How to find $\int x^2e^{x^2}dx$?

Question

How to find $\int x^2e^{x^2}dx$? I tried integration by parts following ILATE rule but it's not working.Please help!! What should I take as first function ? If it's not integrable can you atleast tell how to find the value of definite integration x=1 to x=2 ?

Answer 1

There is no analytical closed form for this integral. It must be evaluated by some numerical procedure, like Gauss-Legendre, for example.

Answer 2

The gaussian error function is defined through: $$\frac{2}{\sqrt{\pi}}\int_{0}^{t}e^{-x^2}\,dx = \text{Erf}(t).$$ In a similar way, the imaginary error function is defined through: $$\frac{2}{\sqrt{\pi}}\int_{0}^{t}e^{x^2}\,dx = \text{Erfi}(t).$$ It follows that: $$ \int_{0}^{t} x^2\,e^{x^2}\,dx = \int_{0}^{t}\frac{x}{2}\left(2x\, e^{x^2}\right)\,dx = \frac{t}{2}\,e^{t^2}-\sqrt{\pi}\;\text{Erfi}(t). $$ An accurate approximation for the integral over $[1,2]$ can be achieved by integrating termwise the Taylor series of $e^{x^2}$:

$$ \int_{1}^{2}e^{x^2}\,dx = \sum_{n\geq 0}\int_{1}^{2}\frac{x^{2n}}{n!}\,dx = \sum_{n\geq 0}\frac{2^{2n+1}-1}{(2n+1)\,n!}.$$ A more accurate approximation can be achieved by exploiting the continued fraction representation of the Dawson's integral.