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