Question

Determine all values of \(p\) for which the improper integral converges. $$ \int_{1}^{\infty} \frac{1}{x^{p}} d x $$

Short answer

The improper integral \(\int_{1}^{\infty} \frac{1}{x^{p}} dx\) converges for \(p>1\) and diverges for \(p\leq1\).

Step-by-step solution

Set up the Integral

The first step is to set up the integral:\[\int_{1}^{\infty} \frac{1}{x^{p}} d x \]. Next this needs to be rewritten as a limit since it's an improper integral:\[\lim_{b\to\infty}\int_{1}^{b} \frac{1}{x^{p}} dx \]

Perform the Integral

Now we need integrate. The integral of \(1/x^p\) regarding dx from 1 to b is: \[\lim_{b\to\infty}\left[\frac{1} {1-p}x^{1-p}\right]_1^b\],which simplifies to : \[\lim_{b\to\infty} \frac{b^{1-p}}{1-p} - \frac{1}{1-p} \]

Evaluate the Limit

To determine the convergence, we need to calculate the limit of this expression as \(b\) approaches infinity. This value is influenced by the exponent of \(b\), which is \(1-p\). If \(1-p > 0\), the term \(b^{1-p}\) will diverge (go to infinity), and if \(1-p < 0\), the term will go to zero. If \(1-p = 0\) (which occurs when \(p=1\)), the term is a constant, and the integral has a particular form that must be handled separately. Thus, we divide the solution into two cases, \(p>1\) and \(p<1\) (excluding the case \(p=1\)). For p < 1, \[\lim_{b\to\infty} \frac{b^{1-p}}{1-p} - \frac{1}{1-p} = \infty\],and for p > 1, \[\lim_{b\to\infty} \frac{b^{1-p}}{1-p} - \frac{1}{1-p} = \frac{-1}{p-1}\]Which is a finite value.

Draw conclusion based on results

Based on the results in Step 3, it can be concluded that the integral \(\int_{1}^{\infty} \frac{1}{x^{p}} dx\) is convergent if \(p>1\) and divergent if \(p\leq 1\)