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
Canonical name	ASufficientConditionForConvergenceOfIntegral
Date of creation	2013-03-22 19:01:13
Last modified on	2013-03-22 19:01:13
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 40A10
Related topic	RatioTest