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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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    sorry, guys. Ehat Rahul Narain says is what I intended to ask. I think one should close this question. Thanks guys.2012-08-25

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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.