Home page for accesible maths 1 1 Further Integration

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

1.54 Example of an improper integral

Example.
01x-sdx=1/(1-s)  0<s<1;\int_{0}^{1}x^{-s}\,dx=1/(1-s)\qquad 0<s<1;

whereas for s1s\geq 1 the integral diverges.

As x0+x\rightarrow 0+, the integrand x-sx^{-s}\rightarrow\infty, so the integral is improper.

Solution. For any s1s\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-s0\delta^{1-s}\rightarrow 0 as δ0+\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 δ0+\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 δ0+\delta\rightarrow 0+.