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.

 

[blue_raver22]

Integrating on an infinite domain can be challenging, but with the use of mathematical tools such as the Bessel function, we can find solutions to complex integrals. In this case, we have an integral involving the Bessel function of order 1 and an exponential function. By using the properties of the Bessel function and some algebraic manipulation, we can simplify the integral to a closed-form expression. This shows the power of mathematical techniques in solving integrals on infinite domains. Furthermore, the result of the integration provides insight into the behavior of the function as the integration limits approach infinity. We can see that the expression converges to a finite value, which can be useful in various applications, such as in physics and engineering. Overall, by integrating this expression, we gain a better understanding of the function and its behavior on an infinite domain.