# What does this particular bessel with a superscript and ordered pair mean?

And the following function

appears. Are there some resources on what this function is? I've looked through the Wolfram documentation, but I couldn't find it. I tried to type it into WolframAlpha, and ChatGPT was useless.

• In the notation of Mathematica and Wolfram Research, if $F:\mathbb R^n\to\mathbb R^m$, $F^{(a_1,\dots,a_n)}$ is taken to mean $$\frac{\partial^{a_1+\dots+a_n}F}{{\partial x_1}^{a_1}\dots{\partial x_n}^{a_n}}$$ Oct 30, 2023 at 1:35

Superscript $$(1,0)$$ means derivative with respect to the first variable in Wolfram's notation. Here $$\mathrm{BesselJ}(\nu,z)$$ is a funcion of two arguments, where $$\nu$$ is the order, so $$\mathrm{BesselJ^{(1,0)}}(\nu,z)$$ = $$\frac{\partial}{\partial \nu}J_{\nu}(z)$$. Hence, $$\mathrm{BesselJ^{(1,0)}}(0, 4\pi) = \frac{\partial}{\partial \nu}J_{\nu}(4\pi)\Big|_{\nu=0}$$
According to DLMF (eq. 10.15.3), $$\frac{\partial}{\partial \nu}J_\nu(z)\Big|_{\nu=n} = \frac{\pi}{2} Y_n(z) + \frac{n!}{2\left(z/2\right)^n} \sum_{k=0}^{n-1} \frac{\left(z/2\right)^k J_k(z)}{k!(n-k)},$$ where $$Y_n$$ is the Bessel Function of the Second Kind. So, for $$\nu = 0$$: $$\frac{\partial}{\partial \nu}J_{\nu}(z)\Big|_{\nu=0} = \frac{\pi}{2} Y_0(z)$$.
I found the following integral in Gradshteyn and Ryzhik's Table of Integrals, Series, and Products (p. 480, formula 3.868.2): $$\int_0^{\infty}\cos\left(a^2x+\frac{b^2}{x}\right)\frac{dx}{x}=-\pi Y_0(2ab)\qquad[a>0, b>0], \tag{1}$$ where $$Y_0$$ is the Bessel function of the second kind and order zero. When $$a=b=\sqrt{2\pi}$$, Eq. $$(1)$$ reduces to your integral: $$\int_0^{\infty}\cos\left(2\pi\left(x+\frac{1}{x}\right)\right)\frac{dx}{x}=-\pi Y_0(4\pi). \tag{2}$$ Therefore, apparently $$\text{BesselJ}^{(1,0)}(0,x)=\frac{\pi}{2}Y_0(x)$$.