How can you prove that a function has no closed form integral? In the past, I've come across statements  along the lines of &quot;function $f(x)$ has no closed form integral&quot;, which I assume means that there is no combination of the operations:

addition/subtraction
multiplication/division
raising to powers and roots
trigonometric functions
exponential functions
logarithmic functions

which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example.
What sort of techniques are used to prove statements like this? What is this branch of mathematics called?

Merged with &quot;How to prove that some functions don't have a primitive&quot; by Ismael:
Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions.
I wonder how we can prove that kind of assertion?
It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ non-constant, the antiderivative
$$f(x)\exp(g(x)) \, \mathrm dx$$
can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution to the differential equation:
$$f = h' + hg$$
$e^{x^2}$ is another classic example of such a function with no elementary antiderivative.
I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes:
https://ksda.ccny.cuny.edu/PostedPapers/liouv06.pdf
Liouville's original paper:

Liouville, J. &quot;Suite du Mémoire sur la classification des Transcendantes, et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients.&quot; J. Math. Pure Appl. 3, 523-546, 1838.

Michael Spivak's book on Calculus also has a section with a discussion of this.
The general pattern is always the same: (1). Study properties of derivatives of closed form functions. (2). If your function $f(x)$ fails one of those properties, then it has no closed form antiderivative.

Here is how to get started: First, consider derivatives of rational functions. Notice that their pole orders at finite points are never 1 (more generally: the residue is 0). If $f(x)$ fails this property, then you know that its antiderivative can not be a rational function. Sounds simple, but the surprising part is that this can be generalized quite easily to logarithmic extensions (handling exponential extensions in Risch algorithm is more involved, and handling algebraic extensions is more involved still).

Take for example a function $f(x) \in \mathbb{C}(x, \ln(x))$. For each irreducible factor $p \in \mathbb{C}(x)[ \ln(x) ]$ in the denominator of $f(x)$, you can define a corresponding residue (which will be algebraic over $\mathbb{C}(x)$). A necessary condition for $f(x)$ to have an elementary anti-derivative is for these residues to be constants  (in the more basic case, $f(x) \in \mathbb{C}(x)$, residues are always constants, and thus, rational functions always have an elementary antiderivative). See "Liouville's theorem (differential algebra)" on wikipedia for more.