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

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?

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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 9:24
Do you mean like $$\int e^{-x^2}dx$$ and why is this tagged [logic] anyway? – Asaf Karagila Aug 24 '12 at 9:27
The question is unclear, but perhaps you mean How can you prove that a function has no closed form integral? – Rahul Aug 24 '12 at 9:27
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

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.