The Condition For An Infinite Integral To Converge

This is the improper integral we want to look at:

$$I = \int_a^\infty f(x) dx.$$

I will show that for the integral to coverge (exist), $f(x)$ must decline faster than $1/x$ for large $x$. (This only works for monotonically decaying functions. Functions like $\frac{\sin x}{x}$ or $\sin(x^2)$ won’t apply. My note is very unrigorous.)

We can split an integral into two parts:

$$I = \int_a^b f(x) dx + \int_b^\infty f(x) dx.$$

Because the first term is a finite integral, saying $I$ exists is equivalent as saying $\int_b^\infty f(x) dx$ exists, for any large $b$.

Notice that

$$\int_b^\infty x^p dx = \begin{cases}\begin{array}{lr} \frac{x^{1+p}}{1+p}|_{b}^{\infty}&, p\neq -1\\ \ln(x)|_{b}^{\infty}&,p= -1 \end{array} \end{cases} .$$

When $p = -1,$ the integral is $\lim_{x \to \infty} \ln(x) - \ln(b),$ which diverge. If something declines slower than $x^{-1}=1/x$, it of course diverge, too!

$$\text{If p $\geq$ -1, } \int_b^\infty x^p dx \,\, \text{diverges}.$$

When $p<-1,$

$$\int_b^\infty x^p dx = \lim_{x \to \infty}\frac{ x^{1+p} - b^{1+p}}{1+p}.$$

Since $p<-1 \Leftrightarrow 1 + p < 0,$

$$\lim_{x \to \infty}x^{1+p} = 0 \implies\int_b^\infty x^p dx = -\frac{b^{1+p}}{1+p} = \text{constant}.$$

We thus proved that, for an integral $\int_a^\infty f(x) dx$ to converge, plus $f(x)$ is monotonically decaying function, we must have

$$f(x) \sim \frac{1}{x^a}, a > 1, \text{for large $x$}.$$

What one cannot conclude without extra hypotheses

One cannot in general deduce that “$f(x)$ decays faster than $1/x$” from the convergence of $\int_a^{\infty} f(x) dx$ alone. For example, if $f$ oscillates (e.g. $f(x)=\sin x/x$) or if $f$ has sparse narrow spikes whose areas sum to a finite total, the integral may converge while $x f(x)$ does not tend to zero.