Definite Integral of $ x^{n} e^{\mathrm{i}x^2}$ I want to find an expression for the integral
\begin{align*}
\int_{x_{1}}^{x_{2}} x^{n} e^{\mathrm{i}x^2}~dx.
\end{align*}
There is one way to use power series expansion
\begin{align*}
\int_{x_{i}}^{x_{2}}x^{n} e^{\mathrm{i}x^2}~dx=\sum_{l=0}^{\infty}\frac{\mathrm{i}^{l}}{l!(n+2l+1)}(x_{2}^{n+2l+1}-x_{1}^{n+2l+1}).
\end{align*}
I am trying to solve it generally in terms of error function or Fresnel integrals. Please suggest some method or formula.
 A: Set
$$
I_n:=\int_{x_1}^{x_2} x^n e^{i x^2} dx.
$$
For $n$ odd you can integrate explicitly. For $n$ even integrate by parts to obtain the recurrence 
$$ 
I_n= \frac{x^{n+1}e^{i x^2}}{n+1}\Bigg\vert_{x=x_1}^{x=x_2}-\frac{2i}{n+1} I_{n+2}. 
$$ This allows you to determine $I_{n+2}$ if you know $I_n$. Now observe that $I_0$ is described  explicitly in terms of  the Fresnel integrals 
$$ S(x)=\int_0^x \sin(t^2) dt,\;\;C(x)=\int_0^x \cos(t^2) dt,
 $$
A: I think this integral is best interpreted in term of the incomplete gamma function. We can consider the indefinite integral
$$I_n=\int x^ne^{ix^2}dx$$
Now let
$$t=-ix^2\\x=\sqrt{it}\\dx=\frac{1}{2}\sqrt{\frac{i}{t}}dt\\x^n=(it)^{n/2}$$
Thus,
$$I_n=\int (it)^{n/2}e^{-t}\frac{1}{2}\sqrt{\frac{i}{t}}dt=\frac{i^\nu}{2}\int t^{\nu-1}e^{-t}dt$$
where $\nu=(n+1)/2$. The integral itself is the very definition of the incomplete Gamma function, $\Gamma(\nu,t)$.
So, the final result you desire is
$$I_n=\int x^ne^{ix^2}dx=-\frac{i^{(n+1)/2}}{2}\Gamma\left(\frac{n+1}{2},-ix^2\right)+C$$
where the final minus sign is due to the change in direction of the integral from $t$ to $-ix^2$.
