Does this integral, related to the Fresnel integrals, have a name or a known solution?

Question

I've been working on a problem in light propagation, where the crux of a derivation boils down to understanding the following integral:

$$\int_{-x_0}^{x_0} e^{i \alpha \sqrt{1-x^2}} dx, $$

with the square root taking the positive imaginary branch when $x_0 > 1$. Unfortunately, I just have not been able to figure out whether this integral has a name, is trivially related to any named integrals, or find a tabulated result! I've asked humans, Wolfram Alpha, and ChatGPT, with no success.

The related Fresnel integrals would allow one to solve an approximation to this integral when $x_0 << 1$

$$e^{i\alpha} \int_{-x_0}^{x_0} e^{- i \frac{\alpha}{2} x^2} dx. $$

But, I also haven't seen this integral appear in tables of integrals related to the Fresnel integrals.

Has anyone come across this integral before, or know of more information about it?

Answer 1

Case 1: $|x_0|\leq 1$

The integral in question can be written in the form of an infinite series that involves Bessel functions. There are multiple ways of arriving to these results. Here, I will compute the integrals

$$I_1(\alpha)=\int_0^{\arcsin x_0}\cos\theta\cos(\alpha\cos\theta),~~~ I_2(\alpha)=\int_0^{\arcsin x_0}\cos\theta\sin(\alpha\cos\theta)$$

which constitute the real and imaginary part of the integral in question. One can proceed by expanding the sine/cosine in a Taylor series, due to the fact that integrals over powers of cosines can be expressed in closed form, in terms of Chebyshev polynomials of the second kind

$$\int_{0}^{\arcsin x_0}(\cos\theta)^{2m+1}d\theta=\frac{x_0}{2^{2m}}\sum_{k=0}^m{2m+1\choose m-k}\frac{U_{2k}\left(\sqrt{1-x_0^2}\right)}{2k+1}$$

$$\int_{0}^{\arcsin x_0}(\cos\theta)^{2m+2}d\theta=\frac{\arcsin x_0}{2^{2m+2}}{2m+2\choose m+1}+\frac{x_0}{2^{2m+1}}\sum_{k=0}^{m}{2m+2\choose m-k}\frac{U_{2k+1}\left(\sqrt{1-x_0^2}\right)}{2k+2}$$

Substituting these into the Taylor series for $I_1, I_2$ and exchanging the order of summation we get the following results:

$$I_1(\alpha)=\sum_{k=0}^\infty \frac{x_0U_{2k}\left(\sqrt{1-x_0^2}\right)}{2k+1}[J_{2k}(\alpha)+J_{2k+2}(\alpha)]$$

$$I_2(\alpha)=\frac{J_1(\alpha)\arcsin x_0 }{2}+\sum_{k=0}^\infty \frac{x_0U_{2k+1}\left(\sqrt{1-x_0^2}\right)}{2k+2}[J_{2k+1}(\alpha)+J_{2k+3}(\alpha)]$$

where $J_n$ are the Bessel functions of order $n$. Case 2 to be done later.

Answer 2

The integral can be simplified a bit: $$I=2\int_{0}^{x_0} e^{i \alpha \sqrt{1-x^2}} dx,$$

I would suggest trying the following change in the power of the exponent:

Instead of

$$\sqrt{1-x^{2}}$$

use $$(1-x)^{\frac{1}{4}}$$

Then

$$I=2\int_{0}^{x_0} e^{i \alpha (1-x)^{\frac{1}{4}}} dx,$$ The last integral can be calculated already in closed form in elementary functions.