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I am looking for a function $f$ with the following properties:

  • $f$ is continuous on $[0,\infty[$
  • $f(0)=1$
  • $f(x)\to0$ as $x\to\infty$
  • $\int_0^{\infty} f(x) \,\mathrm{d}x = \infty$

It is not difficult to find a such a function, for example $f(x) = 1/(x+1)$ would do it. However, I am looking for an example where I can "proof" that the area under the curve (i.e. $\int_0^{\infty} f(x) \,\mathrm{d}x $) goes to infinity for $x \to \infty$ for an audience which doesn't know of how to integrate or differentiate functions.

So a "visual" proof would be sufficient too.

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  • $\begingroup$ Your example function does not comply with your first requirement, i.e. it is discontinuous at $x=-1$. $\endgroup$ – nluigi Jul 13 '17 at 17:43
  • $\begingroup$ Thanks, I added a restriction. $\endgroup$ – Julia Jul 13 '17 at 18:38
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You can use the function $1/(x+1)$ and draw infinitely many rectangles (theorically, of course) of area $1/2$ under the curve.

Each rectangle should have double width and half height of the preceeding one.

Something like this:

enter image description here

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  • $\begingroup$ This is very slick. You can easily transform that into a proof that the harmonic series diverges, which may prove useful for pedagogical purposes. $\endgroup$ – Kevin Jul 13 '17 at 16:51
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    $\begingroup$ @Kevin As far as I know that is the standard way to show the harmonic series diverges... $\endgroup$ – Dason Jul 13 '17 at 17:28
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    $\begingroup$ @Dason: Yes, but it's rarely presented geometrically. $\endgroup$ – Kevin Jul 13 '17 at 17:58
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Start with the trapezoid $T_0$ with vertices $(0,0)$, $(1,0)$, $(1,{1\over2})$, $(0,1)$. It has area ${3\over4}$. Shrink $T_0$ vertically by the factor ${1\over2}$ and extend it horizontally by the factor $2$ to obtain $T_1$ (of the same area as $T_0$), and put $T_1$ flush to the right of $T_0$. Then shrink $T_1$ vertically by the factor ${1\over2}$ and extend it horizontally by the factor $2$ to obtain $T_2$, and put $T_2$ flush to the right of $T_1$, and so on ad infinitum.

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