📂Lemmas

Integrability of 1/x^p

Theorem

The integrability of function $f(x) = \dfrac{1}{x^{p}}$ is as follows:

1. when $x \in (0,1]$, if $p \lt 1$, then $f$ is integrable.

2. when $x \in [1, \infty)$, if $p \gt 1$, then $f$ is integrable.

Explanation

If $x$ is less than $1$, then $p$ must also be less than $1$, and if $x$ is greater than $1$, then $p$ must also be greater than $1$. Just remember this.

Proof

In the case of $x \in (0,1]$

In the case of $p \lt 1$

Since it’s $1 - p \gt 0$, it converges as follows:

$$\begin{align*} \int_{0}^{1} \dfrac{1}{x^p} dx &= \left. \dfrac{1}{-p+1} x^{1-p} \right|_{0}^{1} \\ &= \dfrac{1}{-p+1} \left( 1 - 0 \right) \\ &= \dfrac{1}{-p+1} \lt \infty \end{align*}$$

In the case of $p \gt 1$

Since it’s $p - 1 \gt 0$, it converges as follows:

$$\begin{align*} \int_{0}^{1} \dfrac{1}{x^p} dx &= \left. \dfrac{1}{-p+1} \dfrac{1}{x^{p-1}} \right|_{0}^{1} \\ &= \dfrac{1}{-p+1} \left( 1 - \infty \right) \\ &= \infty \end{align*}$$

In the case of $p = 1$

The integral diverges as follows:

$$\int_{0}^{1} \dfrac{1}{x} dx = \left. \log x \right|_{0}^{1} = \log 1 - (-\infty) = \infty$$

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In the case of $x \in [1, \infty)$

In the case of $p \gt 1$

Since it’s $p - 1 \gt 0$, it converges as follows:

$$\begin{align*} \int_{1}^{\infty} \dfrac{1}{x^p} dx &= \left. \dfrac{1}{-p+1} \dfrac{1}{x^{p-1}} \right|_{1}^{\infty} \\ &= \dfrac{1}{-p+1} \left( \dfrac{1}{\infty^{p-1}} - 1 \right) \\ &= \dfrac{1}{p-1} \lt \infty \end{align*}$$

In the case of $p \lt 1$

Since it’s $1-p \gt 0$, it diverges as follows:

$$\begin{align*} \int_{1}^{\infty} \dfrac{1}{x^p} dx &= \left. \dfrac{1}{-p+1} x^{1-p} \right|_{1}^{\infty} \\ &= \dfrac{1}{-p+1} \left( \infty^{1-p} - 1 \right) \\ &= \infty \end{align*}$$

In the case of $p = 1$

The integral diverges as follows:

$$\int_{1}^{\infty} \dfrac{1}{x} dx = \left. \log x \right|_{1}^{\infty} = \log(\infty) - \log 1 = \infty$$

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