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 ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx}$ 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 $I:(0,\mathrm{\infty })\to \mathbb{R}$ whose mapping is defined by

$$I(t)={\int }_0^{\mathrm{\infty }}\frac{e^{-(tx{)}^2}}{1+x^2}dx$$

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

$$\begin{array}{rl}{I}^{\prime }(t)& ={\int }_0^{\mathrm{\infty }}\frac{e^{-(tx{)}^2}}{1+x^2}\cdot (-2t{x}^2)dx\\ & =-2t{\int }_0^{\mathrm{\infty }}\frac{e^{-(tx{)}^2}(1+x^2)-e^{-(tx{)}^2}}{1+x^2}dx\\ & =-2t{\int }_0^{\mathrm{\infty }}{e}^{-(t^2{x}^2)}dx+2t\cdot I(t)\end{array}$$

Now, if we consider the substitution $x=u/t$ for the left term on the last line, we have
$$\begin{array}{rl}-2t{\int }_0^{\mathrm{\infty }}{e}^{-(t^2{x}^2)}dx& =-2t{\int }_0^{\mathrm{\infty }}\frac{e^{-u^2}}{t}du\\ & =-2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du\end{array}$$
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

$$I^{\prime }(t)-2t\cdot I(t)=-K$$

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

$$I(t)=-K{e}^{t^2}{\int }_0^t{e}^{-s^2}ds+C{e}^{t^2}$$

for some constant $C$. Now, from the definition of $I$, one can verify the following properties.
$$\begin{array}{r}\underset{t\to 0}{lim}I(t)=\frac{\pi }{2};\underset{t\to \mathrm{\infty }}{lim}I(t)=0\end{array}$$
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 $e^{-t^2{x}^2}$ 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

$$\frac{\pi }{2}=\underset{t\to 0}{lim}I(t)=-K{e}^0\cdot {\int }_0^0{e}^{-s^2}dx+C{e}^0=C$$

Next, using the second property, we expect the following to hold.
$$\underset{t\to \mathrm{\infty }}{lim}{e}^{t^2}(-K{\int }_0^t{e}^{-s^2}ds+C)=0$$
Since $e^{t^2}$ tend to infinity as $t\to \mathrm{\infty }$, 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.
$$\begin{array}{rl}0& =\underset{t\to \mathrm{\infty }}{lim}(-K{\int }_0^t{e}^{-s^2}ds+C)\\ & =-K{\int }_0^{\mathrm{\infty }}{e}^{-s^2}ds+C\\ & =-K\cdot \frac{K}{2}+C\end{array}$$
Should this hold, then, we can see that
$$\begin{array}{rl}\underset{t\to \mathrm{\infty }}{lim}{e}^{t^2}(-K{\int }_0^t{e}^{-s^2}ds+C)& =\underset{t\to \mathrm{\infty }}{lim}\frac{{\displaystyle -K{\int }_0^t{e}^{-s^2}ds+C}}{e^{-t^2}}\\ & =\underset{t\to \mathrm{\infty }}{lim}\frac{-K{e}^{-t^2}}{(-2t)\cdot e^{-t^2}}=0\end{array}$$
which is what we wish to hold.

To summarize, the following equations must hold.
$$\begin{array}{r}\frac{\pi }{2}=C;0=-\frac{K^2}{2}+C\end{array}$$
which implies $C=0$ and $K=\sqrt{\pi }$. However, using symmetry, we can see that
$$\begin{array}{r}\sqrt{\pi }=K=2{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-s^2}ds\end{array}$$
which is precisely the Gaussian integral.