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According to the table here, we have

$$ \mathcal{L} \left\{ Y_0(x) \right\} = -\frac{2 \sinh^{-1} s}{\pi \sqrt{s^2 + 1}} $$

where $ Y_0(x) $ is the Bessel function of the 2nd kind, order zero.

How can this be derived? I couldn't find any derivation online. Ideally we should be able to derive it from either the Bessel differential equation, but when I tried it, the singularity at $ x = 0 $ prevents me from writing down $ Y_0(0) $ and $ Y_0'(0) $ in the Laplace transform. Maybe the Frobenius series expansion could be used too.

By reverse engineering, I can see that the transform $ \hat{Y} $ obeys the differential equation

$$ \frac{\mathrm{d} \hat{Y}}{\mathrm{d} s} + \frac{s}{s^2 + 1} \hat{Y} = -\frac{2}{\pi} \cdot \frac{1}{s^2 + 1}. $$

Nick_2440
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    Using the Laplace transform of the Bessel function of the first kind, it follows from the definition of $Y_{\nu}(x)$ that $$\begin{align} \lim_{\nu \to 0}\int_{0}^{\infty} e^{-sx} , Y_{\nu}(x) , \mathrm dx &= \lim_{\nu \to 0} \frac{\frac{(\sqrt{s^{2}+1}-s)^{\nu}}{\sqrt{s^{2}+1}}\cos(\nu \pi)- \frac{(\sqrt{s^{2}+1}-s)^{-\nu}}{\sqrt{s^{2}+1}}}{\sin( \nu \pi )} \ & = \frac{2}{\pi}\frac{1}{\sqrt{s^{2}+1}} , \ln \left(\sqrt{s^{2}+1}-s \right) \ &=-\frac{2}{\pi}\frac{1}{\sqrt{s^{2}+1}} , \ln \left(\sqrt{s^{2}+1}+s \right). \end{align} $$ – Random Variable Feb 08 '25 at 23:09
  • Is that L'Hopital's rule for the limit? – Nick_2440 Feb 08 '25 at 23:39
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    Yes it is. We have the indeterminate form $0/0$. – Random Variable Feb 08 '25 at 23:42

1 Answers1

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I'm going to expand on my comment since I didn't say anything about why $$\lim_{\nu \to 0} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx = \int_{0}^{\infty} e^{-sx} \, Y_{0}(x) \, \mathrm dx. $$

For $s>0$ and $v > -1$, the Laplace transform of the Bessel function of first kind is $$\int_{0}^{\infty} e^{-sx} \, J_{\nu}(x) \, \mathrm dx = \frac{\left(\sqrt{s^{2}+1}-s \right)^{\nu}}{\sqrt{s^{2}+1}}.$$

Using the definition $$Y_{\nu}(x) = \frac{J_{\nu}(x)\cos(\nu \pi)-J_{-\nu}(x)}{\sin(\nu \pi)}, $$

we have $$ \begin{align} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx &= \int_{0}^{\infty} e^{-sx} \, \frac{J_{\nu}(x)\cos(\nu \pi)-J_{-\nu}(x)}{\sin(\nu \pi)} \, \mathrm dx \\ &= \frac{\frac{(\sqrt{s^{2}+1}-s)^{\nu}}{\sqrt{s^{2}+1}}\cos(\nu \pi)- \frac{(\sqrt{s^{2}+1}-s)^{-\nu}}{\sqrt{s^{2}+1}}}{\sin( \nu \pi )} \end{align}$$ for $s>0$ and $|\nu| <1$ ($ \nu \ne 0 $).

And using L'Hopital's rule, we have $$ \begin{align} \lim_{\nu \to 0} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx &= \frac{2}{\pi}\frac{1}{\sqrt{s^{2}+1}} \, \ln \left(\sqrt{s^{2}+1}-s \right) \\ &= -\frac{2}{\pi}\frac{1}{\sqrt{s^{2}+1}} \, \ln \left(\sqrt{s^{2}+1}+s \right) \\ &= -\frac{2}{\pi}\frac{\operatorname{arsinh}(s)}{\sqrt{s^{2}+1}}. \end{align}$$

We now need to argue that $$\lim_{\nu \to 0} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx = \int_{0}^{\infty} e^{-sx} \, Y_{0}(x) \, \mathrm dx, $$ or at least that $$\lim_{\nu \to 0^{+}} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx = \int_{0}^{\infty} e^{-sx} \, Y_{0}(x) \, \mathrm dx. $$

Let $y_{0,1}$ be the first positive zero of $Y_{0}(x)$.

If $0 \le \nu \le \alpha $, then $|Y_{\nu}(x)| \le |Y_{\alpha}(x)|$ on $(0,y_{0,1}]$. See this plot.

And on $(y_{0,1}, \infty)$, $|Y_{\nu}(x)| $ is bounded by a constant, which we'll call $A$.

To apply the dominated convergence theorem, let's break the integral into $$\int_{0}^{y_{0,1}} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx + \int_{y_{0,1}}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx , $$ and let's assume that $0 \le \nu \le 1/2$.

For the integrand of the first integral, we can use $|Y_{1/2}(x)| = \sqrt{\frac{2}{\pi}} \frac{\cos(x)}{\sqrt{x}}$ as the dominating function.

And for the integrand of the second integral. we can use $A e^{-sx}$, $s>0$, as the dominating function.

It then follows that $$\lim_{\nu \to 0^{+}} \int_{0}^{\infty} e^{-sx} \, Y_{\nu}(x) \, \mathrm dx = \int_{0}^{\infty} e^{-sx} \, Y_{0}(x) \, \mathrm dx. $$

Random Variable
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