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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.

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

2 Answers 2

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Using Maclaurin series of exponential function, we will obtain $$ e^{\large x^2}=\sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}. $$ Hence $$ \begin{align} \int_{x=0}^1\ e^{\large x^2}\ dx&=\int_{x=0}^1\ \sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}\ dx\\ &=\sum_{n=0}^\infty\int_{x=0}^1\ \frac{x^{\large 2n}}{n!}\ dx\\ &=\left.\sum_{n=0}^\infty\frac{x^{\large 2n+1}}{(2n+1)\ n!}\right|_{x=0}^1\\ &=\sum_{n=0}^\infty\frac{1}{(2n+1)\ n!}\\ &\approx 1.4626517459. \end{align} $$

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    $\begingroup$ FYI, the summation will yield accuracy $10^{-6}$ for $n=7$. $\endgroup$
    – Tunk-Fey
    May 28, 2014 at 14:53
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Hint: check how the improper integration of $\int{e^{-x^2}}$ happens, and use an $x\rightarrow{ix}$ mapping.

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