Prove that $\int_{0}^{+\infty} e^{-tx}\frac{\sin x}{x} dx$ is differentiable

Question

I'm trying to prove that $$F(t) = \int_{0}^{+\infty} e^{-tx}\frac{\sin x}{x} dx$$ is differentiable in $]0,+\infty [$. First, $F(t)$ is well defined since $F(t) \leq \int_{0}^{+\infty} e^{-tx} dx$ whose integration is finite.

Second, I've tried first to prove that $F(t)$ is continuous using dominated convergence theorem, by that I have to find a domination as an integrable function $g$ so that $$ \lvert e^{-tx}\frac{\sin x}{x} \rvert \leq g(x) $$ I've thought of $g(x) = \lvert \frac{\sin x}{x} \rvert$ but it's not a good candidate since it is not Lebesgue integrable. So I've used directly the definition of continuation, it seems good since: $$ \lvert F(t) - F(t') \rvert \leq \int_{0}^{+\infty} \lvert e^{-tx} - e^{-t'x} \rvert dx \overset{t \to t'}{\to} 0 $$

Third, unfortunately, this direction doesn't help to prove that $F(t)$ is derivable. So, I've came back to the differentiability under domination, it doesn't seem working neither. I've tried to look for a domination of $$ \lvert (e^{-tx}\frac{\sin x}{x})' \rvert = \lvert t e^{-tx}\frac{\sin x}{x}\rvert $$ but cannot found any.

Many thanks for any suggestion.

Answer 1

Define $F:(0,\infty)\rightarrow\mathbb{R}$ by $F(t)=\int_{0}^{\infty}e^{-tx}\frac{\sin x}{x}\,dx$. Let $t_{0}\in(0,\infty)$ be arbitrary. We go to show that the derivative $F'(t_{0})=\int_{0}^{\infty}-e^{-t_{0}x}\sin x\,dx$ . Let $(t_{n})$ be an arbitrary sequence in $(0,\infty)$ such that $t_{n}\neq t_{0}$ and $t_{n}\rightarrow t_{0}$. By Heine's Theorem, to show that $\lim_{t\rightarrow t_{0}}\frac{F(t)-F(t_{0})}{t-t}=\int_{0}^{\infty}-e^{-t_{0}x}\sin x\,dx$, it suffices that $\lim_{n\rightarrow\infty}\frac{F(t_{n})-F(t_{0})}{t_{n}-t_{0}}=\int_{0}^{\infty}-e^{-t_{0}x}\sin x\,dx$.

Since $t_{n}\rightarrow t_{0}>0$ and $t_{n}>0$ for all $n$, there exists $t'$ such that $0<t'<t_{n}$ for all $n=0,1,2,\ldots$. (For, choose $\delta>0$ such that $t_{0}-\delta>0$. Then there exists $N$ such that $t_{n}\in(t_{0}-\delta,t_{0}+\delta)\setminus\{t_{0}\}$ whenever $n\geq N$. Take $t'=\frac{1}{2}\min\left(t_{1},t_{2},\ldots,t_{N-1},t_{0}-\delta\right)>0$). For each $n$, define $\theta_{n}:[0,\infty)\rightarrow\mathbb{R}$ by $\theta_{n}(x)=\frac{e^{-t_{n}x}-e^{-t_{0}x}}{t_{n}-t_{0}}\cdot\frac{\sin x}{x}$. Define $\theta:[0,\infty)\rightarrow\mathbb{R}$ by $\theta(x)=e^{-t'x}$. For each $x\in[0,\infty)$, by Mean-Value Theorem, there exists $\xi_{x,n}$ strictly between $t_{0}$ and $t_{n}$ such that \begin{eqnarray*} & & e^{-t_{n}x}-e^{-t_{0}x}\\ & = & -(t_{n}x-t_{0}x)e^{-\xi_{x,n}x}. \end{eqnarray*} Hence, $\left|\theta_{n}(x)\right|\leq|e^{-\xi_{x,n}x}\sin x|\leq\theta(x)$ because $0<t'\leq\xi_{x,n}$ $\Rightarrow e^{-t'x}\geq e^{-\xi_{x,n}x}$. Moreover, by direct calculation, $\lim_{n\rightarrow\infty}\theta_{n}(x)=-e^{t_{0}x}\sin x$. Note that $\theta$ is integrable, so by Lebesgue Dominated Convergence Theorem, we have that \begin{eqnarray*} & & \lim_{n\rightarrow\infty}\frac{F(t_{n})-F(t_{0})}{t_{n}-t_{0}}\\ & = & \lim_{n\rightarrow\infty}\int_{0}^{\infty}\theta_{n}(x)dx\\ & = & \int_{0}^{\infty}\lim_{n\rightarrow\infty}\theta_{n}(x)dx\\ & = & \int_{0}^{\infty}-e^{-t_{0}x}\sin x\,dx. \end{eqnarray*} We conclude that $F'(t_{0})=\int_{0}^{\infty}-e^{-t_{0}x}\sin x\,dx$.

