Definite Integral of the Form e^-x^2, Gaussian Integral, Euler-Poisson Integral
Theorem
The definite integral over the entire domain of the Gaussian function $f(x) := e^{-x^2}$ is as follows.
$$ \int_{-\infty}^{\infty} e^{-x^2} dx= \sqrt{\pi} $$
Explanation
The physicist Kelvin is said to have remarked that ‘a mathematician is one to whom this integral seems obvious’. It goes by other names such as the Gaussian integral or the Euler-Poisson integral.
It is a shocking integral for high school students, and it is also an enormously important integral, especially in statistics. This is understandable, since within the high school curriculum the antiderivative cannot be found so the computation cannot be carried out, while in the statistics part the probabilities of the normal distribution are implicitly used.
Proof
Strategy: After creating a $y$ that is independent of $x$, we switch to polar coordinates and convert it into an integral over a closed interval. There is said to be a way of proving it even at the high school level by computing the volume of a solid of revolution via the Pappus-Guldinus theorem, but it is essentially the same as this proof, and moreover that method also involves an improper integral, so it can hardly be called high school level.
If we let $\displaystyle I = \int_{-\infty}^{\infty} e^{-x^2} dx$, then it can also be written as $\displaystyle I = \int_{-\infty}^{\infty} e^{-y^2} dy$. Since $x$ and $y$ are independent,
$$ \begin{align*} I^2 =& \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy \\ =& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-( x^2 + y^2 ) } dxdy \end{align*} $$
Switching to polar coordinates,
$$ \begin{align*} I^2 =& \int_{0}^{2 \pi} \int_{0}^{\infty} e^{-r^2 } r dr d\theta \\ =& \int_{0}^{2 \pi} \left[ {{-e^{-r^2}} \over {2}}\right]_{0}^{\infty} d\theta \\ =& \int_{0}^{2 \pi} {{1} \over {2}} d\theta \\ =& \pi \end{align*} $$
Therefore,
$$ I =\sqrt{\pi} $$
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Corollary
Definite Integral over a Half-Line
$$ \int_{0}^{\infty} e^{-x^2} dx= {{\sqrt{\pi}} \over {2}} $$
On the other hand, if the integration range is from $0$ to $\infty$, polar coordinates cannot be used. Of course, from the shape of the Gaussian function, since it is an even function with respect to $x=0$, one can guess without bothering to compute that it becomes half, but since it is an improper integral with an infinitely long range, let us verify it precisely.
Proof
$$ \int_{0}^{\infty} e^{-x^2} dx $$
Here, substituting $x :=-y$,
$$ x \rightarrow 0,\ y \rightarrow 0 \\ x \rightarrow \infty,\ y \rightarrow -\infty \\ x^2=y^2 $$
Since $dx=-dy$,
$$ \int_{0}^{\infty} e^{-x^2} dx = -\int_{0}^{-\infty} e^{-y^2} dy=\int_{-\infty}^{0} e^{-y^2} dy $$
Since the variable of integration has no effect in a definite integral, we can write $\displaystyle \int_{-\infty}^{0} e^{-y^2} dy=\int_{-\infty}^{0} e^{-x^2} dx$, and therefore
$$ \int_{0}^{\infty} e^{-x^2} dx + \int_{-\infty}^{0} e^{-x^2} dx= 2\int_{0}^{\infty} e^{-x^2} dx=\int_{-\infty}^{\infty} e^{-x^2} dx $$
Using the result above,
$$ \int_{0}^{\infty} e^{-x^2} dx=\frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx=\frac{1}{2}\sqrt{\pi} $$
Generalization
The following generalized formula is widely used.
$$ \int_{-\infty}^{\infty} e^{-\alpha x^2} dx= \sqrt{\dfrac{\pi}{\alpha}} $$
See Also
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