Improper integrals and the Gaussian integral

Integrating to infinity

The fundamental theorem (lesson 2.5.3) needs a finite interval and a curve that stays finite on it. An improper integral breaks one of the two. The first kind runs to infinity: $\int_1^\infty \frac{1}{x^2}\,dx$ asks for the area under $\frac{1}{x^2}$ from 1 onwards, for ever.

The plan is to integrate up to a finite right edge $b$, then let $b$ grow (lesson 2.1.3).

Here $\int_1^b \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_1^b = 1 - \frac{1}{b}$, which tends to 1. But $\int_1^b \frac{1}{x}\,dx = \ln b$, which grows without end.

A curve that sinks to 0 can still enclose infinite area. At $x = 10$, $\frac{1}{x^2}$ is 0.01 while $\frac{1}{x}$ is 0.1, and the gap keeps widening. The harmonic series of lesson 1.5.3 diverges in the same way.

Powers: which tails converge

For any power $p \neq 1$, the power rule backwards (lesson 2.5.2) gives

If $p > 1$, the exponent $1 - p$ is negative, so $b^{1-p} \to 0$ and the integral tends to $\frac{1}{p - 1}$. If $p < 1$, $b^{1-p}$ grows without end. The case $p = 1$ is the $\ln b$ above.

Exponential tails

Exponentials sink much faster than any power. $\int_0^b e^{-x}\,dx = 1 - e^{-b}$, which tends to 1.

In general $\int_0^\infty e^{-kx}\,dx = \frac{1}{k}$ for any $k > 0$, using $\int e^{-kx}\,dx = -\frac{e^{-kx}}{k}$ from lesson 2.5.2. For $k \le 0$ the curve does not sink at all, and the integral diverges.

A power times an exponential still converges. By parts (lesson 2.6.2), $\int_0^b x\,e^{-x}\,dx = \big[-(x + 1)e^{-x}\big]_0^b = 1 - (b + 1)e^{-b}$. Since $e^b$ outgrows every power of $b$ (lesson 2.1.3), $(b + 1)e^{-b} \to 0$ and the area is 1.

The finite pieces get there fast: 0.95957232 at $b = 5$, 0.9995006 at $b = 10$ and 0.99999996 at $b = 20$. A higher power changes the value but not the verdict: $\int_0^\infty x^2 e^{-x}\,dx = 2$.

A curve that blows up

The second kind of improper integral has a curve that blows up (lesson 2.1.2) at an end of the interval. In $\int_0^1 \frac{1}{\sqrt{x}}\,dx$, the curve has no value at 0. Start at a small $a > 0$ instead, then let $a$ shrink to 0 from the right.

Here $\int_a^1 x^{-1/2}\,dx = \big[2\sqrt{x}\big]_a^1 = 2 - 2\sqrt{a}$, which tends to 2. But $\int_a^1 \frac{1}{x}\,dx = -\ln a$, which grows without end.

A blow-up inside the interval

Lesson 2.5.3 showed that $\int_{-1}^{1} \frac{1}{x^2}\,dx$ is not $-2$: the curve blows up at 0, inside the interval. Split the integral there into $\int_{-1}^{0}$ and $\int_{0}^{1}$, so each piece has its trouble at an end.

By the rule above with $p = 2$, $\int_0^1 \frac{1}{x^2}\,dx$ diverges, so the whole integral diverges. By contrast, $\int_{-1}^{1} \frac{1}{\sqrt{|x|}}\,dx$ splits into two pieces of 2, so it converges to 4.

Infinite at both ends

An integral over the whole line has two infinite ends. Split it at any point, such as 0, and each half must converge on its own.

$\int_{-b}^{b} x\,dx = 0$ for every $b$, because the two sides cancel. But $\int_0^\infty x\,dx$ diverges, so $\int_{-\infty}^{\infty} x\,dx$ diverges too. Cancelling two infinite areas does not give a number.

The Gaussian integral

The curve $e^{-x^2}$ is a bell: 1 at $x = 0$, falling fast on both sides. Lesson 2.5.2 noted that it has no antiderivative made of powers, roots, exponentials and logs.

The natural guess shows why. The chain rule suggests trying $F = R(x)\,e^{-x^2}$ with $R$ a polynomial; the product rule then gives $F' = (R' - 2xR)\,e^{-x^2}$. For $F' = e^{-x^2}$ we need $R' - 2xR = 1$, but $2xR$ has a higher power of $x$ than $R'$, so they never cancel down to 1.

Mathematicians proved in the nineteenth century that no formula built from familiar functions works, not just this guess. The area function $\int_0^x e^{-t^2}\,dt$ still exists (lesson 2.5.3); it is a new function, computed numerically, like a square root. SymPy's erf from lesson 2.5.2 is this area times $\frac{2}{\sqrt{\pi}}$.

Its total area

The tails converge. For $|x| \ge 1$, $x^2 \ge |x|$, so $e^{-x^2} \le e^{-|x|}$, and exponential tails have finite area. The total area is known exactly.

Check it numerically by integrating from $-b$ to $b$. The missing tails shrink astonishingly fast, because $e^{-x^2}$ is already $1.4 \times 10^{-11}$ at $x = 5$.

So cutting the integral off at a moderate $b$ loses almost nothing. Case study 2.6.5 relies on that.

An extra factor of $x$ changes everything. $x\,e^{-x^2}$ does have an antiderivative, $-\frac{1}{2}e^{-x^2}$, by substitution (lesson 2.6.1). So $\int_0^\infty x\,e^{-x^2}\,dx = \frac{1}{2}$ exactly.

The standard normal density

Stretch the bell sideways by $\sqrt{2}$ and it becomes $e^{-t^2/2}$. Substitute $u = \frac{t}{\sqrt{2}}$, a linear inner function, so $dt = \sqrt{2}\,du$ (lesson 2.6.1). The limits stay at $-\infty$ and $\infty$, and

That is about 2.506628. Dividing the curve by $\sqrt{2\pi}$ gives it total area 1.

Areas under $\varphi$ have no formula either, so they are computed numerically. The band from $-1.96$ to 1.96 holds 0.950004, and the area up to 1 is 0.841345. By symmetry, the area up to 0 is exactly $\frac{1}{2}$.

The area from $-\infty$ up to $x$ is called $N(x)$, the normal cumulative distribution; SciPy computes it as norm.cdf(x). Level 5 (module 5.4) reads these areas as probabilities. Case study 2.6.5 prices an option with them.

Common mistakes

1. An improper integral is a limit. Integrate to a finite edge first, then let it go to infinity or to the blow-up.
2. A curve that sinks to 0 can still have infinite area, like $\frac{1}{x}$ from 1 on.
3. The power tests run opposite ways. Tails $\int_1^\infty$ need $p > 1$; blow-ups $\int_0^1$ need $p < 1$; $p = 1$ fails both.
4. Do not integrate across a blow-up. $\int_{-1}^{1} \frac{1}{x^2}\,dx$ is not $-2$: split at 0, and it diverges.
5. Each half must converge. $\int_{-b}^{b} x\,dx = 0$ does not make $\int_{-\infty}^{\infty} x\,dx$ equal 0.
6. $e^{-x^2}$ has no antiderivative formula, but $x\,e^{-x^2}$ does, by substitution.
7. Match the bell to its area. $e^{-x^2}$ has area $\sqrt{\pi}$; $e^{-x^2/2}$ has area $\sqrt{2\pi}$; $\varphi$ has area 1.