Question

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.

If 0 < p < 1, then${\int }_0^1\frac{1}{x^p}dx$converges to$\frac{1}{1-p}.$

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

The given integral is${\int }_0^1\frac{1}{x^p}dx.$

Step 2. Prove. 

To prove if 0 < p < 1, then ${\int }_0^1\frac{1}{x^p}dx$converges to $\frac{1}{1-p},$ let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649067893443" $(a,b]$but not x = a, then role="math" localid="1649067215041" ${\int }_a^{b}f\left(x\right)dx=\underset{A\to a^{+}}{\lim }{\int }_A^{b}f\left(x\right)dx.$

So,

$$\begin{gathered} {\int }_0^1\frac{1}{x^p}dx \\ =\underset{A\to 0^{+}}{\lim }{\int }_A^1{x}^{-p}dx \\ =\underset{A\to 0^{+}}{\lim }{\left[\frac{x^{-p+1}}{-p+1}\right]}_A^1 \\ =\underset{A\to 0^{+}}{\lim }\left[\frac{1}{1-p}-\frac{1}{1-p}{A}^{1-p}\right] \\ \mathrm{As}0<p<1, \\ =\frac{1}{1-p} \end{gathered}$$

Now, if 0 < p < 1, the pvalue is less than 1and the integral converges on [0, 1].

So, if 0 < p< 1, then $\frac{1}{x^p}<\frac{1}{x}\mathrm{for}x\in [0,1]\mathrm{and}{\int }_0^1\frac{1}{x^p}dx\mathrm{converges}\mathrm{to}\frac{1}{1-p}.$

Hence, the integral is proved.