a sufficient condition for convergence of integral

Suppose that the real function $f$ is positive and continuous on the interval  $[a,\mathrm{\infty })$.  A sufficient condition for the convergence (http://planetmath.org/ConvergentIntegral) of the improper integral

${\displaystyle {\int }_a^{\mathrm{\infty }}}f(x)𝑑x$(1)

is that

(2)

Proof.  Assume that the condition (2) is in .  For an indirect proof (http://planetmath.org/ReductioAdAbsurdum), make the antithesis that the integral (http://planetmath.org/RiemannIntegral) (1) diverges (http://planetmath.org/DivergentIntegral).

Because of the positiveness, we have  ${\int }_a^{\mathrm{\infty }}f(x)𝑑x=\mathrm{\infty }$.  We can use l’Hôpital’s rule (http://planetmath.org/LHpitalsRule):

$$\underset{c\to \mathrm{\infty }}{lim}\frac{{\int }_a^{c}f(x+1)𝑑x}{{\int }_a^{c}f(x)𝑑x}=\underset{c\to \mathrm{\infty }}{lim}\frac{f(c+1)}{f(c)}.$$

Using the http://planetmath.org/node/11373substitution  $x+1=t$  we get

$${\int }_a^{c}f(x+1)𝑑x={\int }_{a-1}^{c-1}f(t)𝑑t={\int }_{a-1}^{a}f(t)𝑑t+{\int }_a^{c}f(t)𝑑t-{\int }_{c-1}^{c}f(t)𝑑t,$$

and dividing this equation by ${\int }_a^{c}f(t)𝑑t$ and taking limits (http://planetmath.org/ImproperLimits) yield ($f$ is bounded!)

$$1>q=\underset{c\to \mathrm{\infty }}{lim}\frac{{\int }_a^{c}f(x+1)𝑑x}{{\int }_a^{c}f(x)𝑑x}=\mathrm{\hspace{0.33em}0}+1-0=\mathrm{\hspace{0.33em}1}.$$

This contradictory result shows that the antithesis is wrong; thus (1) must be convergent (http://planetmath.org/ConvergentIntegral).

Note.  The condition (2) is not necessary for the convergence of (1).  This is seen e.g. in the case of the converging of (2) equals 1.