1
$\begingroup$

Consider the first kind Bessel function $J_0$, one way to define it is $J_0(x)=1/\pi \int_0^\pi \cos(x \sin t)\;dt$.

My question is, $\int_0^n J_0(x)\;dx$ converge when $n$ tends to infinity?

For the graph of Bessel function, see http://en.wikipedia.org/wiki/Bessel_function

1 Answers 1

1

Yes. The infinite oscillatory integral under consideration is well-defined, by considering the asymptotic behavior of $J_n(x)$. To establish the evaluation

$\int_0^\infty J_0(t)\mathrm dt=1$

we treat this as the expression

$\lim_{c\to 0^+} \int_0^\infty \exp(-ct) J_0(t)\mathrm dt$

and then replace the Bessel function with the integral representation

$J_0(x)=\frac1{\pi}\int_0^\pi \exp(ix\cos\,u)\mathrm du$

to yield

$\lim_{c\to 0^+} \frac1{\pi}\int_0^\infty \exp(-ct) \int_0^\pi \exp(it\cos\,u)\mathrm du\mathrm dt$

after which,

$\lim_{c\to 0^+} \frac1{\pi} \int_0^\pi \frac1{c-i\cos\,u}\mathrm du=\lim_{c\to 0^+} \frac1{\sqrt{1+c^2}}=1$

  • 0
    If I remember correctly, this route was also taken in Bowman's book on Bessel functions.2011-11-23