Continuity of Laplace transform

Question

I'm trying to prove the Riemann integral $$\int_0 ^ \infty \frac{\sin(x)} x \ dx = \frac{\pi}{2}$$ using tools that I, as a statistician, am likely not to forget. My general approach is

1. Show that the Laplace transform $\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$ is $-\arctan(t) + \frac \pi 2$ for $t \in (0, \infty)$ by differentiating under the integral (using dominated convergence stuff since the Lebesgue integral exists here) and solving the resulting differential equation.
2. Argue that $\psi(t)$ is continuous at $0$.

I've hit a snag at step 2. So, under what conditions do we have that the Laplace transform of a function is continuous at $t = 0$? Any trivial garuntees that it should be for $\frac {\sin x} x$ in particular? I feel like this would be easier if the function was Lebesgue integrable since when I'm trying to bound $|\psi(0) - \psi(t)|$ I am having a hard time controlling the tail of the integral of $\psi(t)$ for $t$ uniformly in $t$ near $0$.

I'm also finding this somewhat interesting since it looks like $\frac{\sin x} x$ is sitting on a boundary of sorts: we are Lebesgue integrable up to a point, then Riemann integrable, and past that not integrable($t > 0, = 0, < 0$) So a bunch of Lebesgue integrable functions are converging to a function which has an improper Riemann integral but no Lebesgue integral. And yet interchanging limits produces the correct answer anyways when calculating the integral.

EDIT: We already have an answer below that suggests how to do this for $\frac {\sin x} x$ but I'd like to get something on the more general question of under what conditions we have right continuity of the Laplace transform at $0$. Is it always continuous provided that the function has a finite improper Riemann integral?

Answer 1

You can get right continuity at $s=0$ along the real axis, at least: We have $$I(s,n):=\int_{n\pi}^\infty \frac{e^{-st}\sin t}{t}\,dt=\sum_{k=n}^\infty\int_{k\pi}^{(k+1)\pi} \frac{e^{-st}\sin t}{t}\,dt$$ and use the alternating nature of the series to show that the entire sum lies between $0$ and its first term. Thus $I(s,n)\to0$ uniformly in $s\ge0$ as $n\to\infty$. On the other hand, you have unifom convergence of the integrand for $t\in[0,n\pi]$, so you don't even need to invoke Lebesgue theory to get the needed convergence.

Answer 2

I would like to show my way to show that $ \int_0 ^ \infty \frac{\sin(x)} x \ dx=\frac{\pi}{2}$

$\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$

$\psi '(t) := -\int_0 ^ \infty e^{-xt} \sin(x) \ dx$

From Laplace tables we can find sine's Laplace transform=$\frac{1}{t^2+1}$ . if you don't want to use Laplace table, replace $\sin(x)=\frac{e^{ix}-{e^{-ix}}}{2i}$ and find the same result easily from $ \int_0 ^ \infty e^{-ax} \ dx=\frac{1}{a}$.

$\psi '(t) := -\frac{1}{t^2+1} $

$\int \psi '(t)dt=-\int \frac{1}{t^2+1} dt$

$\psi(t) := -\arctan(t)+c $

if $\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$
then $\psi(\infty)=\int_0 ^ \infty \frac{e^{-\infty t} \sin(x)} x \ dx =0$

if $\arctan(\infty)=\frac{\pi}{2}$
then

$c=\frac{\pi}{2}$

$\psi(0) := \int_0^\infty \frac {\sin(x)}{x} dx=-\arctan(0)+c=\frac{\pi}{2}$

Answer 3

You can use an alternate result which is quite nice.

$${\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_0^\infty {{e^{ - st}}\frac{{\sin t}}{t}dt} $$

But

$$\frac{{{e^{ - st}}}}{t} = \int\limits_s^\infty {{e^{ - tm}}dm} $$

So

$$\eqalign{ & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_0^\infty {\left( {\int\limits_s^\infty {{e^{ - tm}}} dm} \right)\sin tdt} \cr & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_s^\infty {\left( {\int\limits_0^\infty {{e^{ - tm}}} \sin tdt} \right)dm} \cr & {\mathcal L}\left\{ {\frac{{\sin t}}{t}} \right\}(s) = \int\limits_s^\infty {\frac{{dm}}{{1 + {m^2}}}} \cr} $$

This means that

$$\int\limits_0^\infty {{e^{ - st}}\frac{{\sin t}}{t}dt} = \int\limits_s^\infty {\frac{{dm}}{{1 + {m^2}}}} $$

Now let $s \to 0$.

