# Laplace transform of Bessel function of order zero

I'm trying to prove that the Laplace transform of the function $$J_0(a\sqrt{x^2+2bx})$$ is $$\frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}$$ as asserted in the eqworld. This formula can also be found in the book "Tables of Integral Transforms" page 207.

My first attempt was to insert $$a\sqrt{x^2+bx}$$ in the series representation of the Bessel function $$J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{2m}$$ and then integrating term by term. However, the Laplace transform of the function $$(x^2+bx)^m$$ is not trivial.

My second attempt was to look at a change of variable in the differential equation of $$J_0$$ $$t^2J_0''(t)+tJ_0'(t)+t^2J_0(t)=0$$ but I was not able to find it.

Thanks for any clue.

• @jack But I accept your answer ! (I check the box "accept the answer" on the left panel). I just add a more complete treatment based on your answer and I add a more rigorous treatment to the end of the proof. Mar 6, 2017 at 20:07
• @jack Ah ok I understand my mistake, I cannot check two answers. As if there is only one possible answer to a question... Sorry for the mistake, It's my first question on stackexchange Mar 6, 2017 at 20:10
• FYI: $J_0$ is just a regular, not modified, Bessel function. Feb 5, 2021 at 14:38

By the substitution $x\mapsto bx$, the problem is equivalent to finding the Laplace transform of $f_c(x)=J_0(c\sqrt{x^2+x})$ or the inverse Laplace transform of $$g_c(p)=\frac{1}{\sqrt{p^2+1}}\,\exp\left(\frac{-c}{p+\sqrt{p^2+1}}\right). \tag{1}$$ It is useful to recall that $\mathcal{L}(J_n(x))(p) = \frac{1}{\sqrt{1+p^2}(p+\sqrt{1+p^2})^n}$. It follows that:

$$\mathcal{L}^{-1}(g_c(p))(x) = \sum_{n\geq 0}\frac{(-c)^n J_n(x)}{n!}=\sum_{n\geq 0}\frac{(-c)^n}{n! 2\pi i^n}\int_{0}^{2\pi}e^{ix\cos\theta}e^{in\theta}\,d\theta \tag{2}$$ and $$\begin{eqnarray*} \mathcal{L}^{-1}(g_c(p))(x) &=& \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ix\cos\theta+ice^{i\theta}\right)\,d\theta\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(i(x+c)\cos\theta-c\sin\theta\right)\,d\theta\\&=&I_0\left(\sqrt{c^2-(x+c)^2}\right)=I_0\left(\sqrt{-x(2c+x)}\right). \end{eqnarray*}\tag{3}$$ The last term equals $\color{red}{J_0(\sqrt{x^2+2cx})}$ as wanted.

A particular thing in Mathematics won't ever cease to amaze me: $a=b$ and $b=a$ are the same thing on a semantic level, however to prove $a\mapsto\ldots\mapsto b$ or $b\mapsto\ldots\mapsto a$ may be asymmetrically extremely hard - almost trivial. I do not know for sure which part of this phenomenon is due to the fact that we are used to have the "initial term" on the left and the "final term" on the right of a blank page (in many Asian cultures it is likely the opposite, it might be interesting to investigate "the other side" of this issue, too), which part is due to the fact that we heavily rely on causal links (even when they are $\Leftrightarrow$, in order to achieve a hierarchical, tree-like structure in our minds). In this particular case, I found easier to compute an inverse Laplace transform rather than a direct Laplace transform, and I think we should be trained to "think backwards" more often. That also reminds me of an interesting question I heard once: if both us and Mathematics are built in the image of God, why do we find so difficult to prove many things? Why mathematical truth is not evident in our eyes?

• Very instructive ! I just don't see how you pass from the second line of the equation (3) to the third line. Mar 2, 2017 at 15:38
• @Serge: by exploiting (again) the integral representation for the $J_0$ or $I_0$ function. Mar 2, 2017 at 15:40
• Yes I understand it is the integral representation you're using. But the integral representation of $J_0(z)$ is $\frac{1}{2\pi} \int_0^{2\pi} e^{i z \cos(t)} dt$. I'm missing a (surely elementary) step. Mar 2, 2017 at 16:08
• @Serge: From the series representation, $I_0(i z)=J_0(z)$. In the central line of $(3)$, it is enough to apply a substitution of the form $\theta\mapsto\theta+\varphi$ (leaving the integration range unchanged by periodicity) to turn such integral into a known integral. Mar 2, 2017 at 16:17
• Ok. I find the solution there link Mar 3, 2017 at 10:42

Here is the complete answer the my first question. Let $J_0$ denote the Bessel function of order zero. We want to show that $$\color{red}{\mathcal{L}(J_0(a\sqrt{x^2+2bx}))} = \frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}$$ with $a$ and $b$ arbitrary real numbers. Denoting $g(x)$ the expected result, we note that it can be expressed as \begin{eqnarray} g(x) & = & \frac{1}{\sqrt{p^2+a^2}} \exp\left\{b\left(p- \sqrt{p^2+a^2}\right) \right\} \nonumber \\ & = & \frac{1}{\sqrt{p^2+a^2}} \exp\left\{b \frac{-a^2}{p + \sqrt{p^2+a^2}} \right\} \\ & = & \frac{1}{\sqrt{p^2+a^2}} \sum_{n=0}^{\infty} \frac{(-b)^n}{n^!} \frac{a^{2n}}{\left(p+\sqrt{p^2+a^2}\right)^n}. \nonumber \end{eqnarray} The Bessel functions of order $n$ have the following Laplace transform $$\mathcal{L}(J_n(ax)) = \frac{a^n}{\sqrt{p^2+a^2} (p+\sqrt{p^2+a^2})^n}.$$ Thus the function $g$ can be written as $g(x) = \sum_{n=0}^{\infty} \frac{(-ab)^n}{n^!} \mathcal{L}(J_n(ax))$. The integral representation of $J_n$ is $$J_n(z) = \frac{1}{2\pi i^n} \int_0^{2\pi} e^{i z \cos(t)+int}\, dt.$$ Using this integral representation, it follows that: \begin{eqnarray} \mathcal{L}^{-1}(g(p))(x) &=& \frac{1}{2\pi} \int_{0}^{2\pi} \exp\left(iax\cos(t)+ibae^{it}\right)\,dt \nonumber \\ & = & \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ia\left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]\right)\,dt \nonumber\\ & = & \frac{1}{2\pi} \sum_{n=0}^\infty \frac{(ia)^n}{n!2^n}\int_{0}^{2\pi} \left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]^n\, dt. \label{eq:integral} \end{eqnarray} Now for each $n$ we can evaluate the integral as $$\int_{0}^{2\pi} \left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]^n\, dt = \left\{ \begin{array}{lll} 0 & \mbox{ if }& n = 2k+1 \\ 2\pi \dbinom{2k}{k} (x+2b)^k x^k& \mbox{ if }& n = 2k \end{array} \right.$$ Substituting, we arrive at the expression $$\mathcal{L}^{-1}(g(p))(x) = \sum_{k=0}^\infty \frac{(-a^2)^k (x^2+2bx)^k}{(k!)^2} = J_0\left(a\sqrt{x^2+2bx}\right).$$