an integrable function which does not tend to zero

In this entry, we give an example of a function $f$ such that $f$ is Lebesgueintegrable on $ℝ$ 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 integrateterm-by-term, then make a change of variable $y=k^{3}x-k^4$ andcompute the answer:

${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}f(x)𝑑x$$={\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 $f(x)$ nottend 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 gettaller, also get narrower in such a way that the total area underthe curve stays finite:

integrable-no-tend-zero.gif

By a variation of our procedure, we can produce a function which isdefined 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

${\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}f(x)𝑑x$$={\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 rationalnumbers. Given a rational number $r$, there are an infinitenumber of ways to express it as a fraction $m/(m+n)$. Foreach of these ways, we have a term in the series which equals 1when $x=r$, hence $f(r)$ diverges to infinity.

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

integrable-everywhere-unbounded_3.gif