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Does $$\int_{0}^{\infty} \frac{\sin(x)}{x} e^{-\alpha x}dx \hspace{0.1cm}, \alpha \in ]0,\infty[$$ converge uniformly?

Using the Dirichlet test:

  1. $\int_0^\infty \frac{\sin(x)}{x}dx = \pi/2$
  2. $e^{-\alpha x}$ is decreasing, bounded and going to $0$.

So it converges uniformly.

Is this ok? Or does it only converge uniformly in $]k,\infty[$ with $k>0$ ?

gt6989b
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Pedro Fernandes
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1 Answers1

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Hint to use the Dirichlet test:

We have $\int_0^c \sin x \, dx$ bounded for all $c > 0$ and independent of $\alpha$ and $\frac{e^{-\alpha x}}{x}$ is monotonically decreasing in $x$ and uniformly convergent to $0$ as $x \to \infty$ for all $\alpha \in [0,\infty)$.

RRL
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  • Thanks again! This subject is really hard for me because I don't think I have the best definitions of criteria and I get confused, and I can't find anything on this on Wikipedia( unbelievable!). I just find theory on the criteria of series, and I have to guess and adapt to integrals. For instance, the criteria I have for the Dirichlet test didn't say that that the function needed to converge uniformly to 0, just that it converges to 0. If you have a pdf or something on this, I appreciate it. I will probably still struggle a little bit, but at least I'll know I'm using the right criteria. – Pedro Fernandes Aug 20 '20 at 19:27
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    .-- and most books now don't really cover this topic well. However The Elements of Real Analysis by Bartle (earlier editions like 2nd) has a good section on improper integrals with a parameter. – RRL Aug 20 '20 at 19:35
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    @PedroFernandes: Actually this is very good. – RRL Aug 20 '20 at 19:37
  • Thank you for the book reference and the link. One question, $|f_{n}(x)|=|\frac{e^{-n x}}{x}|$ doesn't have a supremum right? So why is it uniformly convergent? – Pedro Fernandes Aug 20 '20 at 20:56
  • For $\alpha,x \geqslant 0$ we have $0 \leqslant e^{-\alpha x} \leqslant 1$. Thus, $\left|\frac{e^{-\alpha x}}{x} \right| = \frac{|e^{-\alpha x}|}{x} \leqslant \frac{1}{x}$. The RHS converges to $0$ and does not depend on $\alpha$. Hence, we have uniform convergence. – RRL Aug 20 '20 at 21:01
  • I realize now that you can prove uniform convergence as you did because of Abel's test. This says that $\int_0^\infty f(x, \alpha) g(x, \alpha) , dx$ converges uniformly for $\alpha \in D$ if $\int_0^\infty f(x,\alpha) , dx$ is uniformly convergent and $g$ is monotone (increasing or decreasing) and uniformly bounded for all $x$ and $\alpha \in D$. You have that here with $f(x,\alpha) = \sin x/x$ and $g(x,\alpha) = e^{-\alpha x}$ – RRL Aug 20 '20 at 21:08
  • I'm confused. I thought we had to take the limit of $\alpha$. – Pedro Fernandes Aug 20 '20 at 21:08
  • No we care that $\frac{e^{-\alpha x}}{x} \searrow 0$ as $x \to \infty$ uniformly for $\alpha \in D$. – RRL Aug 20 '20 at 21:10
  • Here is a proof of both tests without the uniform part. They are closely related and the discussion of Abel's test for uniform convergence of an improper integral is even harder to find. – RRL Aug 20 '20 at 21:12
  • Clearly $|e^{-\alpha x}| \leqslant 1$ so it is uniformly bounded for all $x, \alpha \geqslant 0$. Also $e^{-\alpha x_2} \leqslant e^{-\alpha x_1}$ for all $\alpha \geqslant 0$ when $x_2 > x_1 \geqslant 0$. Also $\int_0^\infty \frac{\sin x}{x} , dx$ converges (uniformly) since there is no dependence on $\alpha$ and Abel's test applies. So this can be proved with either test. – RRL Aug 20 '20 at 21:18
  • Thank you so much for your patience! I get that the abel test works now, I was forgetting uniform convergence only applies on integrals/series/functions with parameters, so now that I understand that it becomes clear in my mind. At first, I thought of Abel's test but I was thinking 'is sin(x)/x uniformly convergent?'. Still the Dirichlet test justification is bugging me. Because uniform convergence means that given an $\epsilon>0$ we can take an $n>N$ such that $|f_{n}(x)| < \epsilon \forall x>0$ or am I missing something? Can I get such $N$ ? – Pedro Fernandes Aug 20 '20 at 21:34
  • We can take any N that's it? Because $n>N \implies \frac{e^{-nx}}{x}<\frac{e^{-Nx}}{x} \leq \frac{1}{x} < \epsilon$ is that it? – Pedro Fernandes Aug 20 '20 at 21:40
  • No we need that there exists $X_0 > 0$ independent of $\alpha$ such that for all $x > X_0$ we have $\frac{e^{-\alpha x}}{x} < \epsilon$ for ALL $\alpha \in [0,\infty)$. I think you are confusing this Dirichlet test for uniform convergence of an improper integral with some question of when you can take a limit with respect to $n \to \infty$ under an integral $\int_0^\infty f_n(x) , dx$. – RRL Aug 20 '20 at 21:48
  • I think I may be talking about something else? I was thinking of this https://en.wikipedia.org/wiki/Uniform_convergence because now we are talking about uniform convergence of a function, and not of an integral or series right? What am I confusing? – Pedro Fernandes Aug 20 '20 at 22:00
  • It's because $\alpha$ is not discrete right? – Pedro Fernandes Aug 20 '20 at 22:01
  • I think I never saw a definition of uniform convergence of a function with a continuous parameter, is that what I am confused with? – Pedro Fernandes Aug 20 '20 at 22:02
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    To be clear you are asking first and foremost about the uniform convergence of an improper integral $\int_0^\infty f(x,\alpha) g(x, \alpha) , dx$ for $\alpha \in [0,\infty)$. That means you want to know if $\int_0^b f(x,\alpha) g(x, \alpha) , dx \to \int_0^\infty f(x,\alpha) g(x, \alpha) , dx$ as $b \to \infty$ UNIFORMLY for $\alpha \in [0,\infty)$. The Dirichlet test can be applied if along with the other hypothesis we have $g(x,\alpha) \searrow 0$ as $x \to \infty$ uniformly for all $\alpha \in [0,\alpha)$. It is important that we consider this limit as $x \to \infty$. – RRL Aug 20 '20 at 22:11
  • We dont care about the limit as $\alpha \to \infty$ nor is this a question of the uniform convergence of a sequence of functions $f_n(x) \to f(x)$ as $n \to \infty$ for $x \in D$. Now $g(x,\alpha) \to 0$ uniformly as $x \to \infty$ for $\alpha \in [0,\alpha)$ does imply that $g(n, \alpha) \to 0$ uniformly as $n \to \infty$ ($n \in \mathbb{N}$) for $\alpha \in [0,\infty)$, but the converse of that is not always true. – RRL Aug 20 '20 at 22:15
  • It's clear now! I know exactly what I was confusing, thank you, very enlightening. – Pedro Fernandes Aug 20 '20 at 22:27