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

[mcastillo356]

TL;DR Summary

I'm concerned with the integral itself, but also with the possible notations for the domain.

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

 

[pasmith]

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

 

[mcastillo356]

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

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

 

[mcastillo356]

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

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

Thant you

 

[vanhees71]

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 ;-).

 

[mcastillo356]

Hi, PF

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!

 

[PeroK]

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.

 

[mcastillo356]

Hi, PF

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.

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!

 

[mcastillo356]

Hi, PF

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:

Uniform convergence on any finite interval.

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

Anyhow,

$\displaystyle\int \cos (0x)\,dx = \int 1\,dx = x + C$ is a special case. Otherwise this is an application of $$\frac{d}{dx} \sin ax = a \cos ax$$ 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!

 

[fresh_42]

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 $\int h(x)<\infty .$

 

[mcastillo356]

Hi,

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!

 

[mcastillo356]

Hi, PF

I had to enroll at the UNED to access tutors. Various prejudices have prevented it from doing so. I was also about to ask a student at the University of the Basque Country who was finishing his Bachelor's Degree in Mathematics. I recoiled too.
@fresh_42 last post is quite difficult to me, but the fear of becoming an archaeologist (this thread began on november of 2023) encourages me to take the initiative of, at least, ask a question, the only one for which I can attempt to discuss:

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*}

Fine

Setting $p=0$ results in
$$
\lim_{a \to 0}\dfrac{\sin(aq)}{a}=q.
$$

The aim of setting $p=0$ is to focus on the integral of $\cos(ax)$

$$
\lim_{n \to \infty}\int \cos(a_n x)\,dx=\int \lim_{n \to \infty} \cos(a_n x)\,dx
$$
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 $\int h(x)<\infty .$

$$\int h(x)<\infty$$
makes me wonder. I figure this way the integral:



And in my opinion, range is infinite

Best wishes!

PD: Post without preview

 

[Mark44]

From post #1:

TL;DR Summary: I'm concerned with the integral itself, but also with the possible notations for the domain.

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

It appears that you are treating this indefinite integral as a function of the parameter a, so I'm not really sure what you're asking here.
If $a \ne 0$ then $\int \cos ax\,dx = \frac 1 a \sin(ax) + C$.
If $a = 0$, then $\int \cos ax\,dx = \int \cos(0)\,dx = \int 1\, dx = x + C$.

 

[mcastillo356]

Hi PF, @Mark44 !

From post #1:

It appears that you are treating this indefinite integral as a function of the parameter a, so I'm not really sure what you're asking here.
If $a \ne 0$ then $\int \cos ax\,dx = \frac 1 a \sin(ax) + C$.
If $a = 0$, then $\int \cos ax\,dx = \int \cos(0)\,dx = \int 1\, dx = x + C$.

Well, you got me. I mean, my curiosity runs beyond my mathematical background. It's time for me to take some brief pause to think new ways to focus the thread.
¡Saludos, amigo!
Meet in two days.

 

[mcastillo356]

Hi, PF

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 $\int h(x)<\infty .$

$$\int h(x)<\infty$$

This means

$$cos(a_nx)$$

is bounded above

I'm struggling with this statement. I must say this post is exploratory, not to say it could be erratic.

Well, my attempt to understand it relies in the following quote:

If $a = 0$, then $\int \cos ax\,dx = \int \cos(0)\,dx = \int 1\, dx = x + C$.

$$C\in\mathbb{R}\rightarrow{C\in{(-\infty,\infty)}}$$

I mean, infinity is not included in indefinite integrals, i.e., is eventually bounded above.

There goes this brainstorm.

Best wishes!

 

[Mark44]

This means $\cos(a_nx)$ is bounded above.

Of course. Both the sine and cosine functions are bounded, meaning that they are bounded above as well as below.
$-1 \le \cos(\text{whatever}) \le 1$ and $-1 \le \sin(\text{whatever}) \le 1$

Well, my attempt to understand it relies in the following quote:

If $a = 0$, then $\int \cos ax\,dx = \int \cos(0)\,dx = \int 1\, dx = x + C$.

$$C\in\mathbb{R}\rightarrow{C\in{(-\infty,\infty)}}$$

I mean, infinity is not included in indefinite integrals, i.e., is eventually bounded above.

