# a sufficient condition for convergence of integral

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

 $\displaystyle\int_{a}^{\infty}\!f(x)\,dx$ (1)

is that

 $\displaystyle\lim_{x\to\infty}\frac{f(x\!+\!1)}{f(x)}\;=\;q\;<\;1.$ (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  $\int_{a}^{\infty}\!f(x)\,dx=\infty$.  We can use l’Hôpital’s rule (http://planetmath.org/LHpitalsRule):

 $\lim_{c\to\infty}\frac{\int_{a}^{c}\!f(x\!+\!1)\,dx}{\int_{a}^{c}\!f(x)\,dx}\;% =\;\lim_{c\to\infty}\frac{f(c\!+\!1)}{f(c)}.$

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!)

 $1\;>\;q\;=\;\lim_{c\to\infty}\frac{\int_{a}^{c}\!f(x\!+\!1)\,dx}{\int_{a}^{c}% \!f(x)\,dx}\;=\;0+1-0\;=\;1.$

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.

Title a sufficient condition for convergence of integral ASufficientConditionForConvergenceOfIntegral 2013-03-22 19:01:13 2013-03-22 19:01:13 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 40A10 RatioTest