# Definite Integral of $e^{\large x^2}$

I know there's no elementary antiderivative of $e^{\large x^2}$.

But what if there's a definite integral like

$$\int_0^1e^{\large x^2}\ dx\ ?$$

I tried using basic definite integral property like $\displaystyle\int^a_0f(x)\ dx =\int^a_0f(a-x)\ dx$ but I could see no way out.

• from Mathematica $\frac{1}{2} \sqrt{\pi } \text{erfi}(1)$ can be expressed interm of error function
– S L
May 28, 2014 at 14:08
• May 28, 2014 at 17:19

Using Maclaurin series of exponential function, we will obtain $$e^{\large x^2}=\sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}.$$ Hence \begin{align} \int_{x=0}^1\ e^{\large x^2}\ dx&=\int_{x=0}^1\ \sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}\ dx\\ &=\sum_{n=0}^\infty\int_{x=0}^1\ \frac{x^{\large 2n}}{n!}\ dx\\ &=\left.\sum_{n=0}^\infty\frac{x^{\large 2n+1}}{(2n+1)\ n!}\right|_{x=0}^1\\ &=\sum_{n=0}^\infty\frac{1}{(2n+1)\ n!}\\ &\approx 1.4626517459. \end{align}
• FYI, the summation will yield accuracy $10^{-6}$ for $n=7$. May 28, 2014 at 14:53
Hint: check how the improper integration of $\int{e^{-x^2}}$ happens, and use an $x\rightarrow{ix}$ mapping.