I really don't understand what you're asking here. "infinity is not included in indefinite integrals" -- this doesn't make any sense. If we evaluate an indefinite integral, we get an infinitely large family of antiderivatives, all differing only by a constant. This

Looking at the integral above as a definite integral, we have
If $a = 0$, then $\int_c^d \cos ax\,dx = \int_c^d\cos(0)\,dx = \int_c^d 1\, dx = \left.x\right|_c^d = d - c$.

None of this has anything to do with what you asked about earlier, the domain of an integral.

 

[mcastillo356]

Hi

Of course. Both the sine and cosine functions are bounded, meaning that they are bounded above as well as below.
$-1 \le \cos(\text{whatever}) \le 1$ and $-1 \le \sin(\text{whatever}) \le 1$

Fine

I really don't understand what you're asking here. "infinity is not included in indefinite integrals" --

Neither I. Been looking for some explanation... What did I mean? I've read a couple of times...

If we evaluate an indefinite integral, we get an infinitely large family of antiderivatives, all differing only by a constant.

Better. I personally prefer to leave aside "infinity is not..." I cannot make any sense of it .

Looking at the integral above as a definite integral, we have
If $a = 0$, then $\int_c^d \cos ax\,dx = \int_c^d\cos(0)\,dx = \int_c^d 1\, dx = \left.x\right|_c^d = d - c$.

None of this has anything to do with what you asked about earlier, the domain of an integral.

True. Thanks!

 

[mcastillo356]

Hi, dear PF

$\cdot$ - Missed to write down again domain of the integral, with the right notation:

$$\mathbb R\,-\,\{0\}\Leftrightarrow{\mathbb R\setminus{\{0\}}}$$

$\cdot\,\cdot$ - Range of the integral in terms of $a$ and $C$ parameters

$$\left[\dfrac{-1}{|a|}+C,\dfrac{1}{|a|}+C\right]$$

Could it have any relationship with this quote?:

Uniform convergence on any finite interval.

sentence related to post #5.

Attempt

It might have nothing in common. I'm taking the biparametrical range of the integral, on one hand, and Wikipedia's $\epsilon$ - tube of uniform convergence, on the other.

$\cdot\,\cdot\,\cdot$ - Related to post #10:

Lemma about convergence and integration

Let $f_n\,:\,[a,b]\rightarrow{\mathbb{R}}$ be continous, and $f_n\xrightarrow{uc}f$, then $$\displaystyle\lim_{n \to\infty}{\displaystyle\int_{a}^{b}\,f_{n}(x)}\,dx=\displaystyle\int_{a}^{b}\,f(x)\,dx=\displaystyle\int_{a}^{b}\displaystyle\lim_{n}\,f_n$$

Definition of uniform convergence

$$ \epsilon>0,\,\exists{n_0}/\,\forall{n\geq{n_0}},\,|f_{n}(x)-f(x)|<\displaystyle\frac{\epsilon}{b-a}\,,\,\forall{x\in{[a,b]}}$$

Proof

$$\Bigg|\displaystyle\int_{a}^{b}\,f_{n}(x)\,dx-\displaystyle\int_{a}^{b}\,f(x)\,dx\Bigg|=\Bigg|\displaystyle\int_{a}^{b}\,(f_{n}(x)-f(x))\,dx\Bigg|\leq{\displaystyle\int_{a}^{b}|f_{n}(x)-f(x)|\,dx}\Rightarrow$$

$$\Rightarrow{\Bigg|\displaystyle\int_{a}^{b}\,f_{n}(x)\,dx-\displaystyle\int_{a}^{b}\,dx\Bigg|<\displaystyle\int_{a}^{b}\displaystyle\frac{\epsilon}{b-a}}\,dx=\displaystyle\frac{\epsilon}{\cancel{b-a}}\cancel{(b-a)}=\epsilon$$

Could it have any relationship with this quote?

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 $\int h(x)<\infty .$

Attempt

It might be an alternative proof of the following exchange in the order of integration and limits:

$$\lim_{n \to \infty}\int \cos(a_n x)\,dx=\int \lim_{n \to \infty} \cos(a_n x)\,dx$$

Sorry for the morse-style written post. Hope you fill the gaps, and tell me the mistakes.

Best wishes!