Saturday, June 8, 2019

integration - Integral $int_0^inftyfrac{exp(ialphacos u)-J_0(alpha)}{1+beta u}mathrm{d}u$




I was studying the motion of a particle in a certain magnetic field and one of the quantities that arose was given by the titular integral
$$
F(\alpha,\beta)=\int_0^\infty\frac{\exp(i\alpha\cos u)-J_0(\alpha)}{1+\beta u}\mathrm{d}u
$$

Here $\alpha\in\mathbb{R}$, $\beta>0$ and $J_0$ is a Bessel function.



The integral does exist. I found the following theorem in a book on real analysis:
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $[a,b]$ while $g:\mathbb{R}\rightarrow\mathbb{R}$ is nonnegative, decreasing and continuously differentiable on $[a,b]$. Then $\exists\xi\in[a,b]$:
$$
\int_a^bf(x)g(x)\mathrm{d}x=g(a)\int_a^\xi f(x)\mathrm{d}x

$$

Taking into account the fact that $\exp(i\alpha\cos u)-J_0(\alpha)$ is periodic with period $2\pi$ and that its integral over one period is $0$, we can apply the above theorem seperately to the real and imaginary parts of the original integral to find that for all $a,b>0$:
$$
\left\vert\int_a^b\frac{\exp(i\alpha\cos u)-J_0(\alpha)}{1+\beta u}\mathrm{d}u\right\vert<\frac{6\pi}{1+\beta a}
$$

This implies convergence.



However, I can't find any way to express $F$ in terms of other functions, be it ordinary or special. This wouldn't be too much of a problem if the integrand weren't oscillatory, making numerical evaluation difficult. Therefore, my question is twofold:





  • is there a closed form for $F$, perhaps in terms of special functions,

  • if not, are there at least any efficient methods for numerical computation of $F$?


Answer



You can also directly apply Dirichlet's test (the integral of the numerator is zero when taken over a period, therefore uniformly bounded on all finite intervals).



We have
$$\cos^{2 k} u = 2^{-2 k} \binom {2 k} k +
2^{1 - 2 k} \sum_{1 \leq j \leq k} \binom {2 k} {k - j} \cos 2 j u, \\
\sum_{k \geq 0} \frac {(i \alpha \cos u)^{2 k}} {(2 k)!} =

J_0(\alpha) + \sum_{j \geq 1} \sum_{k \geq j}
2^{1 - 2 k} \binom {2 k} {k - j} \frac {(i \alpha)^{2 k}} {(2 k)!} \cos 2 j u = \\
J_0(\alpha) + 2 \sum_{j \geq 1} (-1)^j J_{2 j}(\alpha) \cos 2 j u.$$

Doing the same calculation for odd indices and integrating termwise, we obtain
$$I(\beta, j) = \int_0^\infty \frac {\cos j u} {1 + \beta u} du =
\frac {\left( \frac \pi 2 - \operatorname{Si}(\omega) \right) \sin \omega -
\operatorname{Ci}(\omega) \cos \omega} \beta,
\quad \omega = \frac j \beta, \\
F(\alpha, \beta) =
2 \sum_{j \geq 1} i^j J_j(\alpha) I(\beta, j).$$


$J_j(\alpha)$ decays superexponentially for a fixed $\alpha$ (roughly speaking, it oscillates when $j < \alpha$ and then decays rapidly).


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