Some questions regarding the integral of cos(ax), "a" not zero

Hi PF,

$$\int \cos ax\,dx,\quad a\in{\mathbb R-\{0\}}\quad x\in{\mathbb{R}}$$

Let's make
$$u=ax,\quad du=adx$$
and apply $$\int \cos u\,du=\sin u+C$$
$$\frac{1}{a}\int \cos ax\,adx=\frac{1}{a}\sin u+C$$
Substituting the definition of u
$$=\frac{1}{a}\sin ax+C$$

Doubts:
(i) Have I written well the integration steps? It is based on a tutorial from YouTube.
(ii) Domain of the integral is right?
(iii) ##\mathbb R-\{0\}\Leftrightarrow{\mathbb R\setminus 0}##?
(iv) Anything missing or to suggest?

Attempt
(i) It's right. A copy and paste from a video to this post.
(ii) Zero must be excluded from the domain; a becomes the denominator of a fraction when evaluating
(iii) I'm sure of the righthanded equivalence; and think I've seen lefthand notation, but I quick search on the textbook I think I read it is not been successful.

Best wishes!

PD: Post without preview

[itex]\displaystyle\int \cos (0x)\,dx = \int 1\,dx = x + C[/itex] is a special case. Otherwise this is an application of [tex]\frac{d}{dx} \sin ax = a \cos ax[/tex] and the Fundamental Theorem.

pasmith said:

[itex]\displaystyle\int \cos (0x)\,dx = \int 1\,dx = x + C[/itex] is a special case. Otherwise this is an application of [tex]\frac{d}{dx} \sin ax = a \cos ax[/tex] and the Fundamental Theorem.

Domain is [itex]\mathbb R[/itex]; Range [itex][-\frac{1}{a},\frac{1}{a}][/itex]. But [itex](-\infty,0)\cup {(0,\infty)}=\mathbb R-\{0\}=\mathbb R \setminus{\{0\}}[/itex], is the domain for [itex]a[/itex]: [itex]\displaystyle\int \cos (ax)\,dx=\displaystyle\frac{1}{a}\sin{(ax)}[/itex]. The question is: I am overstating the relevance of [itex]a[/itex]. Once described the domain, range is an effortless consequence.
Best wishes!

mcastillo356 said:

Domain is [itex]\mathbb R[/itex]; Range [itex][-\frac{1}{a},\frac{1}{a}][/itex]. But [itex](-\infty,0)\cup {(0,\infty)}=\mathbb R-\{0\}=\mathbb R \setminus{\{0\}}[/itex], is the domain for [itex]a[/itex]: [itex]\displaystyle\int \cos (ax)\,dx=\displaystyle\frac{1}{a}\sin{(ax)}[/itex]. The question is: I am overstating the relevance of [itex]a[/itex]. Once described the domain, range is an effortless consequence.
Best wishes!

Indefinite integral. Not possible to describe any domain nor range.

Thant you

Of course for ##a=0## you integrate simply ##f(x)=1##, giving ##F(x)=x+C##. Of course you can get the same result by making ##a \rightarrow 0##, i.e.,
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]=x.$$
It's, of course, most interesting to argue, why taking this limit commutes with integration ;-).

Hi, PF

vanhees71 said:

Of course for ##a=0## you integrate simply ##f(x)=1##, giving ##F(x)=x+C##. Of course you can get the same result by making ##a \rightarrow 0##, i.e.,
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]=x.$$
It's, of course, most interesting to argue, why taking this limit commutes with integration ;-).

Attempt:
Keywords: limits, Big-O notation, ##a\rightarrow 0##, ##\mathcal{O}(a^3)##
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]\Rightarrow{|\sin(a x)|\leq{|x|}}$$
near ##0##
As you can see, no attempt. Some brief explanation of @vanhees71 quote?
Best wishes!

vanhees71 said:

Of course for ##a=0## you integrate simply ##f(x)=1##, giving ##F(x)=x+C##. Of course you can get the same result by making ##a \rightarrow 0##, i.e.,
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]=x.$$
It's, of course, most interesting to argue, why taking this limit commutes with integration ;-).

Uniform convergence on any finite interval.

