac{\alpha}{2} (z_0^{-}+(z_0^{-})^{-1})=\frac{1}{2}(\frac{1-\sqrt{1-\alpha²}}{1}+\frac{\alpha^2}{1-\sqrt{1-\alpha²}})=\frac{1}{2}(\frac{1-2\sqrt{1-\alpha²}+1-\alpha²+\alpha²}{1-\sqrt{1-\alpha²}})=1$$
$$\Rightarrow e^{x(\frac{\alpha}{2} (z_0^{-}+(z_0^{-})^{-1})-1)}=1$$
Then, the result is independent of $x$, which is not possible. Have you seen where is the error?
One error is in finding the residues: $e^{1/z}$ has an essential singularity at $z=0$ which needs to be included when calculating the integral. I would not expect this integral to have a closed form anyway; even the integral of $I_n$ by itself involves a ${}_1F_2$ hypergeometric function.