Bessel函数 $\int_0^x uJ_0(u)\rmd u = xJ_1(x)$

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MathWorld Bessel Function of the First Kind 式 (60) $\int_0^x uJ_0(u)\rmd u = xJ_1(x)$ 如何证?
也在3D diffusion in a cylinder提到过:

To get the probability distribution in $z$, I would integrate over $r$, using the identity $\int_0^uxJ_0(x)\,dx=uJ_1(u)$

也在Fourier transform of projection of spherical cap提到过:

Due to the property that $\int_0^x uJ_0(u) du = xJ_1(x)$, the latter integral can easily be evaluated as

也在Show the Hankel Transform of Rect is Jinc提到过:

Then use the fact that $ \int dx \: x J_0(x) = - x J_0'(x) + C = x J_1(x) + C $

类似题 Fourier-Bessel series of $f(x)=x^2$

hbghlyj

使用recurrence formulas$$\frac{2\nu}{x}J_\nu(x)=J_{\nu-1}(x)+J_{\nu+1}(x)\tag1$$
$$2J'_\nu(x)=J_{\nu-1}(x)-J_{\nu+1}(x)\tag2$$得
\begin{align*}
[xJ_1(x)]'&=J_1(x)+xJ_1'(x)\\&\overset{(2)}=J_1(x)+\frac x2J_0(x)-\frac x2J_2(x)\\&\overset{(1)}=J_1(x)+\frac x2J_0(x)-\frac x2\left[\frac2xJ_1(x)-J_0(x)\right]\\&=xJ_0(x)
\end{align*}