Solution of exponential integral

Question

I know the formula for \begin{align*} \int x^{m} \mathrm{e}^{\mathrm{i} x^{2}}~\mathrm{d}x=\sum_{l=0}^{\infty}\frac{\mathrm{i}^{l}}{l!(m+2l+1)}x^{m+2l+1}. \end{align*} I want to find an expression for the integral \begin{align*} \int x^{m} \mathrm{e}^{\mathrm{i} x^{2}}\mathrm{e}^{-\mathrm{i}k x}~\mathrm{d}x, \end{align*} where $ k $ can be integer or real. Can any one help me?

Answer 1

This is not a full answer, but if $m$ is a positive whole number, than the problem reduces considerably:

$$f(y)=\int e^{ix^2}e^{ixy}\ dx$$

Differentiate both sides $m$ times:

$$f^{(m)}(y)=i^m\int x^me^{ix^2}e^{xy}\ dx$$

and with $y=-k$,

$$f^{(m)}(-k)=i^k\int x^me^{ix^2}e^{-ikx}\ dx$$

So all that's left is for us to solve for $f(y)$. We can see that

$$f(y)=\int e^{i(x^2+xy)}\ dx=e^{-y^2/4}\int e^{iu^2}\ du\\=e^{-y^2/4}(S(u)+C(u)+c)\\=e^{-y^2/4}\left(S\left(x+\frac y2\right)+C\left(x+\frac y2\right)+c\right)$$

which is the Fresnel integral. One should then apply general leibniz rule and Faà di Bruno's formula for the closed form solution for your integral.