# Is there a function such that the integral converges?

I would like to know if there are functions $f:\mathbb{R} \rightarrow \mathbb{R}$ (apart from the trivial one $f=0$) such that

$$\left|\int_{\mathbb{R}^n} f\left(\sum_{i=1}^n x_i\right)dx_1 \cdots dx_n\right|<\infty.$$

Also, whether such class of functions is too narrow.

My feeling is that it is not possible. Any suggestions or ideas?

Thanks! :)

• As a quick idea : $f(x) = \exp( - x^2 )$ should works – Tryss Nov 17 '15 at 20:54
• $f = 0$ comes to mind. – Robert Israel Nov 17 '15 at 20:54
• @Tryss No, that does not. – Robert Israel Nov 17 '15 at 20:57
• No it doesn't because it is the square of the sum and hence, there is a hyperplane where the exponent is 0 and hence the function becomes 1 and diverges... I can't think of a good example, not even $f$ with compact support seems to help. – Martingalo Nov 17 '15 at 20:59
• Martingalo : but the hyperplane is measure zero – Tryss Nov 17 '15 at 21:02

If you interpret the integral as a Lebesgue integral over $\mathbb R^n$, and $n \ge 2$, then you must have $f(x) = 0$ a.e. in order for the integral to converge.
Proof Suppose not. Then for some $\epsilon > 0$ there is a set $A\subset \mathbb R$ of nonzero measure where $|f| > \epsilon > 0$. If $\phi$ is the linear functional $(x_1, \ldots, x_n) \mapsto x_1 + \ldots + x_n$, then $|f(\sum_j x_j)| > \epsilon$ on $\phi^{-1}(A)$ which has infinite $n$-dimensional Lebesgue measure.
If it's an iterated improper (Riemann) integral, then any $f$ with $\int_{-\infty}^\infty f(x)\; dx = 0$ will do.