Is there really no analogue of the derivative product rule for integrals, or we just haven't found one yet?

Question

For a product rule for integrals, I am not talking about integration by parts. That particular formula uses a integral of products inside the formula itself. The derivative product rule does not use a derivative of products inside the formula. I have never seen a book that proves that there is in fact no analogous formula for the integral of a product. Some books prove that there is no formula for the roots of a general 5th degree or higher polynomial function. Can someone prove that there is no such formula for the integral of a product? There is a formula for the integral of a sum, so maybe someone will discover that there is a formula for the integral of a product. I apologize if my notion of formula is not precise enough, but maybe there is a book that precisely defines what a formula is.

Answer 1

I thought there was already such a question here, but I did not find it.

As the comments say, there is no simple formula for $\int f g dx$ in terms of $\int f dx$ and $\int g dx$. There are lots of ways to see this.

(A) $$ \int x\;dx\quad\text{and}\quad \int\frac{1}{x^2}\;dx\quad \text{are rational functions, but}\quad \int\frac{1}{x}\;dx\quad\text{is not} . $$ (B) $$ \int x e^{x^2}\;dx\quad\text{and}\quad \int\frac{1}{x}\;dx\quad \text{are elementary functions, but}\quad\int e^{x^2}\;dx\quad\text{is not} . $$ You come up with what "simple formula" means, then there should be an example like these using that notion of "simple formula".

Answer 2

If we look for any arbitrary $f(x)$ and $g(x)$, there are only 2 combined derivatives that include the product of these function (and their individual derivatives): The quotient rule and the product rule. As its clear we are not dividing by $g^2(x)$, lets focus on the product rule: $$\frac{d}{dx}\left(f(x)g(x)\right) = f'(x)g(x) + f(x)g'(x)$$

The (indefinite) integral represents the anti-derivative of a function, which can be described as: Find a function that if differentiated over $x$ yields the function inside the integral. If Im correctly, you want to find a way that allows us to compute a integral of the following shape: $$\int f(x)g(x)dx$$ as a function without the integrated product of both terms. Sadly, if we look at the definition of the product rule, one might notice 2 things: First there are 2 products and both products contain both $f(x)$ and $g(x)$ or their derivatives. Therefore you are always stuck with the following deduction: $$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x)dx$$ $$f(x)g(x)- \int f(x)g'(x)dx = \int f'(x)g(x)dx$$ Which equals to the integration by parts rule and does not remove the integral of the product in its equation.