Integral of $\cos(\sqrt{x^3})$ As the title explains, I can't seem to get anywhere on this one, so could someone show me how I would proceed with solving this?
$$\int\cos\left(\sqrt{x^3}\right)\,dx$$
I can't shrug off the feeling that I'll end up somewhere in the complex realm. 
 A: Another form of solution is to express the integrand in a power series and integrate term by term. 
$$ \cos(\sqrt{x^3}) = 1 - \frac{x^3}{2!} + \frac{x^6}{4!} - \frac{x^9}{6!} +\ldots, \quad x \ge 0.$$
Because this power series is convergent for all $x \ge 0$, the power series of the resulting integral is also convergent for all $x\ge0$. Hence,
$$\begin{align}
\int \cos(\sqrt{x^3})\,dx &= x - \frac{x^4}{2!4} + \frac{x^7}{4!7} - \frac{x^{10}}{6!10} +\ldots \\
&= \sum_{n=0}^{\infty} \frac{(-1)^nx^{3n+1}}{(2n)!(3n+1)}, \quad x >= 0.
\end{align}
$$
A: Start with
$$
 \cos x = \frac{1}{2} \left( e^{ix} + e^{-ix} \right).
$$
From Mathematica:
$$
\int \cos \left(\sqrt{x^3}\right) \, dx
=
-\frac{x\, \Gamma \left(\frac{2}{3},-i \sqrt{x^3}\right)}{3 \left(-i \sqrt{x^3}\right)^{2/3}}-\frac{x\, \Gamma \left(\frac{2}{3},i \sqrt{x^3}\right)}{3 \left(i \sqrt{x^3}\right)^{2/3}}
=
-\frac{1}{3} x \left(E_{\frac{1}{3}}\left(-i \sqrt{x^3}\right)+E_{\frac{1}{3}}\left(i \sqrt{x^3}\right)\right)
$$
where
$$
E_n (z) = \int_{1}^{\infty}e^{-zt}t^{-n}dt, \quad
\Gamma (a, z) = \int_{z}^{\infty}e^{-t}t^{a-1}dt
$$
No analytic solution exists in terms of more basic functions. 
A quick look at the function $\cos \sqrt{x^{3}}$.

A: This integral has no elementary anti-derivative 
Consider the integral 
$$F(y) = \int^y_0\cos\left(\sqrt{x^3}\right)\,dx$$
By letting $x^{3/2} =t$ which implies $x = t^{2/3}$
$$F(y) = \frac{2}{3}\int^{y^{3/2}}_0t^{\frac{2}{3}-1}\cos\left(t\right)\,dt$$
The  Böhmer integral integral (generalized Fresnel) is defined as 
$$C(x,\alpha) = \int^{\infty}_xt^{\alpha-1}\cos\left(t\right)\,dt$$
This implies 
$$F(y) =\frac{2}{3}\left[\frac{\Gamma(2/3)}{2}-C\left(y^{3/2},\frac{2}{3}\right)\right] $$
