I have the above mentioned integral
$$ \int_{l_1}^{l_2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx $$
which I want to solve. I expect some special functions in its solution, but so far I am out of ideas now.
I have tried the following:
I substitute $t=\sqrt{bx^2+cx+d}$ and get an expression for $x$ that I can insert in the exponential:
$$ t=\sqrt{bx^2+cx+d} $$ $$ \frac{t^2}{b}=x^2+\frac{c}{b}x+\frac{d}{b}=\left (x+\frac{c}{2b}\right )^2 + \frac{d}{b} - \left (\frac{c}{2b}\right )^2 $$ $$ \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}=x $$
The derivative of the substitution t is:
$$ \frac{dt}{dx}=\frac{2bx+c}{2\sqrt{bx^2+cx+d}}=\frac{2bx+c}{2t} $$ inserting everything in the integral gives
$$\int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{t}\frac{2t}{2bx+c}dt=2 \int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{2b\left ( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}\right ) + c}dt $$ This looks nearly like an exponential integral. I would expect a solution involving this kind of integral, but this is where I am stuck right now. Is it possible to give a closed form solution involving special functions?
Too long for a comment (a work in progress) $$ \int \frac{\mathrm{e}^{\pm iax}}{\sqrt{bx^2+cx+d}}dx $$ Focusing on the denominator $$ b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) =b\left[\left(x+\frac{c}{2b}\right)^2+\frac{d}{b}-\left(\frac{c}{2b}\right)^2\right] $$ lets change variables $$ t = \frac{x+\frac{c}{2b}}{\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}}\implies x = -\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t $$ leads to $$ b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) = b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right) $$ therefore we have $$ \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right)\int \frac{\mathrm{e}^{\pm ia\left(-\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t\right)}}{\sqrt{b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right)}}dt $$ let $t=\cosh \theta$ we have $$ t^2-1 = \sinh^2 \theta\\ dt = \sinh \theta d\theta $$ thus we get $$ \frac{\mathrm{e}^{\mp i\frac{ac}{2b}}}{\sqrt{b}}\int_{\bar{l_1}}^{\bar{l_2}} \mathrm{e}^{\pm i\lambda_1\cosh \theta}d\theta $$ where $$ \lambda_1 = a\left[-\frac{c}{2b} + \sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right] $$
I am still working on the last integral.