More on Improper Integrals; $f\ge0$, $\int_a^\infty f\,dx<\infty\Rightarrow \lim_{x\to\infty}f(x)=0?$

In this course we shall mainly focus on improper integrals of nonnegative functions (i.e., a function $f$ such that $f(x)\ge 0$ for each $x$ in the domain). For nonnegative functions, \[\int_\square^\square f\,dx\text{ converges}\iff \int_\square^\square f\,dx <\infty,\] hence sometimes we conclude something converges by just writing $\int_\square^\square f\,dx <\infty$.

An important case (corollary 2 below) needs to be mentioned, since it is impossible to explain everything in tutorial, I decide to provide more detail in this blog (also typing maths in blogger is very easy!).


First of all, let's prove the following basic result:

Fact.  If $f$ is continuous on $[a,b]$, then there is $c\in (a,b)$ such that \[

\int_a^bf(x)\,dx = (b-a)f(c).

\]

Proof. Let $F(x)=\int_a^xf(t)\,dt$, then $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, hence by mean-value theorem, \[

\frac{F(b)-F(a)}{b-a}=F'(c)=f(c),

\] for some $c\in (a,b)$, hence \[

\int_a^bf(t)\,dt=F(b)=F(b)-F(a)=(b-a)f(c).\qed

\]

Theorem 1. If $f:[a,\infty)\to [0,\infty)$ is continuous and $\lim_{x\to\infty}f(x)$ exists, then \[

\int_a^\infty f(x)\,dx <\infty \implies \lim_{x\to\infty}f(x)=0.

\]

Proof. Consider for each $n\in \N$ so large that $n>a$, then by the fact above, consider $f$ on $[n,n+1]$ there is $x_n\in (n,n+1)$ such that (a subscript $n$ appears in $x_n$ because this number will change when $n$ vary), \[

\int_{n}^{n+1}f(x)\,dx = f(x_n).

\] Since \[\limn \int_{n}^{n+1}f(x)\,dx = \limn \brac{\int_a^{n+1}f\,dx - \int_a^nf\,dx} =\int_a^\infty f\,dx -\int_a^\infty f\,dx=0 , \] we have $\limn f(x_n)=0$. Since $\lim_{x\to\infty}f(x)$ exists and $\limn x_n=\infty$ (recall that $x_n>n$), we have \[

\lim_{x\to\infty}f(x)=\limn f(x_n)=0.\qed

\]
From logic we understand that the statement $p\implies q$ is the same as $\text{not }q\implies \text{not }p$, hence:

Corollary 2. If $f:[a,\infty)\to [0,\infty)$ is continuous and $\lim_{x\to\infty}f(x)$ exists, then \[

\lim_{x\to\infty}f(x)\neq 0\implies \int_a^\infty f(x)\,dx=\infty.

\]

Example. Does the improper integral $\dis \int_1^\infty \frac{xe^{1/x}}{1+x}\,dx $ converge?  How about $\dis \int_1^\infty \frac{1}{x}\,dx $?

Solution. Since \[\lim_{x\to\infty} \frac{xe^{1/x}}{1+x}=1\neq 0,\] the improper integral diverges by the corollary.

Although $\dis \lim_{x\to\infty} \frac{1}{x}=0$, since theorem 1 is not an if-and-only-if statement, we cannot conclude $\dis \int_1^\infty \frac{1}{x}\,dx$ converges. Indeed it diverges by straightforward computation.$\qed$

Remark. For nonnegative functions, the information $\int_a^\infty f(x)\,dx<\infty$ alone provides almost no information to $f$. For example:

• We can construct a nonnegative continuous function which does not converge to 0 but $\int_a^\infty f(x)\,dx<\infty$ (we will see this in tutorial note 9). 
• Worst still, nonnegative and improper integrable functions can even be unbounded.
• Don't be confused by the fact in theorem 1, only when $\lim_{x\to\infty}f(x)$ exists we can conclude $\lim_{x\to\infty}f(x)=0$.

If we understand later the convergence test of infinite series, one will see some tests for infinite series can be translated to improper integrals. So the convergence tests for infinite series will provide you more insight to convergence tests for improper integrals.