I'm going to take an example to elaborate my question.
My teacher said that - $\int \sin(x^3)dx$ is unsolvable for all x. Just recently, he said that $\int_{-\pi/6}^{\pi/6} \sin(x^3) = 0$. Now I get why its zero but my question is -
- Indefinite Integration is an anti-derivative process. If a function can be differentiated, then why can't it be integrated so as to get its primitive? Are those functions like some "unsolved mysteries of mathematics" ? Because if we're integrating something then its almost understood that its primitive exists (at least thats what I think).
- What about an expression makes it non-integrable?
- Can we look (or examine it a little bit) and say that the expression is integrable or not? If so, what are the things we should look out for?
Whether an indefinite integral has a closed form is a highly technical area.
That doesn’t mean the indefinite integral doesn’t exist, it just means it doesn’t have a certain form.
Just as some real numbers are irrational, some indefinite integrals exist and can’t be written in a closed form in terms of other “usual” functions. In any event, these functions are integrable.
A useful non-integral example of a similar thing is the Lambert W-function. Given an $x\geq 0,$ $y=W(x)$ if $x=ye^y.$
We can’t write $W$ in terms of “elementary functions.” (Proving that requires some extremely tricky techniques.) But we can compute $W$ numerically, and we can tell specific values, like $W(0)=0, W(e)=1.$ We can prove things about $W.,$ like that it is an increasing function.
Mathematicians do this a lot. We can prove properties of $F(x)=\int\sin(x^3)\,dx,$ even if we can’t write $F(x)$ in a useful way.
There are functions which are actually not integrable, but they are highly strange, and in particular, those functions cannot be written in a closed form, either.