The Gaussian Integral

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In this activity, we will take a look at a rather tricky improper integral.

Perhaps you have lain awake at night, thinking it is impossible to calculate $\int _{-\infty }^{\infty } e^{-x^2}dx$ . After all, we typically evaluate improper integrals by first computing a proper integral and then taking limits. That strategy uses the Second Fundamental Theorem of Calculus, but there is no closed-form expression for $\int e^{-x^2} dx$ . Thankfully, we have other methods at our disposal. But we’ll need to proceed with finesse since we don’t have the heavy machinery of multivariable calculus.

Let’s first define a few functions that will help us out. $\begin{align*} &F(t)=\int_{0}^{t} e^{-x^2}dx \\ &G(t)=\int_{0}^{1} \frac{e^{-t^2(1+x^2)}}{1+x^2} dx \\ &H(t)=F(t)^2+G(t) \end{align*}$ The first function looks relevant, but it’s not clear yet how the other ones will help us. Still, let’s move on by exploring $H$ . $\begin{align*} H(0)= F(0)^2+H(0)=\left[\int_{0}^{0} e^{-x^2}dx \right]^2+\int_{0}^{1} \frac{1}{1+x^2} dx = \left[ \answer[given]{0} \right]^2 + \answer[given]{\pi/4} = \answer[given]{\pi/4} \end{align*}$

Next, let’s find $H'(t)$ . We can find the derivative of $F$ using the . Now even though the limits of integration in $H$ don’t depend on $t$ , the integrand does. So to evaluate $H'(t)$ , you may simply pull the derivative (with respect to $t$ ) inside and treat $x$ like a constant.

$\begin{align*} H'(t) =2F(t)F'(t)+H'(t) =2 \left[ \int_{0}^{t} e^{-x^2}dx \right] \left[\answer[given]{e^{-t^2}}\right]+\int_{0}^{1} \frac{ -2t(1+x^2) e^{-t^2(1+x^2)}}{(1+x^2)} dx \end{align*}$

Try to simplify the answer above to get $\begin{align*} H'(t) &= 2e^{-t^2} \left[ \int_{0}^{t} e^{-x^2}dx \right] -2e^{-t^2} \left[ \int_{0}^{1} t e^{-t^2 x^2} dx \right] \\ &=2e^{-t^2} \left[ \int_{0}^{t} e^{-x^2}dx \right] -2e^{-t^2} \left[ \int_{0}^{1} e^{-(tx)^2} (tdx) \right], \end{align*}$ where the last change has been made to trigger your desire to integrate by substitution. So we set $u(x)=\answer [given]{tx}$ , so that $du=\answer [given]{tdx}$ .
We shouldn’t forget about our limits of integration! When $x=0$ , $u=\answer [given]{0}$ . When $x=1$ , $u=\answer [given]{t}$ .

Putting it all together, we get $\begin{align*} H'(t)=2e^{-t^2} \left[ \int_{0}^{t} e^{-x^2}dx \right] -2e^{-t^2} \left[ \int_{0}^{t} e^{-(u)^2}du \right] \end{align*}$ But $u$ is just a dummy variable, so doesn’t that second integral look a lot like $F(t)$ ? In other words, $H'(t)=0$ . That means our function $H$ is constant; it’s always $\pi /4$ . That’s a shock! So on one hand, $\begin{align*} \lim_{t \to \infty} H(t) = \lim_{t \to \infty} \frac{\pi}{4} = \answer[given]{\pi/4}. \end{align*}$ On the other hand, $\begin{align*} \lim_{t \to \infty} H(t) = \lim_{t \to \infty} \left[ F(t)^2+G(t) \right] = \left[\lim_{t \to \infty} F(t) \right]^2 + \lim_{t \to \infty} G(t). \end{align*}$ But $\begin{align*} \lim_{t \to \infty} G(t) = \lim_{t \to \infty} \int_{0}^{1} \frac{e^{-t^2(1+x^2)}}{1+x^2} dx = \int_{0}^{1} \frac{ \lim_{t \to \infty} e^{-t^2(1+x^2)}}{1+x^2} dx = \int_{0}^{1} 0 dx = \answer[given]{0}. \end{align*}$ This means $\begin{align*} \frac{\pi}{4} = \left[\lim_{t \to \infty} F(t) \right]^2 = \left[\lim_{t \to \infty} \int_{0}^{t} e^{-x^2} dx \right]^2=\left[ \int_{0}^{\infty} e^{-x^2} dx \right]^2. \end{align*}$ We may solve to get $\begin{align*} \int_{0}^{\infty} e^{-x^2} dx = \answer[given]{\sqrt{\pi}/2}. \end{align*}$ And finally, since $e^{-x^2}$ is an even function, $\begin{align*} \int_{-\infty}^{\infty} e^{-x^2} dx = \answer[given]{\sqrt{\pi}}. \end{align*}$ You may sleep soundly tonight.