[SOLVED] Integrating on an infinite domain

[dwsmith]

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 ...

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

$\chi$ $\sigma$

 

[Random Variable]

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}}}$

 

[dwsmith]

@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.