whereas for s≥1s\geq 1 the integral diverges.
As x→0+x\rightarrow 0+, the integrand x-s→∞x^{-s}\rightarrow\infty, so the integral is improper.
Solution. For any s≠1s\neq 1 we have ∫δ1x-sdx=[11-sx1-s]δ1=11-s(1-δ1-s)\int_{\delta}^{1}x^{-s}\,dx={\left[\frac{1}{1-s}x^{1-s}\right]_{\delta}^{1}}\,% {=\frac{1}{1-s}(1-\delta^{1-s})}. If s∈(0,1)s\in(0,1) then δ1-s→0\delta^{1-s}\rightarrow 0 as δ→0+\delta\rightarrow 0+. Thus the integral converges to 11-s\frac{1}{1-s}.
On the other hand, if s>1s>1 then δ1-s→∞{\delta^{1-s}\rightarrow\infty} as δ→0+\delta\rightarrow 0+.
Finally, if s=1s=1 then ∫δ11xdx=[logx]δ1=-logδ→∞\int_{\delta}^{1}\frac{1}{x}\,dx\,{=\left[\log x\right]_{\delta}^{1}}\,{=-\log% \delta\rightarrow\infty} as δ→0+\delta\rightarrow 0+.