 Solution

On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta -1}-x^{n-\beta }$. Their integrals areBoth powers are integrable because $0<\beta <1$. If $\beta <1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their integrals form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.

If $\beta >1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta =1/2$ all summands and the integrand are zero. Thus in every case the integral is absolutely integrable andThe common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.