# 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.

Title an integrable function which does not tend to zero AnIntegrableFunctionWhichDoesNotTendToZero 2014-02-01 3:01:47 2014-02-01 3:01:47 rspuzio (6075) rspuzio (6075) 15 rspuzio (6075) Example msc 26A42 msc 28A25