Hi, PF

vanhees71 said:

Of course for ##a=0## you integrate simply ##f(x)=1##, giving ##F(x)=x+C##. Of course you can get the same result by making ##a \rightarrow 0##, i.e.,
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]=x.$$
It's, of course, most interesting to argue, why taking this limit commutes with integration ;-).

I've been studying this quote for a while. Any link that should lead to a certain clue, and eventually to understand, contextualize, @PeroK explanation, i.e.

PeroK said:

Uniform convergence on any finite interval.

My background is a fuzzy understanding of pointwise and uniform convergence, but I think I could manage with Big-O notation, limits, finite intervals, Riemann Sums, mere definition of topological space, integration...

Am I ready? For sure, I'm willing to give a try.

Best wishes!

Hi, PF

vanhees71 said:

Of course for ##a=0## you integrate simply ##f(x)=1##, giving ##F(x)=x+C##. Of course you can get the same result by making ##a \rightarrow 0##, i.e.,
$$\lim_{a \rightarrow 0} = \frac{1}{a} \sin(a x)=\lim_{a \rightarrow 0} \frac{1}{a} [a x +\mathcal{O}(a^3)]=x.$$
It's, of course, most interesting to argue, why taking this limit commutes with integration ;-).

This quote captured my attention since I read it. Unfortunately, it is a little bit far from my background. It seemed achievable, but no way. The gap I've been incapable to jump has been the following:

PeroK said:

Uniform convergence on any finite interval.

I thought I could manage, but it's not been the case.

Anyhow,

pasmith said:

[itex]\displaystyle\int \cos (0x)\,dx = \int 1\,dx = x + C[/itex] is a special case. Otherwise this is an application of [tex]\frac{d}{dx} \sin ax = a \cos ax[/tex] and the Fundamental Theorem.

I've got this quote to hold the line. With it in mind, I can keep the track. My decision is basically to return to the textbook the place I left a couple of weeks ago.

Thanks, family, best wishes!

mcastillo356 said:

This quote captured my attention since I read it. Unfortunately, it is a little bit far from my background. It seemed achievable, but no way. The gap I've been incapable to jump has been the following:
...
Thanks, family, best wishes!

Not sure I understand what you are asking. We know that ##\dfrac{d}{dx}\sin(ax)=a\cdot \cos(ax).## The fundamental theorem of calculus says ##\int_p^q f(x)\,dx = F(q)-F(p)## if ##\dfrac{d}{dx}F(x)=f(x).## Thus
$$
\int_p^q \cos(ax)\,dx = \dfrac{\sin(aq)}{a}-\dfrac{\sin(ap)}{a}
$$
and taking the limit ##a\to 0## yields
\begin{align*}
\lim_{a \to 0}\int_p^q \cos(ax)\,dx = \int_p^q\lim_{a \to 0}\cos(ax)\,dx=q-p=\lim_{a \to 0}\dfrac{\sin(aq)}{a}-\lim_{a \to 0}\dfrac{\sin(ap)}{a}
\end{align*}
Setting ##p=0## results in
$$
\lim_{a \to 0}\dfrac{\sin(aq)}{a}=q.
$$
Let ##(a_n)_{n\in \mathbb{N}}\stackrel{n\to\infty }{\longrightarrow }0## be a convergent sequence that represents our limit ##a## to zero. Then we may exchange the order of differentiation and integration
$$
\lim_{n \to \infty}\int \cos(a_n x)\,dx=\int \lim_{n \to \infty} \cos(a_n x)\,dx
$$
because:

• ##cos(a_nx)## are measurable (we can integrate them)
• ##x\mapsto cos(a_nx)## converge pointwise
• ##cos(a_nx)## are bounded, i.e. ##|\cos(a_nx)|\leq h(x)=1## and ##\int h(x)<\infty .##

Hi,

fresh_42 said:

Not sure I understand what you are asking.

Yes it is. I've studied the post for a while, but still doubts. I'm plannig on ask them to a UNED (https://en.wikipedia.org/wiki/National_University_of_Distance_Education) teacher. My opinion is that it is a good step (turn to face-to-face and expert help).
Best wishes!