# does this integration converge?

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

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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,

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

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I was trying to do something similar and wondering about the justifications for the individual steps. I gather that you can change the order of integration by Fubini's theorem? I was going to use the series representation instead of the integral representation; I believe I could change the order in that case, too, also by Fubini? Also, I gather that you can change the order of the integration and the limit by the dominated convergence theorem? What I don't understand is why the asymptotic behaviour is enough to ensure convergence of the integral -- doesn't the $o(1)/\sqrt t$ term prevent this? – joriki Nov 22 '11 at 3:11
I remember this argument from Watson's treatise. Unfortunately, I don't have my copy with me, and Google won't let me preview. Let me get back to you on that (unless somebody beats me to it). (The integral is variously associated with either Lipschitz or Parseval.) – J. M. Nov 22 '11 at 3:32
If I remember correctly, this route was also taken in Bowman's book on Bessel functions. – J. M. Nov 23 '11 at 2:47