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## Easy variant of Fubini thm in some special cases

### Easy variant of Fubini thm in some special cases

Mon May 27, 2013 6:11 pm
sos440
Integration Guru

Posts: 124
Location: California, US

(This is a draft, which still need to be completed. If you find something unsatisfactory, please let me know them so that we can improve the quality together.)

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

1. $\displaystyle x^{-\alpha} f^{(-1)}(x) \to 0$ as $\displaystyle x \to 0^{+}$ and $\displaystyle x \to \infty$, and
2. $\displaystyle x^{-\alpha - n} | f^{(-n)}(x)|$ is integrable on $\displaystyle [0, \infty)$.
Then $\displaystyle \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, dx$ exists (as an improper sense) and the following identity holds.

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

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*}
we confirm that $\displaystyle \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, dx$ exists at least as an improper integral and the identity

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

$\displaystyle \int_{0}^{\infty} \frac{\Gamma(\alpha+n)}{x^{\alpha+n}}|f^{(-n)}(x)| \, dx = \int_{0}^{\infty} \int_{0}^{\infty} s^{\alpha+n-1} |f^{(-n)}(x)| e^{-xs} \, dsdx$
is finite and by Fubini's theorem, we can interchange the order of integration to obtain

\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*}
Here, it is easy to confirm that the inner integral $\displaystyle \textstyle \int_{0}^{\infty} f^{(-n)}(x) e^{-sx} \, dx$ is integrable for any $\displaystyle s > 0$. Thus by taking the limit to the integration by part formula

\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*}
we check that

\displaystyle \begin{align*} \int_{0}^{\infty} f(x) e^{-sx} \, dx = s^{n} \int_{0}^{\infty} f^{(-n)}(x) e^{-sx} \, dx. \end{align*}
Therefore putting the results together, we obtain the desired conclusion. ////

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

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, dx = \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} \frac{s^{\alpha-1}}{s^2 + 1} \, ds.$
Plugging $\displaystyle s^2 = x^{-1} - 1$ and applying the beta function identity, it follows that the result is

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, dx = \frac{\pi}{2\Gamma(\alpha)} \csc \left( \frac{\pi \alpha}{2} \right).$
Every integral harbors a story as fascinating as a comic book!