Prove $\int^a_0 x(a^2-x^2)^{\nu-1} I_0(b x)dx= 2^{\nu-1}a^\nu b^{-\nu}\Gamma(\nu)I_\nu(a b)$ How can I prove the following equality?

$$ \int^a_0 x\left(a^2-x^2\right)^{\nu-1} I_0\left(b x\right)dx=
2^{\nu-1}a^\nu b^{-\nu}\Gamma\left(\nu\right)I_\nu\left(a b\right),
$$
  under the condition $a>0$, $b>0$ $\nu>0$, where $I_\nu(\cdot)$ denotes the modified Bessel function of the first kind of order $\nu$.

 A: Easiest way to show this is by series expansion.  First, rescale the integral to get
$$a^{\nu+1} \int_0^1 du \, u (1-u^2)^{v-1} I_0(a b u) $$
Expand the Bessel into a series:
$$a^{\nu+1} \sum_{k=0}^{\infty} \frac{(a b/2)^{2 k}}{k!^2} \int_0^1 du \, u^{2 k+1} (1-u^2)^{\nu-1} $$
The integral is a beta function:
$$\int_0^1 du \, u^{2 k+1} (1-u^2)^{\nu-1} = \frac12 \int_0^1 dv \, v^k (1-v)^{\nu-1} = \frac12 \frac{\Gamma(k+1) \Gamma(\nu)}{\Gamma(k+\nu+1)} $$
The integral is now a sum:
$$\frac12 a^{\nu+1} \sum_{k=0}^{\infty} \frac{(a b/2)^{2 k}}{k!^2} \frac{\Gamma(k+1) \Gamma(\nu)}{\Gamma(k+\nu+1)} = \frac12 \Gamma(\nu) a^{\nu+1} \sum_{k=0}^{\infty} \frac{(a b/2)^{2 k}}{k! \Gamma(k+\nu+1)}  $$
We are almost there.  Now let's multiply and divide by a factor that will bring us our new Bessel:
$$\frac12 \Gamma(\nu) a^{\nu+1} \left (\frac{2}{a b}\right )^{\nu} \sum_{k=0}^{\infty} \frac{(a b/2)^{2 k+\nu}}{k! \Gamma(k+\nu+1)} = 2^{\nu-1} \Gamma(\nu)\frac{a}{b^{\nu}} I_{\nu}(a b)$$
which is the value of the integral.
REMARK
The OP's value for the integral is not correct.  One can tell by simple dimensional analysis.  For instance, if $a$ is in unit of length, then $b$ must be in units of inverse length.  The integral itself is in units of length to the $\nu+1$ power.  The OP's result, however, has units of length to the $2 \nu$ power.
