Interchanging the order of integration (2)

Here I am going to introduce some easy results on some criteria for interchanging the order of integration 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 $f$ be a locally integrable function on $(0, \infty)$. That is, $f$ is a measurable function which is integrable on any compact subset $K \subset (0, \infty)$. We say that $f$ has an antiderivative vanishing at the origin if there exists a continuous function $F$ on $[0, \infty)$ such that $F(0) = 0$ and

$$ \int_{a}^{b} f(t) \, \mathrm{d}t = F(b) - F(a) $$

holds for any $0 < a < b< \infty$. Then the Lebesgue differentiation theorem shows that $F$ is differentiable a.e. with $F' = f$. We will use the notations

$$ f^{(-1)}, \quad D^{-1} f, \quad \cdots $$

to denote $F$. Since any continuous function on $[0, \infty)$ has an antiderivative vanishing at $0$ by the fundamental theorem of calculus, we can inductively define

$$ f^{(-n-1)}(x) := \int_{0}^{x} f^{(-n)}(t) \, \mathrm{d}t, \quad n \geq 1. $$

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

 

Theorem 1. Suppose that $f$ has an antiderivative vanishing at the origin and that there exists $\alpha > 0$ such that

 

1. $x^{-\alpha} f^{(-1)}(x) \to 0$ as $x \to 0^{+}$ and $x \to \infty$, and

2. $\int_{0}^{\infty} x^{-\alpha - n} \left| f^{(-n)}(x) \right| \, \mathrm{d}x < \infty$ for some $n \in \Bbb{N}$.

 

Then the integral $\int_{0}^{\infty} x^{-\alpha} f(x) \, \mathrm{d}x$ exists (as in improper sense) and the following identity holds.

\begin{equation*} \int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, \mathrm{d}x = \int_{0}^{\infty} \frac{s^{\alpha-1}}{\Gamma(\alpha)} \int_{0}^{\infty} f(x) e^{-sx} \, \mathrm{d}x \mathrm{d}s \tag{1} \end{equation*}

 

Proof. Applying the L'hospital's rule to the condition 1, we find that for any $k \geq 0$ we have

$$ \frac{f^{(-k-1)}(x)}{x^{\alpha+k}} \to 0, \qquad \text{as} \ \ x \to 0^{+} \ \ \text{and} \ \ x \to \infty. $$

So taking the limit $(a, b) \to (0, \infty)$ to the integration by parts formula

\begin{align*} \int_{a}^{b} \frac{f(x)}{x^{\alpha}} \, \mathrm{d}x = \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}} \, \mathrm{d}x, \end{align*}

we confirm that $\int_{0}^{\infty} x^{-\alpha} f(x) \, \mathrm{d}x$ exists at least as an improper integral and the identity

$$\int_{0}^{\infty} \frac{f(x)}{x^{\alpha}} \, \mathrm{d}x = \frac{\Gamma(\alpha+n)}{\Gamma(\alpha)} \int_{0}^{\infty} \frac{f^{(-n)}(x)}{x^{\alpha+n}} \, \mathrm{d}x$$

holds. But by assumption,

$$ \int_{0}^{\infty} \int_{0}^{\infty} s^{\alpha+n-1} \left| f^{(-n)}(x) \right| e^{-xs} \, \mathrm{d}s \mathrm{d}x = \int_{0}^{\infty} \frac{\Gamma(\alpha+n)}{x^{\alpha+n}} \left|f^{(-n)}(x) \right| \, \mathrm{d}x < \infty $$

by Fubini's theorem, we can interchange the order of integration to obtain

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

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

\begin{align*} \int_{a}^{b} f(x) e^{-sx} \, \mathrm{d}x = \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} \, \mathrm{d}x, \end{align*}

we check that

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

Therefore putting the results together, we obtain the desired conclusion $\text{(1)}$.

