Can all Integration and differentiation of a real function form be determined?

Question

Can every integration and differentiation on $\mathbb{R}^2$ be determined exactly?

I am curious of this, because I know that there are some integration and differentiation that do not yet have a way to solve them.

However, I also never heard of any theorem that state that there are some integrals and differentiations that cannot be solved.

So, is there any theorem?

I'm sorry, but your question isn't clear. Are you asking if every function defined from a subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ is integrable and differentiable, or if all derivatives/integrales can be expressed in terms of simple functions or about function infinitely-differentiable/integrable? Please clarify what you mean. — andreas.vitikan, Aug 24 '12 at 09:24
Do you mean like $$\int e^{-x^2}dx$$ and why is this tagged [logic] anyway? — Asaf Karagila, Aug 24 '12 at 09:27
The question is unclear, but perhaps you mean How can you prove that a function has no closed form integral? — , Aug 24 '12 at 09:27
You have to clarify what you mean by "determined exactly" and whether you are talking about integral with number values or integrals like the one Asaf mentions in his comment. — Kaveh, Aug 24 '12 at 13:34
sorry, guys. Ehat Rahul Narain says is what I intended to ask. I think one should close this question. Thanks guys. — Tao Mao, Aug 25 '12 at 13:38

score 0 · Answer 1 · answered Dec 16 '14 at 03:11

I don't know where to find the proof, but if you restrict your domain to $\mathbb{R}$, then the following equation is known to have no solution in terms of elementary functions. $$\int e^{x^2}dx$$ In addition, the elliptic integrals (arc-length of an ellipse) do not necessary have solutions in terms of elementary functions. For instance: $$L=\int \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$ Take the equation of an ellipse: $$ax^2+by^2=c^2$$ Where $a\neq b$. If $a=b$, then you have a circle, which has a trivial arc length. Use implicit differentiation to solve: $$2axdx+2bydy=0 \rightarrow axdx=-bydy \rightarrow \frac{dy}{dx}=\frac{-ax}{by}$$ Use the first quadrant, so: $$y=\sqrt{\frac{c^2-ax^2}{b}}$$ So combining equations gives: $$L=\int \sqrt{1+\left(\frac{-ax}{b\sqrt{\frac{c^2-ax^2}{b}}}\right)^2}$$ $$L=\int \sqrt{1+\frac{a^2x^2}{b(c^2-ax^2)}}dx$$ This is the formula for the arc-length of an ellipse, and it has no elementary solution. It is defined as an elliptic integral of the second kind.

Can all Integration and differentiation of a real function form be determined?

1 Answers1