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:

By a variation of our procedure, we can produce a function which is defined almost everywhere on $\mathbb{R}$ , is Lebesgue integrable, but is unbounded is any interval, no matter how small. The trick is to use a pairing function. Define $$ p(m,n) = {(m+n+2)(m+n+3) \over 2} + n + 1 $$ and define $$ f(x) = \sum_{m=1}^\infty \sum_{n=1}^\infty {p(m,n) \over p(m,n)^6 (x-m/n)^2 + 1}. $$ Note that, as $m$ and $n$ range over the naural numbers, $p(m,n)$ ranges over all the natural numbers and, furthermore $p(m,n) = p(m',n')$ only when $m = m'$ and $n = n'$ . Hence, by the same sort of calculation as above, but using the change of variables $y = p(m,n)^3 x - p(m,n)^2 m / n$ , we may conclude that $\int_{-\infty}^{+\infty} f(x) \, dx = \pi^3 / 6$ .

Now, however, we find that $f$ cannot be bounded in any interval, however small. For, in any interval, we can find rational numbers with arbitrarily high denominators. By construction, $f(m/n) > p(m,n) > n$ , so it is impossible to bound $f$ .