Is there a closed form for the following integral:
$$\int_{0}^a\exp\left[\frac{i\pi x^2}{b}\right]\operatorname{sinc}\left(\frac{\pi ax}{b}\right)dx$$
where $i=\sqrt{-1}$ and $\operatorname{sinc}(x)=\frac{\sin{x}}{x}$? I would satisfied with a solution involving easily-computable special functions (such as error functions, Bessel functions, etc.). I looked for this in G&R, but couldn't find anything useful. Any ideas?
I assume $a,b > 0$. With the change of variables $x = a t$ this becomes $$ \dfrac{b}{\pi a} \int_0^1 e^{i \pi a^2 t^2/b} \dfrac{\sin(\pi a^2 t/b)}{t}\; dt$$ Taking $\theta = \pi a^2/b$, we need to compute $$ f(\theta) = \int_0^1 e^{i\theta t^2} \dfrac{\sin(\theta t)}{t} \; dt$$ Of course $f(0) = 0$, and $$ f'(\theta) = i \int_0^1 e^{i\theta t^2} \sin(\theta t)\;t\; dt + \int_0^1 e^{i\theta t^2} \cos(\theta t)\; dt $$ which can be expressed as a rather messy closed-form expression: according to Maple $$ \left( \dfrac{1+i}{16} \right) {{ \left( \sqrt {2\pi }{{\rm e}^{-i \theta/4}}{\rm erf} \left( \left( \dfrac{3-3i}{4} \right) \sqrt {2\theta} \right)\theta^{-1/2}-2i{{\rm e}^{2\,i\theta}}\theta^{-1}+ \sqrt {2\pi }{{\rm e}^{-i\theta/4}}{\rm erf} \left( \left(\dfrac{1-i}{4}\right) \sqrt {2\theta}\right)\theta^{-1/2}-2{{\rm e}^{2\,i\theta}} \theta^{-1} +(2+2i)\theta^{-1} \right) }} $$ I doubt that there is a closed-form antiderivative for this.
We can also expand $f(\theta)$ in a Maclaurin series $$f(\theta) = \sum_{n=1}^\infty a_n \theta^n$$ where $$ a_n = i^{n-1} \sum _{k=0}^{\left\lfloor (n-1)/2\right\rfloor}{\frac {1}{ \left( n-2\,k-1 \right) !\, \left( 2\,k+1 \right) !\, \left( 2\,n-2\, k-1 \right) }} $$ can be expressed in terms of hypergeometric functions.