Answer 2

There is a low-tech solution that is instructive because it uses a method that can cope with amplitudes of slow decay. Exploit the fact that the humps in the sinc function decay steadily and reverse sign with every multiple of $\pi$.

Let $I_n(t)= \int_0^{n \pi} e^{- t x} \frac{ \sin x}{x} dx$ and let $I_{\infty}(t)$ denote the function you wish to show is differentiable. Regard the integral $I_n$ as the sum of $n$ separate integrals, taken over consecutive intervals of length $\pi$ (the separate humps).

You can use the Leibniz alternating series test to show that when $n$ is even,

(i) $ I_{n} (t)\leq I_{\infty}(t) \leq I_{n+1}(t)$ and

(ii) the spread between these two endpoints tends to zero as $n\to \infty$.

Consider next $ I'_n(t)=\int_0^{n \pi} -e^{- t x} \sin x dx$ whose integrand also has alternating humps with steadily decreasing mass (because of the exponential decay factor).

By similar reasoning the spread between the derivatives $I'_{n+1}(t) -I'_n(t)$ tends to zero as $n\to \infty$. Then use a standard result in analysis to show that these Cauchy estimates are sufficient to justify interchanging limit and derivative.

Answer 3

@Danny has shown a solution which evaluates the derivative of $F(t)$ via its definition. Another approach is to uses directly the differentiability under domination, it is suggested by Sorfosh via a very similar answer.

The idea is that the differentiability is a local property, to prove that $F(t)$ is differentiable at a point $t_0$ it's enough to consider a neighbourhood $]t_0 - \epsilon, t_0+\epsilon[ \subset ]0,+\infty[$. In that interval:

$$ \lvert (e^{-tx} \frac{\sin x}{x})' \rvert = e^{-tx} \lvert \sin x \rvert \leq e^{-(t_0 - \epsilon)x} $$

Moreover

$$ \int_0^{+\infty} e^{-(t_0 - \epsilon)x} dx = \frac{1}{t_0-\epsilon} < +\infty $$ then we can apply the theorem of convergence under domination.

Answer 4

The key tool here is the following result which is a consequence of the dominated convergence theorem of Lebesgue. Let $\varphi:X\times(a,b)\rightarrow\mathbb{R}$ ($b$ could be $+\infty$) be such that

1. the function $x\mapsto\varphi(x,t)$ is a Borel function for every $t$,

2. the function $t\mapsto\varphi(x,t)$ is a differentiable in $(a,b)$ for every $x\in X$,

