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I was evaluating this on WolframAlpha: https://www.wolframalpha.com/input?i2d=true&i=Integrate%5BDivide%5Bcos%5C%2840%292%CF%80%5C%2840%29Divide%5B1%2Cx%5D%2B1x%5C%2841%29%5C%2841%29%2Cx%5D%2C%7Bx%2C0%2C%E2%88%9E%7D%5D

And the following function BesselJ^(1,0) (0, 4pi)

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.

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  • $\begingroup$ 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 $$\frac{\partial^{a_1+\dots+a_n}F}{{\partial x_1}^{a_1}\dots{\partial x_n}^{a_n}}$$ $\endgroup$
    – K.defaoite
    Oct 30, 2023 at 1:35

2 Answers 2

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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 $\nu = 0$: $\frac{\partial}{\partial \nu}J_{\nu}(z)\Big|_{\nu=0} = \frac{\pi}{2} Y_0(z)$.

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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 $\text{BesselJ}^{(1,0)}(0,x)=\frac{\pi}{2}Y_0(x)$.

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