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.

  • $\begingroup$ I haven't run the whole calculation because I am too lazy. But basically all you have to do is to start from the given formula, and use the binomial theorem to reformulate $$\frac{1}{(a^2+b^2)^{n+1/2}}=\frac{1}{a^{2n+1}}\cdot \frac{1}{(1+(b/a)^2)^{n+1/2}},$$ then some manipulations will give you back the formula you obtained. Note that $\sqrt{\pi}=\Gamma(\frac{1}{2})$ $\endgroup$
    – b00n heT
    Sep 4, 2016 at 20:36

1 Answer 1


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 $${n-1/2+r \choose r}{\Gamma(n+1/2)}=\frac{(n-1/2+r)\cdots(n+1/2)\Gamma(n+1/2)}{r!}=\frac{1}{r!}\ \Gamma(n+r+1/2)$$


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