Questions: There is at least one value of p such that the integral from 0 to infinity of 1/x^p dx is convergent.

Transcript text: There is at least one value of p such that $\int_{0}^{\infty} \frac{1}{x^{p}} d x$ is convergent.

Solution

Solution Steps

To determine if the integral $\int_{0}^{\infty} \frac{1}{x^{p}} \, dx$ is convergent, we need to analyze the behavior of the integrand at the boundaries $0$ and $\infty$. Specifically, we need to find the values of $p$ for which the integral converges.

For the integral to converge at $0$, the exponent $p$ must be such that the integrand does not diverge as $x$ approaches $0$.
For the integral to converge at $\infty$, the exponent $p$ must be such that the integrand does not diverge as $x$ approaches $\infty$.

Solution Approach

Analyze the behavior of the integrand $\frac{1}{x^p}$ at $x = 0$ and $x = \infty$.
Determine the range of $p$ for which the integral converges.

Step 1: Analyze the Integral

We need to determine the conditions under which the integral \[ \int_{0}^{\infty} \frac{1}{x^{p}} \, dx \] is convergent. This involves analyzing the behavior of the integrand $\frac{1}{x^p}$ at the boundaries $x = 0$ and $x \to \infty$.

Step 2: Behavior at $x = 0$

Consider the integral from $0$ to $1$: \[ \int_{0}^{1} \frac{1}{x^{p}} \, dx \]

For this integral to converge, the exponent $p$ must be such that the integrand does not diverge too quickly as $x$ approaches $0$. Specifically, we need: \[ \int_{0}^{1} \frac{1}{x^{p}} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_{0}^{1} \]

If $p < 1$, then $x^{1-p}$ approaches $0$ as $x \to 0$, and the integral converges.
If $p \geq 1$, then $x^{1-p}$ diverges as $x \to 0$, and the integral diverges.

Step 3: Behavior at $x \to \infty$

Consider the integral from $1$ to $\infty$: \[ \int_{1}^{\infty} \frac{1}{x^{p}} \, dx \]

For this integral to converge, the exponent $p$ must be such that the integrand decays sufficiently fast as $x$ approaches $\infty$. Specifically, we need: \[ \int_{1}^{\infty} \frac{1}{x^{p}} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_{1}^{\infty} \]

If $p > 1$, then $x^{1-p}$ approaches $0$ as $x \to \infty$, and the integral converges.
If $p \leq 1$, then $x^{1-p}$ does not decay fast enough, and the integral diverges.

Final Answer

Combining the conditions from both parts, we find that the integral \[ \int_{0}^{\infty} \frac{1}{x^{p}} \, dx \] is convergent if and only if $0 < p < 1$.

Thus, there is at least one value of $p$ such that the integral is convergent.

\[ \boxed{\text{True}} \]

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