 

Note that, slightly modifying the proof and replacing the role of Fubini's theorem by the Tonelli's theorem, the condition 2 can be replaced by

 

$f^{(-n)} \geq 0$ for some $n \in \Bbb{N}$.

 

The theorem above assumes some decaying behavior of $f$. But in some practical uses, this condition automatically follows and hence need not be checked. The following corollaries discusses the circumstances where it is the case.

 

Lemma 2. Assume there exists $\alpha > 0$ such that $\int_{0}^{\infty} x^{-\alpha} f(x) \, \mathrm{d}x$ exists as in improper sense. Then $f$ has an antiderivative vanishing at the origin and the condition 1 of the Theorem #.

 

Proof. Let $g(x) = x^{-\alpha} f(x)$. Then the condition of the lemma shows that $g$ is locally integrable and

$$ \lim_{\begin{subarray}{l} a \to 0^{+} \\ b \to \infty \end{subarray}} \int_{a}^{b} g(t) \, \mathrm{d}t \quad \text{converges}. $$

If we put $G(x) = \int_{1}^{x} g(t) \, \mathrm{d}t$, then it implies that $G(x)$ converges as either $x \to 0^{+}$ or $x \to \infty$. Thus by letting $\tilde{G}(x) = G(x) - G(0^{+})$, it immediately follows that $\tilde{G}$ is an antiderivative of $g$ vanishing at the origin.

 

Now let $$ F(x) = x^{\alpha} g^{(-1)}(x) - \alpha \int_{0}^{x} t^{\alpha-1} g^{(-1)} (t) \, \mathrm{d}t. $$ Since $t^{\alpha - 1}$ is integrable near $t = 0$ and $g^{(-1)}(t)$ is continuous, $F(x)$ is a well-defined continuous function such that $F(x) \to 0$ as $x \to \infty$. Moreover, it is easy to check that $F'(x) = f(x)$ a.e. This implies that $f$ has the antiderivative vanishing at the origin and $f^{(-1)} = F$.

 

Finally, it remains to show that $f^{(-1)}$ satisfies the condition 1 of the Theorem 1. But this is a direct consequence of the L'hospital's rule applied to $F(x)$. Therefore the proof is complete.

 

As a corollary, we obtain the following statement.

 

Corollary 3. Assume there exists $\alpha > 0$ such that

 

1. $\int_{0}^{\infty} x^{-\alpha} f(x) \, \mathrm{d}x$ exists as in improper sense and

2. $\int_{0}^{\infty} x^{-\alpha - n} \left| f^{(-n)}(x) \right| \, \mathrm{d}x < \infty$ for some $n \in \Bbb{N}$.

 

Then $\text{(1)}$ holds. Also, the condition 2 can be replaced by the following one:

 

$f^{(-n)} \geq 0$ for some $n \in \Bbb{N}$.

 

Remark. In the previous posting, we showed that the condition 2 of Corollary 3 is unnecessary if $\alpha$ is a positive integer.

 

Now we consider some easy applications.

 

Example 4. Let $\alpha \in (0, 2)$ and consider the integral

$$ \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, \mathrm{d}x. $$

It is easy to check that this integral exists as in improper sense. Also, the antiderivative of $\sin x$ is $1 - \cos x$, which vanishes at $0$ and is non-negative. Thus by the Corollary 3,

\begin{align*} \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, \mathrm{d}x &= \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} s^{\alpha - 1} \left( \int_{0}^{\infty} \sin x \, e^{-sx} \, \mathrm{d}x \right) \, \mathrm{d}s = \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} \frac{s^{\alpha-1}}{s^2 + 1} \, \mathrm{d}s \end{align*}

Plugging $s = \tan \theta$ and applying the beta function identity and the Euler's reflection formula, we obtain

$$ \int_{0}^{\infty} \frac{\sin x}{x^{\alpha}} \, \mathrm{d}x = \frac{\pi}{2\Gamma(\alpha)} \csc \left( \frac{\pi \alpha}{2} \right). $$