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This question already has an answer here:

Consider $J_0$ the zeroth order Bessel function. I'm trying to compute the Laplace transform

$$\mathcal{L}[J_0](s) = \int_0^\infty J_0(t) e^{-st}dt,$$

but until now I couldn't find a good way to do it. Substituting the series for $J_0$ doesn't seem to be a good idea. The other thing I thought of, was using the differential equation for $J_0$. That is, we know that

$$t^2J_0''(t)+tJ_0'(t)+t^2J_0(t)=0,$$

so that multiplying by $e^{-st}$ and integrating we have

$$\int_0^\infty t^2 J_0''(t)e^{-st}dt+\int_0^\infty tJ_0'(t)e^{-st}dt+\int_0^\infty t^2J_0(t)e^{-st}dt=0,$$

but I couldn't get very far with this. What I could indeed notice is that the last integral may be rewritten as

$$\int_0^\infty t^2 e^{-st}J_0(t)dt=\int_0^\infty \left(tJ_0(t)\right)\left(-\dfrac{d}{ds}\left(e^{-st}\right)\right)dt=-\dfrac{d}{ds}\left[s\int_0^\infty tJ_1(t)e^{-st}dt\right],$$

but this doesn't seem to help. On the other hand the other two integrals I had no idea to deal with them.

So, how can we compute this Laplace transform in a good way?

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marked as duplicate by Hans Engler, mickep, user147263, Claude Leibovici calculus Oct 18 '15 at 6:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is also done in my answer here in another question. $\endgroup$ – mickep Oct 18 '15 at 4:20
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Honestly I'd probably go the route of using $$J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}$$ With $\alpha = 0$ this boils down to $$J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{2m}$$ and as a Laplace Transform you'd find $$\int_0^\infty J_0(x)e^{-st}\text{d}t = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^22^{2m}}\int_0^\infty x^{2m}e^{-st} \text{d}t$$ The meat of the problem is computing the transform $$\int_0^\infty x^{2m}e^{-st} \text{d}t$$ but that transform is well known to be $\frac{(2m)!}{s^{2m+1}}$, so it all comes down to $$\sum_{m=0}^\infty \frac{(-1)^m}{(m!)^22^{2m}}\cdot \frac{(2m)!}{s^{2m+1}}$$

Edit With some pointers from Weaam, we have $$\sum_{m=0}^\infty \frac{(-1)^m}{(m!)^22^{2m}}\cdot \frac{(2m)!}{s^{2m}} = \frac{1}{\sqrt{1+\left(\frac{1}{s}\right)^2}}$$ via Taylor Series definition. So $$\mathcal{L}\left[J_0\right](s)=\frac{1}{s}\left(\sum_{m=0}^\infty \frac{(-1)^m}{(m!)^22^{2m}}\cdot \frac{(2m)!}{s^{2m}}\right) = \frac{1}{\sqrt{1+s^2}}$$

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    $\begingroup$ @Weaam encorporated into my answer. Thanks for that! I was not expecting such an incredibly concise closed-form solution. $\endgroup$ – graydad Oct 18 '15 at 4:05

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