32. for what values of ( p ) is the integral\n int_{1}^{infty} \frac{1}{x^{p}} d x \nconvergent? evaluate…

Explanation:

Step1: Consider different cases for (p)

Case 1: When (p = 1) $$\int_{1}^{\infty}\frac{1}{x^{p}}dx=\int_{1}^{\infty}\frac{1}{x}dx$$ By the definition of improper integral (\int_{1}^{\infty}\frac{1}{x}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{x}dx) Since (\int\frac{1}{x}dx=\ln x + C), then (\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{x}dx=\lim_{t\rightarrow\infty}(\ln t-\ln1)=\lim_{t\rightarrow\infty}\ln t=\infty)

Case 2: When (p\neq1) $$\int_{1}^{\infty}\frac{1}{x^{p}}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-p}dx$$ Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (\int x^{-p}dx=\frac{x^{-p + 1}}{-p + 1}+C=\frac{x^{1 - p}}{1 - p}+C) Then (\lim_{t\rightarrow\infty}\int_{1}^{t}x^{-p}dx=\lim_{t\rightarrow\infty}\left[\frac{t^{1 - p}}{1 - p}-\frac{1^{1 - p}}{1 - p}\right])

Step2: Analyze the limit for (p\neq1)

If (p>1), then (1 - p<0). So (\lim_{t\rightarrow\infty}t^{1 - p}=\lim_{t\rightarrow\infty}\frac{1}{t^{p - 1}} = 0) (\lim_{t\rightarrow\infty}\left[\frac{t^{1 - p}}{1 - p}-\frac{1}{1 - p}\right]=\frac{0}{1 - p}-\frac{1}{1 - p}=\frac{1}{p - 1})

If (p<1), then (1 - p>0). So (\lim_{t\rightarrow\infty}t^{1 - p}=\infty) (\lim_{t\rightarrow\infty}\left[\frac{t^{1 - p}}{1 - p}-\frac{1}{1 - p}\right]=\infty)

Answer:

The integral (\int_{1}^{\infty}\frac{1}{x^{p}}dx) is convergent when (p>1), and (\int_{1}^{\infty}\frac{1}{x^{p}}dx=\frac{1}{p - 1}) for (p>1)