I am trying to find a close form representation for the following integral:

$$ A(x;a,b,c)= \int_{0}^{x}\frac{\sin\left(a k+b k^{2}\right)+\sin\left(c k-b k^{2}\right)}{k}dk $$ for $0<x \ll \infty$. I can prove that the integral exists and it is asymptotically bounded in $x$. Looking at its asymptotic properties, I also have the hunch that it is related with Sine integral function, error function as well as Fresnel integrals.

If a close form is hard, I would also appreciate an analytic approximation or close form expression of the Taylor series coefficients with respect to $x$ or a recurrence relation that can help to compute it.

  • $\begingroup$ Differentiating A with regard to either a or c yields an expression in terms of Fresnel integrals. $\endgroup$ – Lucian Mar 26 '15 at 17:18
  • $\begingroup$ If the interval of integration would have been R, then, by integrating back with regard to a and c, we could actually have found a closed form expression for the original function in terms of the afore-mentioned Fresnel integrals, since their parametric derivatives would have yielded an elementary trigonometric function. However, as it currently stands, I am less optimistic about such also prospects applying here. $\endgroup$ – Lucian Mar 26 '15 at 17:29
  • $\begingroup$ This king of integrals seem to be "en vogue" here this days... $\endgroup$ – tired Mar 26 '15 at 17:41
  • $\begingroup$ If $ b x<<1$ there will be a nice approximate solution i think.. $\endgroup$ – tired Mar 26 '15 at 17:44

I don't believe there is a closed form for the integral. But the Maclaurin series wrt $x$ can be obtained from the Maclaurin series for the integrand. Expand the two sines. I'll look at just one of the terms. Note that

$$\eqalign{ \dfrac{\sin(ak) \cos(bk^2)}{k} &= \sum_{i=0}^\infty \sum_{j=0}^\infty (-1)^{i+j} \dfrac{a^{2i+1} b^{2j}}{(2i+1)!(2j)!} k^{2i+4j}\cr &= \sum_{m=0}^\infty \sum_{j=0}^{\lfloor m/2 \rfloor} (-1)^{m-j} \dfrac{a^{2m-4j+1} b^{2j}}{(2m-4j+1)! (2j)!} k^{2m}} $$ so that the integral is

$$ \int_0^x \dfrac{\sin(ak) \cos(bk^2)}{k}\; dk = \sum_{m=0}^\infty \sum_{j=0}^{\lfloor m/2 \rfloor} (-1)^{m-j} \dfrac{a^{2m-4j+1} b^{2j}}{(2m-4j+1)! (2j)! (2m+1)} x^{2m+1}$$

The coefficient of $x^{2m+1}$ can be expressed using a hypergeometric function: $$(-1)^m {\frac {{a}^{2\,m+1}}{\left( 2\,m+1 \right)! } {\mbox{$_4$F$_1$}(-\frac{m}2,-\frac{m}2-\frac14,-\frac{m}2+\frac14,-\frac{m}2+\frac12;\,\frac12;\,-64\,{\frac {{b}^{2}}{{a}^{4}}})} } $$


By completing the squares we just need to be able to integrate $\frac{\sin(\alpha x^2)}{x}$ over some interval - the non-integrable part $\frac{\cos(\alpha x^2)}{x}$ will cancel out.

That is not so terrible by exploiting Taylor series and the fact that $\sin z$ is an odd entire function. Moreover,

$$\int_{0}^{+\infty}\frac{\sin(\alpha x^2)}{x}\,dx = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin(\alpha t)}{t}\,dt = \frac{\pi}{4}\operatorname{Sign}(\alpha).$$

  • $\begingroup$ How does this help in case of finite $x$? $\endgroup$ – tired Mar 26 '15 at 19:24

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