1. ## Convergence of integrals

1) Let's assume that the integral of f(x) between a and infinity is convergent and that f(x) is equally continuous in [a, +infinity). Prove that the limit of f(x) when x tends to infinity is zero.

Okay so I assume negatively that the limit of f(x) when x tends to infinity is either different from zero or doesn't exist at all. How do I continue?

2) Let f(x) be a positive and integrable function in [0,t] to every t>0. Let's decide a>0 and define h(t) to be the integral of f(x) between t-a and t+a. t>=a. Prove that if the integral of f(x) between zero and infinity is convergent, then the integral of h(t) between a and infinity is convergent.

Have no idea how to solve this...not even a direction

2. ## Re: Convergence of integrals

Originally Posted by yosilacohen
1) Let's assume that the integral of f(x) between a and infinity is convergent and that f(x) is equally continuous in [a, +infinity). Prove that the limit of f(x) when x tends to infinity is zero.

Okay so I assume negatively that the limit of f(x) when x tends to infinity is either different from zero or doesn't exist at all. How do I continue?
A little question: what exactly do You mean as 'equally continous'?...

Kind regards

$\chi$ $\sigma$

3. ## Re: Convergence of integrals

Originally Posted by yosilacohen
1) Let's assume that the integral of f(x) between a and infinity is convergent and that f(x) is equally continuous in [a, +infinity). Prove that the limit of f(x) when x tends to infinity is zero.

Okay so I assume negatively that the limit of f(x) when x tends to infinity is either different from zero or doesn't exist at all. How do I continue?
I assume that "equally continuous" means uniformly continuous. If it is not true that $f(x)\to0$ as $x\to\infty$, then there must exist some $\delta>0$ and a sequence of points $x_n$ such that $x_{n+1}> x_n+1$ and $f(x_n)>\delta$ (for all n). The uniform continuity will then tell you that there exists $\varepsilon>0$ such that $f(x) >\delta/2$ whenever $|x-x_n|<\varepsilon$ (for some n). Thus there is an infinite sequence of disjoint intervals on each of which the function is larger than $\delta/2$. That means that $\int_0^\infty f(x)\,dx$ cannot converge.

4. ## Re: Convergence of integrals

Originally Posted by yosilacohen
1) Let's assume that the integral of f(x) between a and infinity is convergent and that f(x) is equally continuous in [a, +infinity). Prove that the limit of f(x) when x tends to infinity is zero.

Okay so I assume negatively that the limit of f(x) when x tends to infinity is either different from zero or doesn't exist at all. How do I continue?

2) Let f(x) be a positive and integrable function in [0,t] to every t>0. Let's decide a>0 and define h(t) to be the integral of f(x) between t-a and t+a. t>=a. Prove that if the integral of f(x) between zero and infinity is convergent, then the integral of h(t) between a and infinity is convergent.

Have no idea how to solve this...not even a direction
For Question 1, it might help if you realise that any infinite SERIES can only be convergent if the terms tend to 0. Isn't a definite integral just a sum?

5. ## Re: Convergence of integrals

Originally Posted by Prove It
For Question 1, it might help if you realise that any infinite SERIES can only be convergent if the terms tend to 0. Isn't a definite integral just a sum?
That in itself is not enough to prove this result. In fact, if you merely assume that f is continuous (rather than uniformly continuous) then it is possible for $\textstyle\int_0^\infty f(x)\,dx$ to converge without f(x) tending to 0 as x goes to infinity.

For example, suppose that f(x) is zero except for a narrow spike around each integer value of x. More precisely, suppose that f(x) goes linearly from 0 to 1 in the interval $\bigl[n-\tfrac1{n^2},n\bigr]$, and then goes down linearly from 1 to 0 in the interval $\bigl[n, n+\tfrac1{n^2}\bigr].$ The area under this triangular spike is $1/n^2$, and the total area under the curve is finite because $\textstyle\sum1/n^2$ converges. But f(x) does not tend to 0 because it keeps jumping up to 1 and back again.