# an integrable function which does not tend to zero

In this entry, we give an example of a function $f$ such that $f$ is Lebesgue
integrable^{} on $\mathbb{R}$ but $f(x)$ does not tend to zero as
$x\to \mathrm{\infty}$.

Set

$$f(x)=\sum _{k=1}^{\mathrm{\infty}}\frac{k}{{k}^{6}{(x-k)}^{2}+1}.$$ |

Note that every term in this series is positive, hence we may integrate term-by-term, then make a change of variable $y={k}^{3}x-{k}^{4}$ and compute the answer:

${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}f(x)\mathit{d}x$ | $={\displaystyle \sum _{k=1}^{\mathrm{\infty}}}{\displaystyle {\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}{\displaystyle \frac{kdx}{{k}^{6}{(x-k)}^{2}+1}}$ | ||

$={\displaystyle \sum _{k=1}^{\mathrm{\infty}}}{\displaystyle \frac{1}{{k}^{2}}}{\displaystyle {\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}{\displaystyle \frac{dy}{{y}^{2}+1}}$ | |||

$={\displaystyle \frac{{\pi}^{2}}{6}}\cdot \pi ={\displaystyle \frac{{\pi}^{3}}{6}}$ |

However, when $k$ is an integer, $f(k)>k$, so not only does $f(x)$ not tend to zero as $x\to \mathrm{\infty}$, it gets arbitrarily large. As we can see from the plot, we have a sequence of peaks which, as they get taller, also get narrower in such a way that the total area under the curve stays finite:

By a variation of our procedure, we can produce a function which is
defined almost everywhere on the interval $(0,1)$, is Lebesgue integrable,
but is unbounded^{} on any subinterval, no matter how small. For instance,
define

$$f(x)=\sum _{m=1}^{\mathrm{\infty}}\sum _{n=1}^{\mathrm{\infty}}\frac{1}{{(m+n)}^{6}{(x-m/(m+n))}^{2}+1}.$$ |

Making a computation similar to the one above, we find that

${\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}f(x)\mathit{d}x$ | $={\displaystyle \sum _{m=1}^{\mathrm{\infty}}}{\displaystyle \sum _{n=1}^{\mathrm{\infty}}}{\displaystyle {\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}{\displaystyle \frac{dx}{{(m+n)}^{6}{(x-m/(m+n))}^{2}+1}}$ | ||

$={\displaystyle \sum _{m=1}^{\mathrm{\infty}}}{\displaystyle \sum _{n=1}^{\mathrm{\infty}}}{\displaystyle \frac{1}{{(m+n)}^{3}}}{\displaystyle {\int}_{-\mathrm{\infty}}^{+\mathrm{\infty}}}{\displaystyle \frac{dy}{{y}^{2}+1}}$ | |||

$=\pi {\displaystyle \sum _{k=1}^{\mathrm{\infty}}}{\displaystyle \frac{k-1}{{k}^{3}}}$ |

Hence the integral is finite.

Now, however, we find that $f$ cannot be bounded in any interval,
however small. For, in any interval, we can find rational
numbers. Given a rational number $r$, there are an infinite^{}
number of ways to express it as a fraction $m/(m+n)$. For
each of these ways, we have a term in the series which equals 1
when $x=r$, hence $f(r)$ diverges to infinity^{}.

To help in understanding this function, we have made a slide show which shows partial sums of the series. As before, the successive peaks become narrower in such a way that the arae under the curve stays finite but, this time, instead of marching off to infinity, they become dense in the interval.

Title | an integrable function which does not tend to zero |
---|---|

Canonical name | AnIntegrableFunctionWhichDoesNotTendToZero |

Date of creation | 2014-02-01 3:01:47 |

Last modified on | 2014-02-01 3:01:47 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 15 |

Author | rspuzio (6075) |

Entry type | Example |

Classification | msc 26A42 |

Classification | msc 28A25 |