TITLE: How to calculate $\int_{}^{}e^{x^2}dx$
QUESTION [0 upvotes]: This is from math contest it was asked like that 
how do you calculate this integrale 
$$\int_{}^{}e^{x^2}dx$$. 
My thoughts:

I tired integration by part 
Integration by substitution

REPLY [2 votes]: As someone mentioned, there is no solution to this expressible with elementary functions. However that is kind of a boring answer really, so I'm gonna try and give something more exciting.
You could try rewriting the function like this:
$$\int e^{x^2} dx = \int \underset{\overbrace{\text{integrate}}}{\left(2x e^{x^2}\right)} \underset{\overbrace{\text{differentiate}}}{\left(\frac{1}{2x}\right)} dx$$
with integration by parts, integrating the numerator ($2x$ is the internal derivative, right?), differentiating the denominator and then iterate. You would then be getting an expression with a negative exponent power series multiplied with the original function $P(x^{-1})e^{x^2}$, where $$P(x) =\sum_{k=0}^\infty p_kx^k$$

If you are fine with power series you might also be satisfied with integrating the original function's Taylor expansion term-wise, which would just be substituting $x^2$ into $x$ for exponential function power series $$\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}, /\text{ sub. } x\rightarrow x^2 /, \exp(x^2) = \sum_{k=0}^\infty \frac{x^{2k}}{k!}$$

A third technique would be considering the Hermite basis for the Fourier transform which can be used to show that Fourier transform can reach arbitrary close to square integrable functions. This basis is in the consecutive derivatives of the function $\exp(-x^2)$, so it's actually only true if we rotate your function and look along the imaginary axis and then differentiate. Although the details of actually doing this would probably be a bit technical.
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