Hint for $\int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{\left(\sqrt{a^2+b^2}-a\right)^v}{b^v\sqrt{a^2+b^2}}$

Question

I'm looking for a hint of how to show that: \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{\left(\sqrt{a^2+b^2}-a\right)^v}{b^v\sqrt{a^2+b^2}},\qquad \Re v>-1,\quad a>0,\quad b>0, \end{align*} where $J_v$ is the Bessel function of the first kind of order $v$.

Two of my approaches, which didn't lead me to the result, are:

• (The motivation for this approach is this) Use the integral representation: \begin{align*} J_v(z)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i\left(v\theta-z\sin\theta\right)}d\theta-\frac{\sin v\pi}{\pi}\int_0^\infty e^{-z\sinh t-vt}dt,\qquad \Re z>0, \end{align*} apply Fubini's Theorem, and after a substitution I arrived at: \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{1}{\pi i}\oint_{|z|=1}\frac{z^v}{bz^2+2az-b}dz-\frac{2\sin v\pi}{\pi}\int_1^\infty\frac{u^{-v}}{bu^2+2au-b}du, \end{align*} which I couldn't show that has de desired closed-form.
• Use the integral representation: \begin{align*} J_v\left(z\right)=\frac{1}{2\pi i}\int_{C} e^{z(t-t^{-1})/2}t^{-v-1}dt,\qquad |\arg z|<\frac{\pi}{2}, \end{align*} where $C$ is a path from $(-\infty,\delta)$, around the origin counterclockwise and back to $(-\infty,-\delta)$; that is, a reflection of a Hankel contour about the line $\Re z=0$. Using this integral representation I arrived at \begin{align*} \int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{-1}{\pi i}\oint_D\frac{z^v}{bz^2+2az-b}du, \end{align*} where $D$ is the inversion of $C$ over the circle $|z|=1$; that is, the path resulting from $C$ under the mapping $z\to\frac{1}{z}$. I couldn't evaluate this last integral either.

Answer 1

To evaluate \begin{equation} I=\int_0^\infty e^{-ax}J_v\left(bx\right)\,dx \end{equation} we can use the Schläfli-Sommerfeld Integral (which is equivalent to the representation used in the first try of the OP) \begin{equation} J_{\nu}\left(z\right)=\frac{1}{2\pi i}\int_{\infty-\pi i}^{\infty+\pi i}e^{z\sinh t-\nu t}\,dt \end{equation} The contour consist of three sides of a rectangle with vertices at $\infty-i\pi,-i\pi,i\pi,\infty+i\pi$. By changing the order of integration, \begin{align} I&=\frac{1}{2\pi i}\int_{\infty-\pi i}^{\infty+\pi i}e^{-\nu t}\,dt\int_0^\infty e^{-x(a-b\sinh t)}\,dx\\ &=\frac{1}{2\pi i}\int_{\infty-\pi i}^{\infty+\pi i}\frac{e^{-\nu t}}{a-b\sinh t}\,dt \end{align} The opposite of this integral can be evaluated by the residue theorem, considering the contour composed of staight lines: from $\infty+i\pi$ to $i\pi$, then to $-i\pi$ and finally returning to infinity at $\infty-i\pi$. The only pole in the contour is single and situated at $\sinh^{-1}\frac{a}{b}$, with a residue \begin{equation} -\frac{e^{-\nu \sinh^{-1}\frac{a}{b}}}{b\cosh\left(\sinh^{-1}\frac{a}{b}\right)} \end{equation} From the logarithmic representation of $\sinh^{-1}$, we can express \begin{align} I&=\frac1b\frac{1}{\sqrt{1+\frac{a^2}{b^2}}}\frac{1}{\left( \sqrt{1+\frac{a^2}{b^2 }}+\frac{a}{b}\right)^\nu}\\ &=\frac{ \left(\sqrt{a^2+b^2}-a)\right)^\nu}{b^\nu\sqrt{a^2+b^2}} \end{align} as expected.