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As we all know that integral of the function $e^{-x^2}$ is a non elementary function in the form of an error function. But can we calculate it's values at particular points for example the integral from $-1$ to $1$ in a numerical value.

Also say if we can not then van we construct an elementary function using the difference between $2$ values equal to the area under the the curve and finding the area manually?

Aarush Saharan
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    Generally, we can't calculate its values; we can only approximate them. – JonathanZ Apr 12 '24 at 14:56
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    FYI, there are points for which the value will be nice. For example, there exist values for $a$ and $b$ (in fact, infinitely many, even uncountably many) such that the integral from $a$ to $b$ is exactly $\frac{1}{2}.$ However, those values for $a$ and $b$ will likely not be expressible in any nice closed form manner. (I don't know if it's known that no such values for $a$ and $b$ exist, as this is different from the indefinite integral having no closed form, but it's almost certainly the case that no such values for $a$ and $b$ are known, otherwise such values would probably be widely known.) – Dave L. Renfro Apr 12 '24 at 15:41

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First please check out: Is there really no way to integrate $e^{-x^2}$?

There is no form of the integral in terms of elementary function. But you can use infinite series: $$ \int_a^b e^{-x^2}dx = \int_a^b\sum_{n=1}^\infty \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=1}^\infty\frac{(-1)^n}{n!}\int_a^bx^{2n}dx=\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)n!}\left(a^{2n+1}-b^{2n+1}\right) $$