3. $\frac{\partial\varphi}{\partial t}(x,t)$ is dominated on $X\times(a,b)$ by an integrable function $\psi(x)$,

then $F(t):=\int_{X}\varphi(x,t)\,dx$ with $t\in(a,b)$, is differentiable for every $t\in(a,b)$ and we have $$ \frac{dF}{dt}(t)=\int_{X}\frac{\partial\varphi}{\partial t}(x,t)\,dx. $$ In the considered case we have

$$ \varphi:\underbrace{\left[0,+\infty\right)}_{X}\times\underbrace{\left(0,\infty\right)}_{(a,b)}\longrightarrow\mathbb{R},\qquad\varphi(x,t)=\frac{\sin x}{x}e^{-tx}. $$ To show that $F$ is differentiable over $\mathbb{R}_{*}^{+}$ we need to reason as follows. Let $t_{0}\in\mathbb{R}_{*}^{+}$, then \begin{equation}\label{eq:1} \left|\frac{\partial}{\partial t}\frac{\sin x}{x}e^{-tx}\right|=\left|\sin xe^{-tx}\right|\leqslant e^{-t_{0}x}\qquad\forall(x,t)\in\left[0,+\infty\right)\times\left[t_{0},+\infty\right).\qquad(1) \end{equation} This means that $F$ is differentiable on every interval $\left[t_{0},+\infty\right)$ for every $t_{0}\in\left(0,+\infty\right)$, i.e. it is differentiable over $\left(0,+\infty\right)$.

N.B. we can not consider directly in (1) $\left(0,+\infty\right)$ because in this case we can not get a dominating function which is integrable over $\left[0,+\infty\right)$ (indeed, for every $x\in\left[0,+\infty\right),$ $\sup_{t\in\left(0,+\infty\right)}\left|\sin xe^{-tx}\right|=\left|\sin x\right|$).

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Another possible way to establish the asked differentiability property is related to the fact that the considered integral can be interpret as the Laplace transform of the function $\frac{\sin x}{x}$, i.e., setting $f(x)=\frac{\sin x}{x}$, $$ F(t)=\mathcal{L}[f(\cdot)](t):=\int_0^{+\infty} \frac{\sin x}{x}e^{-tx}\,dx. $$ There is a general result guaranteeing that $F$ is differentiable for every $t\in\mathbb{R}^+_*$. It suffices for example that there exists a $n\in\mathbb{N}_*$ such that $\lim_{x\rightarrow+\infty}\frac{f(x)}{x^n}=0$. This requirement guarantees that we can differentiate under the sign of integral. Indeed it is know that (a consequence of the dominated Lebesgue theorem) if $\varphi:X\times(a,b)\rightarrow\mathbb{R}$.

Answer 5

You can also follow an elementary approach here based on a well-known theorem for interchanging the derivative and limit of a sequence of differentiable functions. Theorem 7.17 in Principles of Mathematical Analysis by Rudin is

Suppose $(F_n)$ is a sequence of functions, differentiable on $[a,b]$ and such that $(F_n(t_0))$ converges for some point $t_0 \in [a,b]$. If $(F_n')$ converges uniformly on $[a,b]$, then $(F_n)$ converges uniformly on $[a,b]$ to a function $F$, and for all $t \in [a,b]$ we have $F'(t) = \lim_{n \to \infty}F_n'(t)$.

In this case, we have

$$F_n(t) := \int_{0}^{n} e^{-tx}\frac{\sin x}{x} dx, \quad F'_n(t) = -\int_{0}^{n} e^{-tx}\sin x dx$$

Consider any point $t_0 \in [a,b] \subset (0,\infty)$. The improper integral is convergent (by the Dirichlet test) and

$$\lim_{n \to \infty}F_n(t_0) = \int_{0}^{\infty} e^{-t_0x}\frac{\sin x}{x} dx$$

Since $| e^{-tx} \sin x| \leqslant e^{-ax}$ for all $t \in [a,b]$ and $\int_0^\infty e^{-ax} \, dx = \frac{1}{a} < \infty$, it follows by the Weierstrass M-test that $(F_n')$ is uniformly convergent on the interval and applying the theorem we get for all $t \in [a,b]$

$$F'(t) = \lim_{n \to \infty} F_n'(t) = -\int_0^\infty e^{-tx} \sin x \, dx$$

This holds for every $t \in (0,\infty)$ since the choice of $t_0$ and the interval $[a,b]$ was arbitrary.