0

I want to calculate the inverse Laplace transform of Bessel function: $$J_{as}(x)=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m+as}}{m!Γ(m+as+1)}=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m}(\frac{x}{2})^{as}}{m!Γ(m+as+1)}=\sum_{m=0}^\infty\frac{(-1)^{m}(\frac{x}{2})^{2m}e^{asln(\frac{x}{2})}Γ(-m-as)sin\pi(1+m+as)}{m!\pi}$$ if I know Laplace's reversal of the multiplication of the exponential function and the sin function, I also know that Laplace's reversal of the gamma function, thus I can solve with convolution theorem my problem .your question this is, what is the inverse Laplace transform of gamma function that is raised in question?

Dana
  • 29
  • 6
  • Laplace's inverse of the sinus function alone does not exist. – Dana Nov 27 '18 at 22:48
  • Why do you think so ? What about $e^{cs}$ and $e^{s^2/n^2} e^{cs}$ ? – reuns Nov 27 '18 at 22:49
  • I have to calculate the features you mentioned. – Dana Nov 27 '18 at 22:58
  • But my question this is:, if I use the reflection formula, the gamma function that was raised in the question, what is Laplace's inverse? – Dana Nov 27 '18 at 23:00
  • The idea is that $\sin(\pi s)$ grows way too much to be integrable with $e^{st}$ in any way. But, we can regularize it, looking at $f_n= \mathcal{L}^{-1}[\sin(\pi s)e^{s^2/n^2}]$ which is perfectly defined, then let $n \to \infty$. The objet that $f_n$ converges to is called an analytic functional, it is even more complicated than distributions (such as the Dirac delta). – reuns Nov 27 '18 at 23:01
  • I want to know that what is the inverse Laplace of $Γ(-a-s)$that is raised of reflection formula? – Dana Nov 27 '18 at 23:08
  • In fact, I want to obtain the inverse Laplace of the Bessel function, which includes a gamma function in the fraction denominator and a power function in the form of a fraction. The power function in the form of a fraction can be written as an exponential function of copper. Now my problem is the gamma function in the denominator. – Dana Nov 27 '18 at 23:17
  • Do you understand why my comment answers negatively ? Did you look at $f_n= \mathcal{L}^{-1}[\sin(\pi s)e^{s^2/n^2}]$ ? Use the change of variable $u =\sin \tau$ (or $\sinh \tau$) to put the integral representations of the Bessel functions in the form of Fourier-Laplace transforms. – reuns Nov 28 '18 at 00:34
  • I don't understand your purpose. the inverse Laplace transform from the Sin function that was raised of gamma function in denominator, with the power function in the numerator , is solvable. My problem, is gamma function in numerator . – Dana Nov 28 '18 at 06:46
  • I edit my question. – Dana Nov 28 '18 at 06:47

1 Answers1

1

$$J_\alpha(x) = \frac{1}{\pi} \int_0^\pi \cos(\alpha\tau - x \sin\tau)\,d\tau - \frac{\sin \alpha\pi}{\pi} \int_0^\infty e^{-x \sinh t - \alpha t} \, dt$$

Apply the change of variables $u = \sin \tau, v = \sinh \tau$ to obtain that $J_\alpha(x)$ is a Laplace transform, whose inverse Laplace transform is thus obvious by the inversion theorem.

As $e^{i \pi s}$ and $\sin(\pi s)$, $\frac{1}{\Gamma(s)}$ grows way too fast on vertical lines to be the Laplace transform of a function.

Its regularized inverse Laplace transform is an analytic functional (which tell us an expression for $\mathcal{L}^{-1}[\frac{F(s)}{\Gamma(s)}]$ in term of $\mathcal{L}^{-1}[F(s)]$ assuming both are well-defined). See how it works for $\lim_{n \to \infty} \mathcal{L}^{-1}[e^{i \pi s}e^{s^2/n^2}]$

reuns
  • 77,999