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# 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
$x\rightarrow\infty$.

Set

$f(x)=\sum_{{k=1}}^{\infty}{k\over k^{6}(x-k)^{2}+1}.$ |

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:

$\displaystyle\int_{{-\infty}}^{{+\infty}}f(x)\,dx$ | $\displaystyle=\sum_{{k=1}}^{\infty}\int_{{-\infty}}^{{+\infty}}{k\,dx\over k^{% 6}(x-k)^{2}+1}$ | ||

$\displaystyle=\sum_{{k=1}}^{\infty}{1\over k^{2}}\int_{{-\infty}}^{{+\infty}}{% dy\over y^{2}+1}$ | |||

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

However, when $k$ is an integer, $f(k)>k$, so not only does $f(x)$ not tend to zero as $x\rightarrow\infty$, 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

$f(x)=\sum_{{m=1}}^{\infty}\sum_{{n=1}}^{\infty}{1\over(m+n)^{6}(x-m/(m+n))^{2}% +1}.$ |

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

$\displaystyle\int_{{-\infty}}^{{+\infty}}f(x)\,dx$ | $\displaystyle=\sum_{{m=1}}^{\infty}\sum_{{n=1}}^{\infty}\int_{{-\infty}}^{{+% \infty}}{dx\over(m+n)^{6}(x-m/(m+n))^{2}+1}$ | ||

$\displaystyle=\sum_{{m=1}}^{\infty}\sum_{{n=1}}^{\infty}{1\over(m+n)^{3}}\int_% {{-\infty}}^{{+\infty}}{dy\over y^{2}+1}$ | |||

$\displaystyle=\pi\sum_{{k=1}}^{\infty}{k-1\over k^{3}}$ |

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.

## Mathematics Subject Classification

26A42*no label found*28A25

*no label found*

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