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So, I am confused as to how I would evaluate the integral: $\int_s^\infty J_n(t)e^{-t}dt$

Of course, I would a step by step explanation that shows how I would compute this integral.

Yes, this integral was loosely inspired by the Laplace Transform, but the main difference is the fact that the results of this seed ($\mathcal V${$f(t)$}($s$)=$\int_s^\infty f(t)e^{-t}dt$) are different from the Laplace transform's seed.

For example: if I take the $\mathcal{V}${$1$}($s$), I get $e^{-s}$, as opposed to $\frac1s$ that the Laplace transform gives me.

Therefore, many of the results are different from the Laplace Transform.

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  • $\begingroup$ You can try integral representation of $J_n$ and then swap order of integration. $\endgroup$ – N74 Aug 26 '16 at 17:53
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    $\begingroup$ Possible duplicate of Laplace transform of the Bessel function of the first kind $\endgroup$ – N74 Aug 26 '16 at 17:55
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    $\begingroup$ @N74: this is a bit different, and a bit more difficult, since the left endpoint of the integration range is a generic $s$ and not $0$. $\endgroup$ – Jack D'Aurizio Aug 26 '16 at 21:04
  • $\begingroup$ Anyway, we may compute the integral over $(0,+\infty)$ as $\frac{1}{\sqrt{2}(1+\sqrt{2})^n}$ through the linked question, then compute the integral over $(0,s)$ by expanding $J_n(t)$ as its Taylor series centered at $t=0$. $\endgroup$ – Jack D'Aurizio Aug 26 '16 at 21:07
  • $\begingroup$ I updated the text, giving more detail as to what is going on. $\endgroup$ – AstroFox Aug 26 '16 at 22:11
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I'll post it as an answer though it is incomplete, but too long for a comment: $$ \begin{align} \int_{s}^{\infty} J_{n}(bx) e^{-ax} \, dx &= \frac{1}{2 \pi} \int_{s}^{\infty} \int_{-\pi}^{\pi} e^{i(n \theta -bx \sin \theta)} e^{-ax} \, d \theta \, dx \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \int_{s}^{\infty} e^{i n \theta} e^{-(a+ib \sin \theta)x} \, dx \, d \theta \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta}}{a + ib \sin \theta}e^{-(sa+isb\sin \theta)} \, d \theta \\ &= \frac{e^{-sa}}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta} e^{-isb\sin \theta}}{a + ib \sin \theta} \, d \theta \\ &= \frac{e^{-sa}}{2 \pi} \int_{|z|=1} \frac{z^{n}e^{-{sb \over 2} \left(z-\frac{1}{z} \right) }}{a+\frac{b}{2} \left(z-\frac{1}{z} \right)} \frac{dz} {iz} \\ &= \frac{e^{-sa}}{i \pi} \int_{|z|=1} \frac{z^{n}e^{-{sb \over 2} \left(z-\frac{1}{z} \right) }}{bz^{2}+2az-b} dz \end{align}$$

Now you just have to evaluate the two residues and you're done.

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  • $\begingroup$ How would one calculate residue? I know that I may be asking a lot, but I do wish for this kind of documentation so that I can understand how this works. EDIT: I also noticed that you placed $e^{-at}$ rather than $e^{-t}$ - is this intentional? $\endgroup$ – AstroFox Aug 26 '16 at 23:08
  • $\begingroup$ First verify all my steps are correct, then you can eventually post another question. It is not an hard task, you can solve it easily with the taylor expansion for $e^{1/x}$, but I'm not good at explainations, so you'll need help by someone else. $\endgroup$ – N74 Aug 26 '16 at 23:14
  • $\begingroup$ For the residue look here ( there's something similar):math.stackexchange.com/questions/1598846/… . I used the $a$ coefficient to have something similar to the link I posted in my comment (I copied most of that answer). $\endgroup$ – N74 Aug 26 '16 at 23:29

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