Answer 4

Here is a general result that answers (2) in the OP:

Theorem: Suppose $f\in L^{loc}_1([0,\infty)$ (i.e., $f\mathbb{1}_{[a,b]}\in L_1([0,\infty)$ for all $0\leq a\leq b<\infty$), and suppose that $I=\int^\infty_0f$ converges (in the sense that $\lim_{A\rightarrow\infty}\int^A_0f$ exists). Then

1. For any $s>0$, $I(s)=\int^\infty_0e^{-st}f(t)\,dt$ converges (that is $I(s):=\lim_{A\rightarrow\infty}\int^A_0e^{-s t}f(t)\,dt$ exists as a real number).
2. $\lim_{s\rightarrow0}I(s)=I$.

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Comments:

1. Very often, $f$ is Riemann integrable on every closed interval $[a,b]\subset[0,\infty)$ and $I=\int^\infty_0f$ is defined as a Riemann improper integrable. The assumption in the Theorem above generalizes this situation.
2. In the context of the OP, $f(t)=\frac{\sin t}{t}\mathbb{1}_{\{t\neq0\}}+\mathbb{1}_{\{t=0\}}(t)$ satisfies the condition of the Proposition since for any $n\in\mathbb{N}$, $$\int^{\pi (n+1)}_0\frac{\sin t}{t}\,dt=\sum^n_{k=0}(-1)^k\int^{(k+1)\pi}_{k\pi}\frac{|\sin t|}{t}\,dt$$ defines an alternating series with main term $a_k=\int^{(k+1)\pi}_{k\pi}\frac{|\sin t|}{t}\,dt=\int^{\pi}_0\frac{\sin t}{k\pi+t}\,dt\searrow0$ as $k\rightarrow\infty$. Notice that $f_s(t):= e^{-s t}f(t)$ is in fact Lebesgue integrable for any $s>0$, which is a much stronger condition than needed.

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Proof of Theorem: Fix $s>0$. Since $f$ is (Lebesgue) integrable in any bounded interval in $[0,\infty)$ then so is $t\mapsto e^{-st}f(t)$. By the second mean value theorem of Lebesgue integrals (see here for example), for any $0<A<B$, there exists $x_0=x_0(A,B)\in[A,B]$ such that $$\int^B_A e^{-s t} f(t)\,dt= e^{-As}\int^{x_0}_A f(t)\,dt$$ Since $I$ is convergent, for any $\varepsilon>0$, there is $M>0$ such that \begin{align} \Big|\int^p_q f(t)\,dt\Big|<\varepsilon/2\quad\text{whenever}\quad M\leq p\leq q\tag{1}\label{one} \end{align} As $0<e^{-As}\leq1$, we have that for $B\geq A\geq M$, \begin{align} \Big|\int^B_A e^{-sA}f(t)\,dt\Big|=e^{-sA}\Big|\int^{x_0}_Af(t)\,dt\Big|<\frac{\varepsilon}{2}\tag{2}\label{two} \end{align} This implies that $J(m):=\int^m_0e^{-st}f(t)\,dt$ is Cauchy and so, $\lim_{m\rightarrow\infty}J(m)$ exists (as a real number). This proves (1). Observe that \eqref{two} is uniform in $s>0$.

To prove (2) let $M$ large enough so that \eqref{one} holds. Then, for $0\leq M\leq A\leq B$ \begin{align} \Big|\int^B_0 f(t)\,dt-\int^B_0 e^{-st}f(t)\,dt\Big|&\leq \Big|\int^A_0 f(t)(1-e^{-st})\,dt\Big|+\Big|\int^B_Af(t)\,dt\Big|+\Big|\int^B_Ae^{-st}f(t)\,dt\Big|\\ &<\int^A_0 |f(t)|(1-e^{-st})\,dt+\varepsilon \end{align} Letting $B\rightarrow\infty$ yields $$|I-I(s)|\leq\int^A_0 |f(t)|(1-e^{-st})\,dt+\varepsilon$$ As $|f(t)|(1-e^{-st})\mathbb{1}_{[0,A]}(t)\leq |f(t)|\mathbb{1}_{[0,A]}(t)\in L_1([0,\infty)$ and $\lim_{s\rightarrow0} f(t)(1-e^{-st})\xrightarrow{s\rightarrow0}f(t)$ pointwise, $$\limsup_{s\rightarrow0}|I-I(s)|\leq\varepsilon$$ by dominated convergence. Part (2) of the theorem follows immediately.