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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?

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up vote 6 down vote accepted

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 $$ \text{erfi}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{t^2}\ dt $$

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That is nice. Thanks! – George Dec 3 '12 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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Observe that $$\left(\int_0^1 e^{x^2}dx\right)\cdot\left(\int_0^1 e^{y^2}dy\right)=\int_0^1\int_0^1 e^{x^2+y^2}dxdy.$$ Now, if you change to polar coordinates, you can get a closed form answer. After finding this answer, you will have to take the square root since the left-hand side above is actually your integral squared.

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This may not be so easy to evaluate in polar coordinates since it is a rectangular region... – Morgan Sherman Dec 3 '12 at 22:33
My apologies, I was thinking of $e^{-x^2}$. – Clayton Dec 3 '12 at 22:38
... on $(0,\infty)$ or $(-\infty,\infty)$ – Robert Israel Dec 3 '12 at 22:40

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