Proof of the Gaussian Integral Using the Wallis Integral

Proposition

The following holds:

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$$

Proof

Lemma

For $x\ne 0$, the following inequalities hold:

$$1-x^2<e^{-x^2}<\frac{1}{1+x^2}$$

Proof of the lemma

For $x\ne 0$,

$$e^x>1+x$$

In fact, the right-hand side is the Taylor expansion truncated after the second term of the left-hand side, representing the tangent line of $y=e^x$ at $x=0$.

Since $y=e^x$ is convex (as $e^x>0$ even after two derivatives), $y=e^x$ lies above the tangent line $y=1+x$.

Therefore, $e^x>1+x$ holds.

By substituting $-x^2$ for $x$ in this inequality, we get $1-x^2<e^{-x^2}$.

Also, substituting $x^2$ for $x$ and taking the reciprocal gives $e^{-x^2}<\frac{1}{1+x^2}$.

Substitution in the Wallis Integral

Consider the Wallis integral

$$I_m={\int }_0^{\frac{\pi }{2}}{\sin }^{m}xdx$$

Substitution $y=\cos x$

By subsituting $y=\cos x$ for $I_{2n+1}$,

$$I_{2n+1}={\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\sin xdx={\int }_1^0(1-y^2{)}^n(-\frac{dy}{dx})dx={\int }_0^1(1-y^2{)}^{n}dy={\int }_0^1(1-x^2{)}^{n}dx$$

Substitution $y=\cot x$

By subsituting $y=\cot x$ for $I_{2n+1}$,

$$I_{2n-2}={\int }_0^{\frac{\pi }{2}}{\sin }^{2n}x\frac{1}{{\sin }^{2}x}dx={\int }_{\mathrm{\infty }}^0\frac{1}{(1+y^2{)}^n}(-\frac{dy}{dx})dx={\int }_0^{\mathrm{\infty }}\frac{1}{(1+y^2{)}^n}dy={\int }_0^{\mathrm{\infty }}\frac{1}{(1+x^2{)}^n}dx$$

Evaluation of the Gaussian Integral

Let

$$I={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx$$

By substituting $x=\sqrt{n}y$,

$$I={\int }_0^{\mathrm{\infty }}{e}^{-n{y}^2}\sqrt{n}dy=\sqrt{n}{\int }_0^{\mathrm{\infty }}{e}^{-n{x}^2}dx$$

From the lemma,

$$\sqrt{n}{\int }_0^1(1-x^2{)}^{n}dx<\sqrt{n}{\int }_0^1{e}^{-n{x}^2}dx<I<\sqrt{n}{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2{)}^n}$$

Since the leftmost and rightmost terms are in the same form as the Wallis integral after substitution, we ultimately have:

$$\sqrt{n}{I}_{2n+1}<I<\sqrt{n}{I}_{2n-2}$$

As seen from Wallis' Formula,

$$\underset{m\to \mathrm{\infty }}{lim}\sqrt{m}{I}_m=\sqrt{\frac{\pi }{2}}$$

so,

$$\sqrt{n}{I}_{2n+1}=\frac{\sqrt{n}}{\sqrt{2n+1}}\sqrt{2n+1}{I}_{2n+1}\to \frac{1}{\sqrt{2}}\sqrt{\frac{\pi }{2}}=\frac{\sqrt{\pi }}{2}(n\to \mathrm{\infty })$$

$$\sqrt{n}{I}_{2n-2}=\frac{\sqrt{n}}{\sqrt{2n-2}}\sqrt{2n-2}{I}_{2n-2}\to \frac{1}{\sqrt{2}}\sqrt{\frac{\pi }{2}}=\frac{\sqrt{\pi }}{2}(n\to \mathrm{\infty })$$

Therefore,

$$I=\frac{\sqrt{\pi }}{2}$$

Adjustment of the Integration Interval

Since the integrand is an even function,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=2I=\sqrt{\pi }$$

Thus, the result is proved.