Integrability in $RR$

[Camillo]

Find the values $alpha > 0 $ and $beta > 0 $ such that the function $ u $ defined by :

$u(x)= k^alpha $ if $x in I_k $; ( $ k=1,2,....) $ ; $u(x) = 0 $ if $x !in uu_(k>=1)I_k $ , where $I_k =(k,k+k^(-beta)) $

is integrable in $RR$ .

It is worth to note that, for some of the values found for $alpha, beta $ , the function is integrable, but $u $ does not $ rarr 0 $ for $ x rarr +oo$.

[Camillo]

Quite correct

[Eredir]

"Camillo":Find the values $alpha > 0 $ and $beta > 0 $ such that the function $ u $ defined by :

$u(x)= k^alpha $ if $x in I_k $; ( $ k=1,2,....) $ ; $u(x) = 0 $ if $x !in uu_(k>=1)I_k $ , where $I_k =(k,k+k^(-beta)) $

is integrable in $RR$ .

It is worth to note that, for some of the values found for $alpha, beta $ , the function is integrable, but $u $ does not $ rarr 0 $ for $ x rarr +oo$.

We observe that $u(x)$ is piecewise continuous, so the integral in $[-a,a]$ exists.
We need to establish for which values of $\alpha$ and $\beta$ this integral is convergent.

The function $u(x)$ is also piecewise constant, so the integral reduces to a series
$int_(RR)u(x)dx=\sum_(k=1)^(\infty)k^(\alpha)(k+k^(-\beta)-k)=\sum_(k=1)^(\infty)k^(\alpha-\beta)$
and this series converges if $\alpha<\beta-1$.