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Define $F:[0,\infty)\to\mathbb{R}$ by $$F(t)=\int_{0}^{\infty}e^{-tx}\dfrac{\sin x}{x}\,dx.$$ Prove that $F$ is continuous at $t=0$.

I already proved that $$F'(t)=-\dfrac{1}{1+t^2}, \quad t>0.$$ I don't know if this result could be useful for the proof of the continuity at 0.

Ángela Flores
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1 Answers1

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Recall that for a sequence of continuous functions, if we have uniform convergence $F_n(t) \to F(t)$ as $n \to \infty$ for $t \in [a,b]$ then $F$ is continuous.

This improper integral is convergent and we have for each $t$,

$$\lim_{n\to \infty}F_n(t) = \lim_{n\to \infty}\int_{1/n}^{n}e^{-tx}\dfrac{\sin x}{x}\,dx = \int_{0}^{\infty}e^{-tx}\dfrac{\sin x}{x}\,dx = F(t)$$

Since the integrand is continuous for $(x,t) \in [1/n,n]\times[0,1]$ we have for each $n$ that $F_n$ continuous for $t \in [0,1]$. If we can show that the convergence is uniform, then it follows that $F$ is continuous on $[0,1]$ and, in particular, at $t= 0$.

In this case, the uniform convergence of the improper integral follows from Abel's test. See here and here.

The conditions are met here since $x \mapsto e^{-tx}$ is bounded and monotone for $(x,t) \in [0,\infty) \times [0,1]$ and the improper integral,

$$\int_0^ \infty \frac{\sin x}{x} \, dx, $$

is convergent and, thus, uniformly convergent for all $t \in[0,1]$ trivially since there is no dependence on $t$.

Proof that $F_n$ is continuous on $[0,1]$

The integrand $f(x,t) = e^{-tx}\dfrac{\sin x}{x}$ is uniformly continuous on the compact set $[1/n,n]\times [0,1]$. Take any $t_0 \in [0,1]$. For any $\epsilon > 0$ there exists $\delta > 0$ independent of $x,t$ such that if $|t- t_0| < \delta$, we have

$$|f(x,t) - f(x,t_0)| < \epsilon/(n-1/n)$$

Thus,

$$|F(t) - F(t_0)| \leqslant \int_{1/n}^n|f(x,t) - f(x,t_0)| \, dx < (n-1/n)\epsilon/(n - 1/n) = \epsilon$$

RRL
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  • How can you see that F_n is continuous at t=0? It's possible to see that it is continuous for t>0. I think is the same problem with the original integral. – Ángela Flores May 17 '19 at 01:32
  • But can you assure that this extension coincides with the definition given for F? – Ángela Flores May 17 '19 at 01:38
  • What does that actually mean? First we want to establish that $f(x,t) = e^{-tx}\dfrac{\sin x}{x}$ for $x > 0$ is continuous. That is obvious -- all the component functions are continuous. For $x=0$, the integrand as written is not defined but we can replace it with the extension $f(0,t) = e^{-tx}$ since this does not change the value of the integral. At this point I don't care about $F$. I'm just showing that an integral of $f(x,t)$ over a compact set is continuous. If you really want to be precise we could look at this as an improper integral at both ends. – RRL May 17 '19 at 01:43
  • I'll change the lower limit of the integral for you in the answer. – RRL May 17 '19 at 01:44
  • I can understand why the argument follows. My problem here is that you can prove easily the continuity of F_n(t) for t>0 (or F) (using dominated convergence theorem, for example. Here is when you use the continuity of the integrand), but not the continuity at t=0. I don't know if the continuity of the integrand in t=0 for every fixed x is enough to say that F is, too. – Ángela Flores May 17 '19 at 01:54
  • I purposely did not use the dominated convergence theorem for the reason you mentioned. We can't find an integrable dominating function when $t=0$. The integral of $\sin(x)/x$ is conditionally convergent and that of $|\sin(x)/x|$ is divergent. You need this approach. The convergence of this integral is uniformly convergent on the closed interval including $t=0$. That is actually a harder part to understand. You are telling me that $F_n(t)$ is not continuous because of something about $F$ -- but $F$ is irrelevant here. – RRL May 17 '19 at 02:01
  • Do you disagree that the function $f(x,t) = e^{-tx}\dfrac{\sin x}{x}$ is continuous on $[1/n,n]\times[0,1]$. Even when $t=0$ we have $f(x,0) = \dfrac{\sin x}{x}$ which is continuous. – RRL May 17 '19 at 02:05
  • I agree. But I can't see how this continuity implies the continuity of $\int f(x,t),dx$. in $[0,1]$. – Ángela Flores May 17 '19 at 02:08
  • That is a basic theorem: If $f:[a,b]\times[c,d] \to \mathbb{R}$ is continuous then $F(t) = \int_a^bf(x,t) , dx$ is continuous on $[c,d]$. This works because we are dealing with compact intervals and a continuous function on a compact set is uniformly continuous. Do you want to see more of the proof? Most anyalisys books have this theorem. – RRL May 17 '19 at 02:14
  • Thanks! Now I can see it. – Ángela Flores May 17 '19 at 02:21
  • Yes. The main problem now is to find the right statement of the Abel test. – Ángela Flores May 17 '19 at 02:45
  • @ÁngelaFlores: I added the details on Abel's test for uniform convergence of an improper integral with a parameter. The second link indicates how it is proved if you are interested. – RRL May 17 '19 at 03:47