Bessel函数Laplace变换

hbghlyj

$$J_0(x)=\frac{2}{π}\int_0^{π/2}\cos(x\cos θ)d θ$$
求证
$$\int_0^∞ J_0(x)e^{-a x}d x=\frac{1}{\sqrt{1+a^2}}\quad(a>0)$$

hbghlyj

由Fubini定理\begin{align*}&\frac{2}{π}\int_0^∞\int_0^{π/2}\cos(x\cos θ)e^{-a x}d θd x\\&=\frac{2}{π}\int_0^{π/2}\int_0^∞\cos(x\cos θ)e^{-a x}d xd θ&\because\small\int_0^∞\cos(x\cos θ)e^{-a x}d x=ℒ\{\cos(x\cos θ)\}(a)=\frac{a}{a^2+\cos^2θ}\\&=\frac{2}{π}\int_0^{π/2}\frac{a}{a^2+\cos^2θ}d θ\\&=\frac{2a}{π}\int_0^{π/2}\frac{\sec^2θ}{a^2\sec^2θ+1}d θ&x=\tanθ\\&=\frac{2a}{π}\int_0^∞\frac1{a^2(x^2+1)+1}dx\\&=\frac{2a}{π(1+a^2)}\int_0^∞\frac1{{a^2\over a^2+1}x^2+1}dx\\&=\frac1{\sqrt{1+a^2}}
\end{align*}又见math.stackexchange.com/questions/1217543