MHB Integrating on an infinite domain

[Dustinsfl]

How can I integrate this expression:
\[
\int_0^{\infty} \mathcal{J}_1(kR)e^{-kz}dk = \frac{1}{R} \left[1 - \frac{z^2}{\sqrt{R^2 + z^2}} \right]
\]
where \(\mathcal{J}_1\) is the Bessel function of order 1.

 

[chisigma]

How can I integrate this expression:
\[
\int_0^{\infty} \mathcal{J}_1(kR)e^{-kz}dk = \frac{1}{R} \left[1 - \frac{z^2}{\sqrt{R^2 + z^2}} \right]
\]
where \(\mathcal{J}_1\) is the Bessel function of order 1.

It is remarkable the fact that in the right term the variable z, that exists in the left term, doesn't exists (Dull) ...

Any way... a general formula does exist...

$\displaystyle \mathcal{L} \{ a^{n}\ J_{n} (a\ t)\} = \frac{(\sqrt{s^{2}+ a^{2}} - s)^{n}}{\sqrt{s^{2}-a^{2}}}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[polygamma]

You can evaluate the Laplace transform of the Bessel function of the first kind of positive integer order by using the integral representation

$ \displaystyle J_{n}(bx) = \frac{1}{\pi} \int_{0}^{\pi} \cos(n \theta -bx \sin \theta) \ d \theta = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i(n \theta - b x \sin \theta)} d \theta $$ \displaystyle\int_{0}^{\infty} J_{n}(bx) e^{-ax} \ dx = \displaystyle \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\pi}^{\pi} e^{i(n \theta -bx \sin \theta)} e^{-ax} \ d \theta \ dx$

$ = \displaystyle \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$

$ = \displaystyle\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}{\pi i} \int_{|z|=1} \frac{z^{n}}{bz^{2}+2az-b} \ dz$

$ = \displaystyle \frac{1}{\pi i} \int_{|z|=1} \frac{z^{n}}{b(z-z_{1})(z-z_{2})} \ dz $

where $\displaystyle z_{1} = -\frac{a}{b} + \frac{\sqrt{a^{2}+b^{2}}}{b}$ and $\displaystyle z_{2} = -\frac{a}{b} - \frac{\sqrt{a^{2}+b^{2}}}{b}$Only $z_{1}$ is inside the unit circle.So $ \displaystyle \int_{0}^{\infty} J_{n} (bx) e^{-ax} \ dx = \frac{1}{\pi i} 2 \pi i \ \text{Res} \left[ \frac{z^{n}}{bz^{2}+2az-b}, z_{1} \right]$

$ \displaystyle = \lim_{z \to z_{1}} \frac{z^{n}}{bz+a} = \frac{(\sqrt{a^{2}+b^{2}}-a)^{n}}{b^{n}\sqrt{a^{2}+b^{2}}}$

 

[Dustinsfl]

@RandomVariable,
I actually read your post on math.SX because ChiSigma's has a typo with the minus. His typo had my googling lapace transform of the Bessel Eq and I found yours on SX prior to your post here.