In my real analysis course I was given this exercise:

Calculate $\displaystyle{\int_0^1e^{x^2}dx}$.

What I did was to write $\displaystyle{e^{x^2}=\sum_{n=0}^\infty\dfrac{x^{2n}}{n!}}$ and conclude that $\displaystyle{\int_0^1e^{x^2}dx=\sum_{n=0}^\infty\dfrac{1}{n!(2n+1)}}$. I still don't know if that is correct. My question is:

Is my answer correct? In any case do we know the exact value of $\displaystyle{\sum_{n=0}^\infty\dfrac{1}{n!(2n+1)}}$? Is there another way calculating this integral?


2 Answers 2


Yes, your series is correct.

$e^{x^2}$ does not have an elementary antiderivative. Your integral can be written as $\dfrac{\sqrt{\pi}}{2} \text{erfi}(1)$, where the special function erfi is defined as $$ \text{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2}\ dt $$

  • $\begingroup$ That is nice. Thanks! $\endgroup$
    – George
    Dec 3, 2012 at 22:48

The answer expressed as a summation is correct. The justification of interchanging integral and summation is justified by the fact that the power series expansion of $e^{x^2}$ converges uniformly in $[0,1]$.


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