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I want to show that $$ \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx = \frac{(\sqrt{a^{2}+b^{2}}-a)^{n}}{b^{n}\sqrt{a^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(a) >0 , \, b >0 ),$$ where $J_{n}(x)$ is the Bessel function of the first kind of order $n$.

But the result I get using an integral representation of $J_{n}(bx)$ is off by a factor of $ \displaystyle \frac{1}{b}$, and I don't understand why.

$$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx &= \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\pi}^{\pi} e^{i(n \theta -bx \sin \theta)} e^{-ax} \, d \theta \, dx \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \int_{0}^{\infty} e^{i n \theta} e^{-(a+ib \sin \theta)x} \, dx \, d \theta \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta}}{a + ib \sin \theta} \, d \theta \\ &= \frac{1}{2 \pi} \int_{|z|=1} \frac{z^{n}}{a+\frac{b}{2} \left(z-\frac{1}{z} \right)} \frac{dz} {iz} \\ &= \frac{1}{i\pi} \int_{|z|=1} \frac{z^{n}}{bz^{2}+2az-b} \, dz \end{align}$$

The integrand has simple poles at $\displaystyle z= -\frac{a}{b} \pm \frac{\sqrt{a^{2}+b^{2}}}{b}$.

But only the pole at $\displaystyle z= -\frac{a}{b} + \frac{\sqrt{a^{2}+b^{2}}}{b}$ is inside the unit circle.

Therefore,

$$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx &= \frac{1}{i \pi} \, 2 \pi i \ \text{Res} \left[ \frac{z^{n}}{bz^{2}+2az-b}, -\frac{a}{b} + \frac{\sqrt{a^{2}+b^{2}}}{b} \right] \\ &= {\color{red}{b}} \ \frac{(\sqrt{a^{2}+b^{2}}-a)^{n}}{b^{n}\sqrt{a^{2}+b^{2}}} . \end{align}$$

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Everything is correct up until your computation of the residue. Write $$bz^2+2az-b=b(z-z_+)(z-z_-)$$ where $$z_\pm=-\frac{a}{b}\pm\frac{\sqrt{a^2+b^2}}{b}$$ as you have determined. Now, $${\rm Res}\Bigg(\frac{z^n}{b(z-z_+)(z-z_-)};\quad z=z_+\Bigg)=\lim_{z\to z_+} (z-z_+)\frac{z^n}{b(z-z_+)(z-z_-)}$$ and here you get the desired factor of $1/b$.

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  • $\begingroup$ I had mistakenly written down that $bz^{2}+2az-b=(z-z_{+})(z-z_{-})$ $\endgroup$ – Random Variable Mar 10 '13 at 21:00

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