Lipshitz Integral for $a=0$

Question

I knew that this, $$\displaystyle{\int_0^\infty e^{-ax}J_0(bx)dx=\frac{1}{\sqrt{a^2+b^2}}},$$ holds for $a>0$ but, in an exercise from Arfken, it said that this holds for $a\geq0$.

How can I prove that?

Answer 1

Let's employ the following integral representation of $J_0$

$$ J_0(bz)=\frac{2}{\pi}\int_0^1\frac{\cos(bz t)}{\sqrt{1-t^2}}dt $$

which can be derived from the standard integral representation by an easy change of variables. Now we want to calculate (assuming $b>0$)

$$ I(b)=\int_0^{\infty} dzJ_0(bz)= \frac{2}{\pi}\int_0^{\infty} dz \int_0^1\frac{\cos(bz t)}{\sqrt{1-t^2}}dt $$

Changing the order of integration is not straightforwardly justified because the resulting integral over $t$ obviously doesn't converge in the usual sense. To make this step rigourous requires an interpretation in the sense of tempered distributuions and permitts us to use $\int_{\mathbb{R}} e^{i x y}dx =2\pi \delta(y)$ where $\delta(y)$ is Diracs delta distribution.

$$ I(b)=2\int_0^1\frac{\delta(bt)}{\sqrt{1-t^2}}dt=\frac{1}{b}\int_{-b}^b\frac{\delta(t)}{\sqrt{1-(t/b)^2}}dt=\frac{1}{b} $$

where have used the symmetry of the integrand in the last step.

Answer 2

Another way. We know that: $$ J_0(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\exp\left(i x \sin(t)\right)\,dt \tag{1}$$ hence by assuming $a>0$ we may apply Fubini's theorem to get: $$ \begin{eqnarray*}\int_{0}^{+\infty}e^{-ax}J_0(x)\,dx &=& \frac{1}{2\pi}\int_{-\pi}^{\pi}\int_{0}^{+\infty}\exp\left(i x \sin(t)-ax\right)\,dx\,dt\\&=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{dt}{a-i\sin(t)}\\&=&\frac{1}{\pi}\int_{0}^{\pi}\frac{a\,dt}{a^2+\sin^2(t)}\\&=&\frac{2a}{\pi}\int_{0}^{\pi/2}\frac{dt}{a^2+\sin^2(t)}\\&=&\frac{2a}{\pi}\int_{0}^{+\infty}\frac{du}{a^2(1+u^2)+u^2}\\&=&\color{red}{\frac{1}{\sqrt{1+a^2}}}\tag{2}\end{eqnarray*}$$ through the substitution $t=\arctan(u)$. Now $J_0(x)$ is not a Lebesgue-integrable function over $\mathbb{R}^+$, since $|J_0(x)|$ decays like $\frac{1}{\sqrt{x}}$, but it is improperly Riemann-integrable over $\mathbb{R}^+$. So assuming that the given identity is stated "in the Riemann way", $(2)$ holds also for $a=0$. The two-parameters identity of yours can be deduced from $(2)$ through a simple change of variable.