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I ran into this question looking at "obvious theorems that are false".

Link: https://math.stackexchange.com/a/827393

Someone gave the example of

If you have a continuous function $$f:\mathbb{R} \rightarrow [0,\infty)\, \text{ such that } \int_{-\infty}^\infty f(x)\,dx < \infty, \text{ then } \lim_{x \to \pm \infty}f(x)=0$$

I am having trouble understanding why this is false. If you have a function that is strictly non negative than wouldn't the only possible way of having an integral that doesn't go to $\infty$ is if your function gets very close to zero.

Michael Hardy
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user262291
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    The problem is that it can usually be nearly zero but have narrow spikes all the way out. Infinitely many triangles of height 1 and width $1/n^2$ have finite total area. Note however that such triangles get sharper and sharper... – Ian Oct 07 '16 at 22:08
  • I was wondering can you give me an explicit example of a function that doesn't satisfy the statement above? – user262291 Oct 07 '16 at 22:11
  • Write down a formula for a continuous function whose graph looks like the triangles I described, with their bases being on the x axis. – Ian Oct 07 '16 at 22:13
  • I haven't run into any functions that behave like that. Would you mind giving me the function? – user262291 Oct 07 '16 at 22:15
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    One such example is $f(x)=|\sin x|^{x^2}$. If you actually want to prove the desired behavior, then it will be easier to construct your own example—qualitatively similar to this $f(x)$—that's easier to manipulate for proofs, as Ian has suggested. However, just graphing this $f(x)$ will give you a visual idea of the phenomenon at play here. – Greg Martin Oct 07 '16 at 22:21
  • I was wondering how you would go about integrating this function? Also what does this integral converge to? – user262291 Oct 07 '16 at 22:26

2 Answers2

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I community wiki'd this since this is basically what Ian suggested in the comments.

Without loss of generality, restrict ourselves to integrating on the positive real line (we can "evenly" extend this to $\mathbb{R}$ if desired, since the integral will still be finite).

Fix a convergent, positive series $ \sum a_n$ and define a bijection which assigns to each $a_i$ a triangle of area $a_i$. Choose heights such that the $\lim \sup$ of the heights is positive. Then consider a function

$f:[0, \infty) \to [0, \infty)$ such that the graph of this function is the triangles corresponding to each $a_i$ side to side, properly ordered, possibily with "plateaus of zero" in between. Such a function satisfies the required conditions.

It's tedious and unnecessary to find an explicit formula. If that matters a lot to you, why not try yourself, now that you have an idea what the graph looks like?

MathematicsStudent1122
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Well using Ian's commment of a triangle function:

Let $n_x \le |x| < n_x+1 $.

For $x$. Let $f(n_x) = 0$. Let $f(n_x + (1/2)^{n_x+1}) = 1$. Let $f(n_x + (1/2)^{n_x}) = 0$.

For $n_x < |x| < n_x + (1/2)^{n_x+1}$, let $f(x) = \frac{|x| -x_n}{(1/2)^{n_x+1}}$

For $n_x + (1/2)^{n_x+1}< |x| < n_x + (1/2)^{n_x}$, let $f(x) = \frac{(n_x + (1/2)^{n_x})-|x|}{(1/2)^{n_x+1}}$

For all else $f(x) = 0$.

$$\int_{n_x}^{n_x+1}f(x)\,dx = \frac 1 2 \cdot 1\cdot \left(\frac 12 \right)^{n_x} = \left(\frac 12\right)^{n_x + 1}$$

$$\int_0^\infty f(x) \, dx = \sum \left(\frac 12\right)^k = 1$$

$$\int_{-\infty}^\infty f(x) \, dx =2$$

But $f(x)$ doesn't converge however, as for any $N$ there will be $x,y > N$ where $f(x) =1$ and $f(y) = 0$.

Michael Hardy
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fleablood
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