For what values of $\alpha$, does this integral converge? [closed]

real-analysis convergence-divergence calculus limits definite-integrals

This integral has potential problems at both $x\to 0$ and at $x\to\infty$.

  • As the OP correctly points out, the integral over the range $x\in[\epsilon,\infty)$ converges (absolutely) for $\alpha>1$. For $\alpha=1$, it is the well known sine-integral that converges as an improper Riemann integral. For $\alpha <1$ the integral diverges.

  • Over the range $x\in [0,\epsilon]$, the integrand is approximately given by $$\frac{\sin x}{x^\alpha} = x^{1-\alpha } [1 +O(x)]$$ by Taylor's theorem. In order that the integral converges, we need that $1-\alpha > -1$ or equivalently $\alpha <2 $.

In conclusion, the integral converges for $1\leq \alpha <2$; with absolute convergence except for the value $\alpha=1$.

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