Definite Integration of the form e^-x^2, Gaussian Integral, Euler-Poisson Integral
Theorem
The Gaussian function $f(x) := e^{-x^2}$’s integral over the entire domain is as follows.
$$ \int_{-\infty}^{\infty} e^{-x^2} dx= \sqrt{\pi} $$
Description
Physicist Kelvin is said to have left the remark that “one who finds this integral obvious is a mathematician”. It is also known by other names such as Gaussian integral, or Euler-Poisson integral.
It’s a shocking integration for high school students and especially crucial for statistics. That’s because, while you can’t find a primitive function within the high school curriculum rendering the calculation impossible, the probability of normal distribution is tacitly used in the statistics part.
Proof
Strategy: Create $x$ and $y$ independent of each other, then convert to polar coordinates to turn it into an integral over a closed interval. There is a method said to be understandable at a high school level, through the Pappus-Guldin theorem, to prove by calculating the volume of solids of revolution, but essentially, this proof is the same and includes improper integrals, making it difficult to be deemed as high school level.
If we denote by $\displaystyle I = \int_{-\infty}^{\infty} e^{-x^2} dx$ then it can also be represented by $\displaystyle I = \int_{-\infty}^{\infty} e^{-y^2} dy$. Since $x$ and $y$ are independent,
$$ \begin{align*} I^2 =& \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy \\ =& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-( x^2 + y^2 ) } dxdy \end{align*} $$
Converting 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
Improper integral over a half-line
$$ \int_{0}^{\infty} e^{-x^2} dx= {{\sqrt{\pi}} \over {2}} $$
If the integration range is from $0$ to $\infty$, polar coordinates can’t be used. However, looking at the shape of the Gaussian function, one can guess without calculating that it would be halved for $x=0$ since it is an even function, but since it’s an improper integral over an infinitely long range, let’s verify it accurately.
Proof
$$ \int_{0}^{\infty} e^{-x^2} dx $$
By substituting like $x :=-y$,
$$ x \rightarrow 0,\ y \rightarrow 0 \\ x \rightarrow \infty,\ y \rightarrow -\infty \\ x^2=y^2 $$
Since $dx=-dy$,
$$ \int_{0}^{\infty} e^{-x^2} dx = -\int_{0}^{-\infty} e^{-y^2} dy=\int_{-\infty}^{0} e^{-y^2} dy $$
The integration variable does not affect the definite integral, so it can be written as $\displaystyle \int_{-\infty}^{0} e^{-y^2} dy=\int_{-\infty}^{0} e^{-x^2} dx$, 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 above results,
$$ \int_{0}^{\infty} e^{-x^2} dx=\frac{1}{2}\int_{-\infty}^{\infty} e^{-x^2} dx=\frac{1}{2}\sqrt{\pi} $$
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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