Integral $\int \sin(x^3) {\rm d}x$ I'm new here :)
I'd like to know why do apps like maple and wolframalpha freak out at this integral please(see pic). It's supposed to be a basic one, is it not?

Thanks,
Andre
 A: Primitives of $\sin{(x^3)}$ are written in function of the so-called Exponential Integral, defined as:
$$E_n ( x ) = \displaystyle\int_1^{\infty }\dfrac{e^{-xt}}{t^n}\,\mathrm{d}t$$
for some suitable value of $n$. According to Mathematica:
$$ \int \sin{(x^3)} \, \mathrm{d}x = \frac{i x}{6}  \left(E_{\frac{2}{3}}\left(-i x^3\right)-E_{\frac{2}{3}}\left(i x^3\right)\right)$$
where $i^2=-1$. I think this is the best you can do on finding a closed form for your integral (in terms of a non elementary function). Hope you find this useful.
Here's a plot of the result in the range $0 \leq x \leq 5$. 

A: Welcome to the wonderful world of antiderivatives. While differentiation has all these nice fun rules, antidifferentiation mostly comes down to either hoping that it happens to match the output of one of those nice fun rules. Failing that, the next best option is usually to define a new function as "the antiderivative of that function I was given" and then call that the answer.
(This is also a standard method of "solving" second-order differential equations. E.g., the solutions of the Bessel equation are the Bessel functions, which are defined by being the solutions of the Bessel equation.)
The nice thing is that once you've defined a few of these special antiderivative functions, you can often start expressing other antiderivatives in terms of them by suitable manipulation of the integral. This is what Mathematica is doing here. In terms of the function
$$
E_n(z) = \int_1^\infty \frac{e^{-zt}}{t^n}dx
$$
it gives
$$
\int\sin(x^3) = \frac{i}{6}x\left[E_{2/3}(-ix^3) - E_{2/3}(ix^3)\right]
$$
Now it might be unclear how this is true. This is how you get from the right side to the left
\begin{multline}
\frac{ix}{6}\left[E_{2/3}(-ix^3) - E_{2/3}(ix^3)\right] = \frac{ix}{6}\left(\int_1^\infty\frac{e^{ix^3t}}{t^{2/3}} - \int_1^\infty\frac{e^{-ix^3t}}{t^{2/3}}dt\right) \\= -\frac{x}{3}\int_1^\infty\frac{\sin(x^3 t)}{t^{2/3}}dt =-\int_x^\infty\sin(u^3)du
\end{multline}
where the last equality used the subsitution $t\rightarrow (u/x)^3$.
Now if you're wondering "Wait, how would I know to use that substitution and that there's a special function of that form?", the answer is mostly just experience in working with this sort of thing. Regardless of method, antidifferentiating is primarily about pattern matching to existing results. That's why integral tables are a thing but derivative tables generally aren't.
A: You can find a Taylor series for $\sin(x^3)$ and integrate that.
Since the series for $\sin(x)$ is
$$\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
Substituting $x^3$ for $x$, the series for $\sin(x^3)$ is
$$\sum_{n=0}^\infty\frac{(-1)^nx^{6n+3}}{(2n+1)!}$$
Integrating that yields:
$$\sum_{n=0}^\infty\frac{(-1)^nx^{6n+4}}{(6n+4)(2n+1)!}$$
A: Another line of input to add to the answers here: There was a good post on MathOverflow called "Why is differentiating mechanics and integrating art?":
https://mathoverflow.net/questions/66377/why-is-differentiating-mechanics-and-integration-art
This is a multi-faceted question with a number of different parts. The most upvoted answer basically says that differentiation has enough rules that let you break down pretty much any expression, but integration does not. To me though that's not entirely satisfactory since it doesn't really detail what the "deep" reason is why integration doesn't have those rules but differentiation does. Other answers provide other pieces of the answer: differentiation is a local operation, but integration is global. Integration involves taking into account the behavior at a distant point from the point the integral is evaluated at, but differentiation is only about what is around that particular point. Another one, and perhaps the easiest, is just that some sets are closed under one operation but not its inverse: consider the natural numbers: the number 2 is easily squareable, but its square root is an irrational number. The elementary functions are the same way viz. differentiation and integration.
Regarding your specific integral, you should think about what it would look like if it "could" be integrated. In particular, were the integrand $3x^2 \sin(x^3)$, then integration is simple, by substitution. You can use the "reverse chain rule" of integration by substitution to get back to an antiderivative. But if I leave off the $3x^2$, that is no longer possible: we just have $\sin(x^3)$, and we cannot:
a) use linearity, as it's not a sum (involutary inverse of differentiation linear laws)
b) use integration by parts, as it's not a product (the only possible choice is to take $\sin(x^3)$ as $u$, then $dv$ must be $dx$, but that doesn't help the situation any if you try it) (inverse of differentiation product rule)
c) use partial fractions, as it's not a rational function,
d) use trigonometric integrals as it has a power inside the trig function.
In other words, none of the elementary integrating methods, which are essentially reverses of differentiation rules, will work. (FWIW trig integrals are essentially just further application of parts and substitution so are not really anything "new".)
To understand Wolfram's answer you need two things. One is the function it used: $\Gamma(a, x)$ which is called the "incomplete gamma function". The gamma function is a generalization of the factorial $n!$ to real numbers $x!$. Bernoulli first posed that problem, then Euler gave the solution, back in the 1700s. The gamma function is given by
$$\Gamma(s) = \int_{0}^{\infty} e^{-t} t^{s-1} dt$$
then $x! = \Gamma(x+1)$, while the incomplete gamma function is a further generalization, derived later by someone else whose name this author doesn't know (if it is known at all), given by
$$\Gamma(s, x) = \int_{x}^{\infty} e^{-t} t^{s-1} dt$$
This incomplete gamma function provides, essentially, a new differentiation rule:
$$\frac{d}{dx} \Gamma(s, x) = e^{-x} x^{s-1}$$
which turns out to be sufficient to solve the problem. The other component you will need is something known as Euler's formula, which is a key result of complex variables. Yes, to deal with this problem you have to go outside the real numbers into the complex numbers. Euler's formula is
$$e^{ix} = \cos(x) + i \sin(x)$$
and also note that from this, we have
$$e^{-ix} = \cos(x) - i \sin(x)$$
Now you can derive some formulas for $\cos$ and $\sin$ in terms of $e^{ix}$ and $e^{-ix}$. That is, we can convert sines and cosines to exponentials, which are what appear in the incomplete gamma function. I will leave it for the thread starter, to put these pieces together and find Wolfram's solution by hand.
