Prove: $\int\limits_0^\infty\mathrm J_n(bx)x^ne^{-ax}\,\mathrm dx=\frac{(2b)^n\cdot \Gamma(n+1/2)}{\sqrt\pi(a^2+b^2)^{n+1/2}}$ I'm trying to prove the following but I'm stuck. Can someone help?

$$\int\limits_0^\infty\mathrm J_n(bx)x^ne^{-ax}\,\mathrm dx=\frac{(2b)^n\cdot \Gamma(n+1/2)}{\sqrt\pi(a^2+b^2)^{n+1/2}}$$

My attempt:
By definition of Bessel function, we have,
$$\mathrm J_n(bx)=\sum_{r=0}^\infty (-1)^r\frac{(bx/2)^{2r+n}}{r!\cdot\Gamma(n+r+1)}\\ \implies \mathrm J_n(bx)x^ne^{-ax}=\frac{b^n}{2^n}\sum_{r=0}^\infty (-1)^r\frac{b^{2r}}{2^{2r}\cdot r!\cdot\Gamma(n+r+1)}x^{2r+2n}e^{-ax}\\ \implies \int\limits_0^\infty\mathrm J_n(bx)x^ne^{-ax}\,\mathrm dx=\frac{b^n}{2^n}\sum_{r=0}^\infty (-1)^r\frac{b^{2r}}{2^{2r}\cdot r!\cdot\Gamma(n+r+1)}\int\limits_0^\infty x^{2r+2n}e^{-ax}\,\mathrm dx$$
Now, the integral on the RHS of the above equation is the Laplace transform of $x^{2r+2n}$ with parameter $a$ which is $\dfrac{\Gamma(2r+2n+1)}{a^{2r+2n+1}}$, so we have,
$$\int\limits_0^\infty\mathrm J_n(bx)x^ne^{-ax}\,\mathrm dx=\frac{b^n}{2^n\cdot a^{2n+1}}\sum_{r=0}^\infty (-1)^r\frac{b^{2r}\cdot\Gamma(2r+2n+1)}{2^{2r}\cdot r!\cdot\Gamma(n+r+1)\cdot a^{2r}}$$
After this, I'm stuck.
 A: First off: your calculations are indeed correct. You only need to be patient and carefully rewrite your result. For that we'll start from the right hand side.
As I suggested in the comment, writing$$\frac{1}{(a^2+b^2)^{n+1/2}}=\frac{1}{a^{2n+1}}\cdot \frac{1}{(1+(b/a)^2)^{n+1/2}}.$$
Using the binomial theorem on this last expression gives:
$$\frac{1}{(a^2+b^2)^{n+1/2}}=\frac{1}{a^{2n+1}}\cdot \sum_{r=0}^\infty(-1)^r{n-1/2-r \choose n-1/2}(b/a)^{2r}.$$
Thus, comparing coefficients, we have to show that
$${n-1/2+r \choose r}\frac{(2b)^n\Gamma(n+1/2)}{\sqrt{\pi}}=\frac{b^n}{2^n}\frac{\Gamma(2r+2n+1)}{2^{2r}\cdot r!\cdot\Gamma(n+r+1)} \tag{1}$$
Now let's use the formula:
$$\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\ \Gamma(z)\Gamma(z+1/2)$$
to rewrite
$$\Gamma(2r+2n+1)=(2r+2n)\Gamma(2r+2n)=\frac{2^{2(n+r)-1}}{\sqrt{\pi}}\ \Gamma(n+r)\Gamma(n+r+1/2)$$
so that 
$$\frac{\Gamma(2r+2n+1)}{\Gamma(n+r+1)}=\frac{2^{2(n+r)}}{\sqrt{\pi}}\ \Gamma(n+r+1/2)$$
Thus the whole right hand side equals
$$\frac{(2b)^n}{r!\sqrt{\pi}}\ \Gamma(n+r+1/2).$$
Plugging this back in $(1)$ and clearing terms that occur on both sides, reduces our task to showing  that
$${n-1/2+r \choose r}{\Gamma(n+1/2)}=\frac{1}{r!}\ \Gamma(n+r+1/2),$$
but this is indeed true as by the very definition of the binomial coefficient, the left hand side equals
$${n-1/2+r \choose r}{\Gamma(n+1/2)}=\frac{(n-1/2+r)\cdots(n+1/2)\Gamma(n+1/2)}{r!}=\frac{1}{r!}\ \Gamma(n+r+1/2)$$
