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[SOLVED] Integrating on an infinite domain

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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}}}$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
@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.