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.

  • $\begingroup$ So you want to relate $\int fg\,dx$ and $\int f\,dx,\int g\,dx$? $\endgroup$ – PNDas Nov 23 '20 at 12:33
  • $\begingroup$ Typically the integral of a product can be wildly different from the product of the integrals so I doubt such a formula would exist (e.g. take $ f(x) = \text{sin} (x), g(x) = (1+x^2)^{-1} $). You may be interested in the field of differential Galois theory which broadly tells you when an integral can be expressed in closed form, but again this is dependent on how exactly you choose to define 'closed form'. $\endgroup$ – backstrapp Nov 23 '20 at 12:40
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    $\begingroup$ As a note, if it did exist, it would have to be able to tell you about cases that both do and don't work. For instance, $\int x^2 \cos(x^3)$ can be done but $\int x \,\cos(x^3)$ can't (at least not without adding in other functions like the gamma function). Technically there is the Risch algorithm, but my understanding is that it is hundreds of pages long. $\endgroup$ – johnnyb Nov 23 '20 at 12:48
  • $\begingroup$ There is this, which isn't exactly integration by parts and does use $\int f$ and $\int g$ on the right-hand side (except they call them $f$ and $g$, because their left-hand side is written $\int f'g'$), but it still has you integrating products on the right-hand side. But then many of the derivative rules (including the product rule) invoke new derivatives on the right that were not on the left hand side. $\endgroup$ – David K Nov 23 '20 at 13:19

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".

  • $\begingroup$ By simple formula, I mean a composition of elementary functions in terms of the integrals of f and g. $\endgroup$ – user107952 Nov 23 '20 at 20:00

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.


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