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Integrals and derivated are related as an indefinite integral is an anti-derivative, however, I found it always perplexing that derivatives can be always found easily (well, at least the ones I saw during the past years) and there are so many ways to set a very challenging integral or even one that can't be solved at all (example when the integrand is $e^{x^2}$), is there any mathematical explanation behind this or it is just a "that is how things are" type of thing?

I tried going back to the definition of both and couldn't find anything helpful.

Sergio
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  • There is no "that is how things are" in math. Look at this answer. – azif00 Dec 24 '20 at 06:12
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    This is, to some extent, an artifact of the course you are in and how you think about math via concrete formulas. Talk to a professional mathematician who works in analysis and they'll tell you that integrals are way easier (nicer behavior) than derivatives! When they have to integrate they are happy, but when they have to differentiate they get nervous (sort of). – KCd Dec 24 '20 at 06:12
  • An analogy: can you explain why fractions, as a class of numbers, are better behaved under certain functions (adding, multiplying) but get wrecked under others (square roots, logarithms)? – KCd Dec 24 '20 at 06:14
  • We have that $(f \circ g)' (x) = f'(g(x)) g'(x)$. Consequently, one has the antiderivative identity $\int f'(g(x)) g'(x) = f \circ g$, which is useful, but not as useful as the chain rule. Unfortunately, there is no simple expression for $\int f \circ g$ in terms of $f$ and $g$. – Charles Hudgins Dec 24 '20 at 06:15
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    Maybe a spiritual answer: Finding anti-derivative is an "inverse" problem, and often these inverse problems are harder. For example, let $f(x) = x^x sin(x^2)$, if we want to know what $f(3)$ is, we have an instruction to do that. However if we want to know what $x$ such that $f(x) = 3$, then that's harder. Another such forward-inverse pair is factorization: If we want to compute $3\cdot 5\cdot 13\cdot 23$, that's not hard. But to find what are the factors to $2^{2020} -1$, then it may take some work. (As another forward-inverse instance is the relation between probability and statistics, etc.) – bonsoon Dec 24 '20 at 06:16
  • The derivative of a function is direct under derivation rules. An integration is not so simple. What do you want to know? – abiessu Dec 24 '20 at 06:17
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    @KCd While I agree with the sentiment, I think you've overstated things a bit. Derivatives are certainly easier to program a computer to do for functions that involve only arithmetic, exponentiation, logs, and trig functions. As far as I know, for example, Mathematica still pulls many of its integrals from tables compiled by overzealous (Russian?) mathematicians. Even very simple integrals can be devastatingly hard to write down symbolically (without simply defining a function to be the antiderivative). Not so with derivatives. – Charles Hudgins Dec 24 '20 at 06:18
  • @azif00 but how does this explain my main question? thank you for the link though it is useful – Sergio Dec 24 '20 at 06:34
  • @KCd by nicer behavior do you mean more predictable or what exactly? – Sergio Dec 24 '20 at 06:35
  • @GerryMyerson yes it is, thank you! Do I just delete my question? – Sergio Dec 24 '20 at 06:37
  • I mean the integral of a small function is small (that is important in analysis). The derivative of a small function could be huge, like $\sin(100 x)/10$. – KCd Dec 24 '20 at 06:37
  • You could delete, Sergio. Or, since there have already been some useful comments here, you could just wait it out and let users vote to close as a duplicate. – Gerry Myerson Dec 24 '20 at 06:40
  • @GerryMyerson turns out I can vote to close it myself then proceed to doing it, done. – Sergio Dec 24 '20 at 06:44
  • They aren't easier. It's just a question of naming functions. Take any power series expansion or Fourier series, the integral is equally easy as derivative. – mathreadler Dec 24 '20 at 06:44
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    Because doing infinite sums is harder than taking a single limit. – Allawonder Dec 24 '20 at 08:24

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