Can a positive integrand oscillate fast enough so that the integral is finite?

[Charles49]

Homework Statement

If $$f(x)>0$$ is continuous for all $$x\ge0$$ and the improper integral $$\int_0^{\infty}f(x) dx$$ exists, then $$\lim_{x\rightarrow\infty}f(x)=0.$$

2. Relevant

I think this assertion is false. A counterexample can be constructed along the following lines of thought: Suppose a function oscillates between 0 and 2. Then the integral computes a positive area under the curve. However by increasing the frequency, the oscillations can be made so rapid that each peak is too sharp and too close to another peak for the increment in area along the $x$-axis to avoid becoming negligible.

The Attempt at a Solution

An example is the function $$f(x)=1+\sin e^x.$$ Clearly, $$f>0$$ because $$0<1+\sin(x)<2.$$ Also integrating $$f(x)-1$$,
\begin{align*}
\int_0^\infty\sin e^x dx&= \int_1^\infty \frac{\sin t}{t}\,dt \\
&=\frac{\pi}{2} - \text{Si}(1)\\
&<\infty
\end{align*}
where $$\text{Si}(x)$$ is the sine integral function. However,
\begin{equation*}
\lim_{x\rightarrow\infty}\sin(e^x)\ne0.
\end{equation*}

The problem is can such an integrand be constructed?

 

[DryRun]

$$-1 \le \sin e^x \le 1<br /> \\0 \le 1+ \sin e^x \le 2$$I would have suggested using the pinching theorem to find if the middle term converges or diverges, but the latter is not applicable in this case, as the limits of the two boundaries are not equal. In this case, f(x) diverges: $$0 \le \lim_{n \to \infty} (1+ \sin e^x) \le 2$$But maybe I'm wrong.

 

[micromass]

An example is the function $$f(x)=1+\sin e^x.$$ Clearly, $$f>0$$ because $$0<1+\sin(x)<2.$$

This is false. The function f becomes 0 many times. Not that it is very hard to fix this problem.

 

[gopher_p]

Your intuition is correct regarding the validity of the claim. The easiest-to-construct counterexamples aren't "nice" functions, though. I like to keep a nice stock of "spiky" functions on hand to disprove these kinds of claims.

For the record, there is a continuous function $f$ defined on $[0,\infty)$ with $f(x)>0$ for all $x$ such that $\int_0^{\infty}f(x)dx$ converges while $\limsup_{x\rightarrow\infty}f(x)=\infty$.

 

[Ray Vickson]

Your intuition is correct regarding the validity of the claim. The easiest-to-construct counterexamples aren't "nice" functions, though. I like to keep a nice stock of "spiky" functions on hand to disprove these kinds of claims.

For the record, there is a continuous function $f$ defined on $[0,\infty)$ with $f(x)>0$ for all $x$ such that $\int_0^{\infty}f(x)dx$ converges while $\limsup_{x\rightarrow\infty}f(x)=\infty$.

For the record, what is this function?

RGV

 

[gopher_p]

For the record, what is this function?

RGV

Define $g$ piecewise as follows:

$g(x)=0$ for $x$ in any interval of the form $(n+\frac{1}{n^3},n+1)$ for $n=1, 2, 3, ...$ as well as on $[0,1)$.

Let $x_n$ be the midpoints of intervals of the form $[n, n+\frac{1}{n^3}]$ for $n=1,2,3,...$. Put $g(n)=g(n+\frac{1}{n^3})=0$, and $g(x_n)=n$. Connect the dots with lines on the remainder of $[n, n+\frac{1}{n^3}]$. So we have a triangular spike of height $h_n=n$ and base $b_n=\frac{1}{n^3}$.

Then $g$ is (obviously?) continuous and non-negative. $\int_0^{\infty}g(x)dx=\sum_1^{\infty}\frac{1}{2}b_nh_n=\sum_1^{\infty}\frac{1}{2n^2}=\frac{\pi^2}{12}$. And $\limsup_{x\rightarrow\infty} g(x)\geq \lim_{n\rightarrow\infty}g(x_n)=\infty$.

Let $h(x)=e^{-x}$, put $f=g+h$. $f$ has the desired properties.

 

[Charles49]

gopher_p:

That was very clever! Thanks a lot!

 

[gopher_p]

gopher_p:

That was very clever! Thanks a lot!

I take no credit for the idea. Like I said before, the spiky functions are pretty useful as counterexamples to naive notions of what we think should happen.

As a follow-up exercise, replace continuity of $f$ with uniform continuity. Now is the statement true?

Or keep the original setup, and ask yourself if it must be true that $\liminf_{x\rightarrow\infty} f(x)=0$.

 

[Steely Dan]

This is false. The function f becomes 0 many times. Not that it is very hard to fix this problem.

Does that not affect the answer? If every part of the curve must cover some area, my intuition says the total area covered must be infinite. You could make it oscillate fast enough so that the function is effectively equal to some $\epsilon > 0$ for all $x$, but the integral of that constant small number diverges.

 

[Charles49]

gopher_p, thanks for the suggestions. By the way, this question came up in a proseminar packet which have to master before I can qualify for PhD.

Also Steely_Dan's objection to my previous incorrect example makes sense. Infinite epsilons do add up and the integral diverges. So instead of adding 2, I took the absolute value. Now does this integral diverge:

$$
\int_{0}^\infty\lvert\sin(e^x)\rvert dx?
$$

 

[Steely Dan]

That is still problematic, because the sine function will always have zeros at some locations, so it doesn't satisfy the requirement that the function must be strictly positive.