## Interchanging the order of integration (On progress)

2011. 2. 20. 22:57
Here we are going to present some conditions that enables us to interchange the order of integration which is often not covered by classic Fubini's Theorem or its variants. We assume readers are familiar with basic theory of Lebesgue-Stieltjes integration. (Refer to this article for the definition of terms that are used throughout this post)

1. Preliminaries

We start with two definitions that slightly extends the concept of improper Riemann integral and antiderivative, respectively.

Let $f$ be a locally integrable function defined on the set $\mathbb{R}^{+} = (0, \infty)$ of positive real numbers, $g$ be a complex-valued function on $\mathbb{R}^+$, and $I$ be a complex number. Suppose it happens that $g$ is of bounded variation on $I_n$ and $\textstyle \int_{I_n} f \, dg$ converges to $I$ whenever $(I_n)$ is an increasing sequence of closed bounded intervals $I_n \in \mathbb{R}^{+}$ with $I_n \nearrow \mathbb{R}^{+}$. Then we say $f$ is improperly integrable with respect to $g$ on $\mathbb{R}^+$, or simply, $f \, dg$ is improperly integrable on $\mathbb{R}^+$. Also, we call $I$ the (improper) integral of $f \, dg$ on $\mathbb{R}^+$ and denote this fact by

$I \ = \ \int_{0^+}^{\infty} f \, dg.$

If $g$ is the identity function, then the Lebesgue-Stieltjes integral reduces to the usual Lebesgue integral, and we may drop $dg$. Then it is clear that every integrable function on $\mathbb{R}^+$ is also improperly integrable with same integral value.

We make another convention that, if $a, b > 0$ and $f \chi_{(0, a]}$ (or $f \chi_{[b, \infty)}$, resp.) is improperly integrable with respect to $g$ on $\mathbb{R}^+$, we often denote its integral as $\textstyle \int_{0^+}^{a} f \, dg$ (or $\textstyle \int_{b}^{\infty} f \, dg$, resp.).

Next, we extend the concept of antiderivative. A function $\phi : [0, \infty) \to \mathbb{C}$ is called an $\mathcal{L}$-measure if
(a) it is right-continuous on its domain,
(b) it is of bound-variation on any closed interval $[a, b]$ with $0 < a < b < \infty$, and
(c) $\phi(0) = 0$ and $\lim_{t\to\infty} \phi (t)$ converges to a finite limit.
We denote $\textstyle \lim_{t\to\infty} \phi (t)$ as $\phi(\infty)$, if no confusion arises.

Several important classes of functions fall into the category of $\mathcal{L}$-measures. Some examples are immediate:

1.1. Example. Let $f$ be an improperly integrable function on $\mathbb{R}^+$. Then the function $\phi$, defined as $\textstyle \phi(t) = \int_{0^+}^{t} f$ for $t > 0$ and $\phi(0) = 0$, is an $\mathcal{L}$-measure. Its verification is straightforward.

1.2. Example. Let $(a_n)_{n=1}^{\infty}$ be a sequence of complex numbers such that $\textstyle \sum_n a_n$ converges (either absolutely or conditionally). Then $\textstyle \phi(t) = \sum_{n \leq t} a_n$ defined for $t \geq 0$ is an $\mathcal{L}$-measure.

Now we define the Laplace-Stieltjes transform of an $\mathcal{L}$-measure $\phi$ as

$\mathcal{L}_\phi (s) \ = \ \int_{0^+}^{\infty} e^{-st} \, d\phi(t).$

For this definition to make sense, we must check that $e^{-st} \, d\phi(t)$ is improperly integrable for an appropriate domain of $s$. We claim that this is the case when $s = 0$ or $\Re (s) > 0$. The former is immediate by (a) and (c) of the definition of $\mathcal{L}$-measure, so assume $\Re (s) > 0$. Let $\tilde{\phi}$ be a function which coincides with $\phi$ on the set of continuity of $\phi$ and is regularized at each discontinuity of $\phi$. Simply, this means that $\tilde{\phi}$ is defined as

$\tilde{\phi}(t) \ = \ \lim_{h \to 0} \frac{\phi(t + h) + \phi(t - h)}{2}.$

Then for $0 < a < b < \infty$, integration by parts for Lebesgue-Stieltjes integral yields

$\int_{a}^{b} e^{-st} \, d\phi(t) \ = \ e^{-sa}\tilde{\phi}(a) - e^{-sb}\tilde{\phi}(b) + s \int_{a}^{b} \phi(t) e^{-st} \, dt...$

where the last integral is a Lebesgue integral. Since $|\phi(t)e^{-st}|$ is integrable on $[0, \infty)$, taking $(a, b) \nearrow (0, \infty)$ and applying Dominated Convergence Theorem completes the proof of our claim. We also obtain an identity which deserves its own right:

$\mathcal{L}_{\phi} (s) \ = \ s \int_{0}^{\infty} \phi(t) e^{-st} \, dt. \qquad (\Re (s) > 0)$

We make additional note that, if $\phi$ is as in Example 1.1, then $d\phi(t) = f \, dt$ and the Laplace-Stieltjes transform reduces to the ordinary Laplace transform.

2. Main Theorem

Now we state our main theorem:

2.1. Theorem. Let $\phi$ be an $\mathcal{L}$-measure, and $F = \mathcal{L}_{\phi}$ be its Laplace transform. Then
(a) $F(s)$ is analytic for $\Re (s) > 0$ with
$F^{(n)}(s) = (-1)^{n} \int_{0^+}^{\infty} t^n e^{-st} \, d\phi(t)$
for $n > 0$,
(b) $F(s) \to \phi(\infty)$ as $s \to 0$ in Stolz angle,
(c) $F(s) \to 0$ as $s \to \infty$ in Stolz angle,
(d) $s^n F^{(n)}(s) \to 0$, either as $s \to 0$ or as $s \to \infty$, in Stolz angle for $n > 0$.

Assuming this theorem, we obtain

2.2. Corollary. Let $\phi$ and $F$ be as in Theorem 2.1. Then for $n \geq 0$

$\phi(\infty) = \frac{1}{n!} \int_{0^+}^{\infty} \int_{0^+}^{\infty} t^{n+1} u^{n} e^{-tu} \, d\phi(t)du,$

and for $\alpha > 0$,

$\phi(\infty) = \frac{1}{\alpha} \int_{0^+}^{\infty} \int_{0^+}^{\infty} (t^{2} + \alpha^2) \sin (\alpha u) \; e^{-tu} \, d\ph...$

This corollary can be regarded as changing the order of integration, since the equality clearly holds if $d\phi(t)du$ is replaced by $dud\phi(t)$. As an instance, this provides a very simple way to calculate the famous Dirichlet integral

$I = \int_{0}^{\infty} \frac{\sin x}{x} \, dx.$

We choose $\textstyle \phi(t) = \int_{0}^{t} (\sin x / x) \, dx$, as in Example 1.1. Then $I = \phi(\infty)$, hence Corollary 2.2.(a) with $n = 0$ yields

$\begin{eqnarray*}\int_{0}^{\infty} \frac{\sin t}{t} \, dt& = & \int_{0}^{\infty} \int_{0^+}^{\infty} t e^{-ut} \, d\p...$