Suppose f(x) is a piecewise continuous function defined for [a,∞).
If an inproper integral ∫[a,∞)f(x)dx coverges, what can we say on a limit(x→∞) f(x)?
Since we assume the inproper intgegral caoverges, we can say there is some c such that
inf(x>c)f(x)c)f(x)>=0
otherwise ∫[c,∞)f(x)dx>=inf(x>c)f(x)×(∞-c)＝∞ or ∫[c,∞)f(x)dxc)f(x)×(∞-c)＝-∞ that contradicts the assuption.
So we can say
1, if limit(x→∞) f(x) converges,it converges to 0,
2, limit(x→∞) f(x) can't diverge.
What else can we say?
For example, If limit(x→∞) f(x) oscillates around 0 without convergence perpetually, is it possible that the improper integral ∫[a,∞)f(x)dx converges?
And how about on others?
Best regards
For real finite solution the integrand must converge in the given limits. See Rudin's books on functional analysis. See,
http://math.stackexchange.com/questions/1148527/necessary-condition-for-the-convergence-of-an-improper-integral
https://en.wikipedia.org/wiki/Improper_integral#Convergence_of_the_integral
Perpetual oscillation is possible. Let us imagine a smooth nonnegative function f(x) defined on [0, ∞) that it is unimodal in each integral interval [N, N+1] and the integral has just 1/N2 there. Then the improper integral converges. This function is highly flexible.
If we take a sharper bell curve which has the maximum N on the interval [N, N+1], the function is divergent.
To make the function oscillate alternatively around zero is also possible.
Fujimoto's second observation is not very exact. We should instead say that
2'. limit(x→∞) f(x) = ∞ is impossible.
Dear Pandey
Thanks for your reply.
I couldn't see any reference to the limit of the integrand in two materials you cited.
And I could see no reference to the integration in Rudin's book.
https://59clc.files.wordpress.com/2012/08/functional-analysis-_-rudin-2th.pdf
Best regards
Dear Shiozawa,
Thanks for your reply.
I agree with your answer.
If f(x)=1 for x ∊[N,N+1/N^2],N>=1 otherwise 0,
f(x) oscilates perpetually and the inproper integeral converges.
On my second observation, you are right.
Although my intention was "limit(x→∞) f(x) = ∞ is impossible",
I misused an inexact term "divergence".
Mathematically the situation where the integrand diverges in spite of the convergence of the improper integral is quite likely.
But in my current problem, it seems to be quite unlikely.
Let me explain a little bit my problem.
D(t) is a debt on future time t and r is constat interest rate.
∫(0 to T) rD(t)exp(-rt)dt is a present value of the interest flows over [0,T] and D(T)exp(-rT) is a a present value of the redemption.
In economic analysis any choose of a specific end time of the horizon(in this case T) is arbitrary, so I want to analyse an infinite time horizon framework.
We need the present value of the perpetual interest flows to converge in order to have a finite value(otherwise debtor will bankrupt).
Here the problem comes whether we can neglect the present value of the redemption under the convergence of the present value of perpetual interest flows.
if we apply the above example to the debt behaviour, the future debt can exist in the shorter period as much as in the future.
This seems to be an unlikely situation because it means the future becomes quite different from the earlier period.
Best regards
An example relevant to the above discussion is
\begin{equation}
\int_0^{\infty} \sin(x^2/2)\,dx=\sqrt{\pi}/2
\end{equation}
Jog
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