p-Integral

Tags

Calculus

Cegep/2

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211 words

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2 minutes

Improper integral with the form

$${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}\mathrm{d}x$$

Proof

Determine $p$ such that the integral ${\int }_1^{\mathrm{\infty }}\frac{1}{x^p}\mathrm{d}x$ is at the boundary of convergence.

When $p=1$, we know ${\int }_1^{\mathrm{\infty }}\frac{1}{x}\mathrm{d}x$ is divergent.

When $p\ne 1$,

$$\begin{array}{rl}{\int }_1^{\mathrm{\infty }}\frac{1}{x^p}\mathrm{d}x& =\underset{t\to \mathrm{\infty }}{lim}{\int }_1^t\frac{1}{x^p}\mathrm{d}x\\ & ={\underset{t\to \mathrm{\infty }}{lim}\frac{x^{-p+1}}{-p+1}|}_1^t\\ & =\underset{t\to \mathrm{\infty }}{lim}\left(\frac{1}{1-p}\right)(t^{1-p}-1^{1-p})\end{array}$$

Consider $\underset{t\to \mathrm{\infty }}{lim}{t}^{1-p}$.
When $p<1$, $1-p>0$, thus $t^{1-p}\to \mathrm{\infty }$, making the integral diverge.
When $p>1$, $1-p<0$, thus $t^{1-p}\to 0$, making the integral converge to

$$\begin{array}{rl}\underset{t\to \mathrm{\infty }}{lim}\left(\frac{1}{1-p}\right)(t^{1-p}-1^{1-p})& =-\frac{1}{1-p}\end{array}$$

Since the integral diverges for $p\le 1$ and converges for $p>1$, $p=1$ is the boundary between convergence and divergence.

$◻$

Contributors

Ajitani Hifumi

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