A not-so-rigorous proof of the Gaussian integral


A not-so-rigorous proof of the Gaussian integral

The usual proof for the value of the Gaussian integral involves the use of polar coordinates. In this post, I will attempt to prove that the Gaussian integral $\displaystyle \int_{-\infty}^{\infty}e^{-x^2}\; dx$ has a value of $\sqrt \pi$ with a carefree usage of other methods. I assume this can actually be made rigorous, but as of the moment I do not have the time to read on the material behind the theorems I will use below.

Consider the function $I:(0,\infty)\to \mathbb R$ whose mapping is defined by

\begin{equation} I(t) = \int_{0}^{\infty}\frac{e^{-(tx)^2}}{1+x^2}\; dx \end{equation}

Taking the derivative of $I$, using differentiation under the integral sign, one obtains

\begin{align*} I'(t) &= \int_{0}^{\infty}\frac{e^{-(tx)^2}}{1+x^2}\cdot (-2tx^2)\; dx \\[10pt] &= -2t \int_{0}^{\infty} \frac{e^{-(tx)^2}(1+x^2)-e^{-(tx)^2}}{1+x^2}\; dx \\[10pt] &= -2t \int_{0}^{\infty} e^{-(t^2x^2)} \; dx + 2t\cdot I(t) \end{align*}

Now, if we consider the substitution $x =u/t$ for the left term on the last line, we have
\begin{align*} -2t \int_{0}^{\infty} e^{-(t^2x^2)} \; dx &= -2t \int_{0}^{\infty} \frac{e^{-u^2}}{t}\; du \\[10pt] &= -2\int_{0}^{\infty} e^{-u^2}\; du \end{align*}
which implies that this term is actually independent of $t$, i.e., a constant. Let us denote this term by $-K$. Rearranging our earlier expression, we obtain the linear homogenous first-order differential equation

\begin{equation} I'(t) -2t\cdot I(t) = - K \end{equation}

The solutions to equations of this form are well-studied, and it can be shown through known methods that $I$ must be of the form

\begin{equation} I(t) = -Ke^{t^2}\int_0^te^{-s^2}\; ds + Ce^{t^2} \end{equation}

for some constant $C$. Now, from the definition of $I$, one can verify the following properties.
\begin{align*} \lim_{t\to 0}I(t) = \frac{\pi }{2}; \;\; \lim_{t\to\infty} I(t) = 0 \end{align*}
For a brief explanation, we assume for $I$ to behave fairly well near $0$, so that the limit of $I(t)$ as $t$ approaches $0$ can be computed by its evaluation at $t=0$. Meanwhile, since $e^{-t^2x^2}$ tends to approach $0$ as $t$ grows large, then we expect $I(t)$ to tend to $0$ as well.

Using the first property, we obtain

\begin{equation} \frac{\pi}{2} = \lim_{t\to 0}I(t) = -Ke^0\cdot \int_0^0e^{-s^2}\;dx +Ce^0 = C \end{equation}

Next, using the second property, we expect the following to hold.
\begin{equation} \lim_{t\to\infty}e^{t^2}\left(-K\int_0^te^{-s^2}\; ds + C\right) = 0 \end{equation}
Since $e^{t^2}$ tend to infinity as $t\to\infty$, a necessary way for this limit to possibly exist is for the enclosed expression to tend to $0$, that is, for the following to be true.
\begin{align*} 0 &= \lim_{t\to\infty}\left(-K\int_0^te^{-s^2}\; ds + C\right) \\[10pt] &= -K\int_0^\infty e^{-s^2}\; ds + C \\[10pt] &= -K\cdot \frac{K}{2} + C \end{align*}
Should this hold, then, we can see that
\begin{align*} \lim_{t\to\infty}e^{t^2}\left(-K\int_0^te^{-s^2}\; ds + C\right) &= \lim_{t\to\infty}\frac{\displaystyle -K\int_0^te^{-s^2}\; ds + C}{e^{-t^2}} \\[10pt] &= \lim_{t\to\infty} \frac{-Ke^{-t^2}}{(-2t)\cdot e^{-t^2}} = 0 \end{align*}
which is what we wish to hold.

To summarize, the following equations must hold.
\begin{align*} \frac{\pi}{2} = C; \;\; 0 = -\frac{K^2}{2}+C \end{align*}
which implies $C = 0$ and $K = \sqrt \pi$. However, using symmetry, we can see that
\begin{align*} \sqrt \pi = K = 2\int_0^\infty e^{-u^2}\; du = \int_{-\infty}^{\infty}e^{-s^2}\; ds \end{align*}
which is precisely the Gaussian integral.







Undergraduate mathematics student @UP Diliman