Here I am going to introduce some easy results on some criteria for interchanging the order of integrations which are not covered by the classical Fubini theorem. Though both statements and proofs are weak and easy, it often reduces our burden to large extent.

Let \(\displaystyle f\) be a locally integrable function on \(\displaystyle [0, \infty)\). That is, \(\displaystyle f\) is a measurable function which is integrable on \(\displaystyle [0, R]\) for any \(\displaystyle R > 0\). We denote the antiderivative of \(\displaystyle f\) by \(\displaystyle \textstyle f^{(-1)}(x) := \int_{0}^{x} f(t) \, dt\).

Since \(\displaystyle f^{(-1)}\) is again locally integrable, we can inductively define \(\displaystyle \textstyle f^{(-n-1)}(x) := \int_{0}^{x} f^{(-n)}(t) \, dt.\)

Any function that appears in the sequel will be considered to be locally integrable unless stated otherwise.

- Theorem 1. Assume there exists \(\displaystyle \alpha > 0\) and \(\displaystyle n \in \Bbb{N}\) such that
- \(\displaystyle x^{-\alpha} f^{(-1)}(x) \to 0\) as \(\displaystyle x \to 0^{+}\) and \(\displaystyle x \to \infty\), and
- \(\displaystyle x^{-\alpha - n} | f^{(-n)}(x)|\) is integrable on \(\displaystyle [0, \infty)\).

- \(\displaystyle \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, dx = \int_{0}^{\infty} \frac{s^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{\infty} f(x) e^{-sx} \, dx ds\)

Proof. Applying the L'hospital's rule to the condition 1, we find that for any \(\displaystyle k \geq 0\) we have \(\displaystyle x^{-\alpha-k} f^{(-k-1)}(x) \to 0\) as \(\displaystyle x \to 0^{+}\) and \(\displaystyle x \to \infty\). So taking the limit \(\displaystyle (a, b) \to (0, \infty)\) to the integration by parts formula

- \(\displaystyle \begin{align*} \int_{a}^{b} \frac{f(x)}{x^{\alpha}} \, dx = \sum_{k=0}^{n-1} \frac{\Gamma(\alpha+k)}{\Gamma(\alpha)} \left[ \frac{f^{(-k-1)}(x)}{x^{\alpha+k}} \right]_{a}^{b} + \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)} \int_{a}^{b} \frac{f^{(-n)}(x)}{x^{\alpha+n}} \, dx, \end{align*}\)

- \(\displaystyle \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, dx = \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)} \int_{0}^{\infty} \frac{f^{(-n)}(x)}{x^{\alpha+n}} \, dx.\)

- \(\displaystyle \int_{0}^{\infty} \frac{\Gamma(\alpha+n)}{x^{\alpha+n}}|f^{(-n)}(x)| \, dx = \int_{0}^{\infty} \int_{0}^{\infty} s^{\alpha+n-1} |f^{(-n)}(x)| e^{-xs} \, dsdx\)

- \(\displaystyle \begin{align*} \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, dx &= \int_{0}^{\infty} \int_{0}^{\infty} \frac{s^{\alpha+n-1}}{\Gamma(\alpha)} f^{(-n)}(x) e^{-xs} \, dsdx \\ &= \int_{0}^{\infty} \frac{s^{\alpha+n-1}}{\Gamma(\alpha)} \left( \int_{0}^{\infty} f^{(-n)}(x) e^{-sx} \, dx \right) \, ds. \end{align*}\)

- \(\displaystyle \begin{align*} \int_{a}^{b} f(x) e^{-sx} \, dx = \sum_{k=0}^{n-1} s^{k} \left[ f^{(-k-1)}(x) e^{-sx} \right]_{a}^{b} + s^{n} \int_{a}^{b} f^{(-n)}(x) e^{-sx} \, dx, \end{align*}\)

- \(\displaystyle \begin{align*} \int_{0}^{\infty} f(x) e^{-sx} \, dx = s^{n} \int_{0}^{\infty} f^{(-n)}(x) e^{-sx} \, dx. \end{align*}\)

Let us consider some examples:

Example. Let \(\displaystyle \alpha \in (0, 2)\). For \(\displaystyle f(x) = \sin x\), we have \(\displaystyle f^{(-1)}(x) = 1 - \cos x = \tfrac{1}{2} x^{2} + O(x^4)\). This allows us to easily check that both the condition 1 and 2 of the Theorem 1 are satisfied with \(\displaystyle n = 1\). So it follows from Theorem 1 that

- \(\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, dx = \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} \frac{s^{\alpha-1}}{s^2 + 1} \, ds.\)

- \(\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, dx = \frac{\pi}{2\Gamma(\alpha)} \csc \left( \frac{\pi \alpha}{2} \right).\)