Question

Determine all values of \(p\) for which the improper integral converges. $$ \int_{0}^{1} \frac{1}{x^{p}} d x $$

Short answer

The integral converges for values of \( p < 1 \), and diverges for \( p >= 1 \).

Step-by-step solution

Set the integral

The given improper integral is \(\int_{0}^{1} \frac{1}{x^{p}} d x \). This integral is undefined at \( x = 0 \). We can overcome this difficulty by defining the integral as a limit: \( \lim_{a \to 0+} \int_{a}^{1} \frac{1}{x^{p}} d x \).

Evaluate the integral

Next step is to compute the antiderivative of the integrand. The antiderivative of \( \frac{1}{x^{p}} \) is \( \frac{x^{1-p}}{1-p} \) provided that \( p \neq 1 \). Using the Fundamental Theorem of Calculus, we evaluate the limits at 1 and \( a \): \( \lim_{a \to 0+} [\frac{1^{1-p}}{1-p} - \frac{a^{1-p}}{1-p} ] \), which simplifies to \( \lim_{a \to 0+} \frac{1}{1-p} - \lim_{a \to 0+} \frac{a^{1-p}}{1-p} \). The first limit is a constant and doesn't affect the convergence, but the convergence will be determined by the second limit.

Determine the convergence

In the expression \( \lim_{a \to 0+} \frac{a^{1-p}}{1-p} \), note that the limit as \( a \) approaches 0 of \( a^{1-p} \) is 0 for \( 1-p > 0 \) or equivalently \( p < 1 \), and is infinite for \( 1-p < 0 \) or equivalently \( p > 1 \). Therefore, for \( p < 1 \), the integral converges, and for \( p > 1 \), the integral diverges. If \( p = 1 \), we can directly compute the integral \( \int_{0}^{1} \frac{1}{x} d x \) as \( \ln|x| \), which also diverges at \( x = 0 \). Hence, the integral converges only for \( p < 1 \).