# Why does the integral of cos(x) from negative infinity to positive infinity diverge?

• June 11th 2013, 06:40 AM
blaisem
Why does the integral of cos(x) from negative infinity to positive infinity diverge?
Hello everyone.

I understand that sin(x) is an odd function and therefore its integral over a symmetrical domain is equal to zero. In class today, I was told that this is true even for boundaries from negative infinity to positive infinity. This is because for every positive area there is a corresponding negative area traced out by the function, so the integral cancels.

As an even function, cos(x) is not symmetrical about the origin, so this does not apply. However, when I look at the graph of cos(x) piecewise from negative infinity to negative pi/2, and pi/2 to infinity, it appears to be repeatedly symmetrical about various points on the x-axis. Another way of looking at it is expressing cos(x) = sin(x + pi/2). It appears that I can *convert* an even function into an odd function?

Why can't I use this reasoning to find a value for the integral of cos(x) from negative to positive infinity? Namely, I'd like to assume that areas in negative x cancel with each other, and areas in positive x cancel with each other, so that I can rearrange the integral:

$\int_{-\infty}^{\infty}cos{x}\,dx =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}cos{x}\,dx$

Thank you for reading my question and any advice you might be able to give! :)
• June 11th 2013, 07:09 AM
topsquark
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by blaisem
Hello everyone.

I understand that sin(x) is an odd function and therefore its integral over a symmetrical domain is equal to zero. In class today, I was told that this is true even for boundaries from negative infinity to positive infinity. This is because for every positive area there is a corresponding negative area traced out by the function, so the integral cancels.

As an even function, cos(x) is not symmetrical about the origin, so this does not apply. However, when I look at the graph of cos(x) piecewise from negative infinity to negative pi/2, and pi/2 to infinity, it appears to be repeatedly symmetrical about various points on the x-axis. Another way of looking at it is expressing cos(x) = sin(x + pi/2). It appears that I can *convert* an even function into an odd function?

Why can't I use this reasoning to find a value for the integral of cos(x) from negative to positive infinity? Namely, I'd like to assume that areas in negative x cancel with each other, and areas in positive x cancel with each other, so that I can rearrange the integral:

$\int_{-\infty}^{\infty}cos{x}\,dx =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}cos{x}\,dx$

Thank you for reading my question and any advice you might be able to give! :)

$\int_{-\infty}^{\infty} cos(x)~dx = \lim_{a \to \infty} \int_{-a}^a cos(x)~dx$
does not exist because $\lim_{x \to \infty}cos(x)$ does not exist.

A similar argument is true for sin(x).

-Dan
• June 11th 2013, 10:08 AM
blaisem
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver

Does this mean that:

$\int_{-\infty}^{\infty}sin{x}\,dx$

also doesn't converge?
• June 11th 2013, 10:15 AM
Plato
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by blaisem
Does this mean that:
$\int_{-\infty}^{\infty}sin{x}\,dx$
also doesn't converge?

Yes, it does mean that.

Because $\int_{0}^{\infty}sin{x}\,dx$ does not converge neither does $\int_{-\infty}^{\infty}sin{x}\,dx$.

Now $\forall B\,~\int_{-B}^{B}sin{x}\,dx=0$.
• June 11th 2013, 11:31 AM
HallsofIvy
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by topsquark
$\int_{-\infty}^{\infty} cos(x)~dx = \lim_{a \to \infty} \int_{-a}^a cos(x)~dx$

A slight clarification. That is the "Cauchy principal value" which may exist even if the integral itself does not (and is equal to the integral if the integral converges).
The correct integral is $\int_{-\infty}^\infty cos(x) dx= \lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b cos(x) dx$.

The difference is not really important here because both the integral and the "Cauchy principal value" diverge but, because sin(x) is an "odd" function, $\int_{-a}^a sin(x) dx= \left[- cos(x)\right]_{-a}^a= 0$ for all a so that the "Cauchy principle value" is 0 while the integral $\int_{-\infty}^\infty sin(x) dx= \lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b cos(x) dx$ does not exist, as topsquark said.
• June 11th 2013, 12:10 PM
blaisem
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by Plato
Yes, it does mean that.

Because $\int_{0}^{\infty}sin{x}\,dx$ does not converge neither does $\int_{-\infty}^{\infty}sin{x}\,dx$.

Now $\forall B\,~\int_{-B}^{B}sin{x}\,dx=0$.

Now in Plato's post is written $\int_{-b}^{b}sinx\,dx$= 0 (HallsofIvy used a instead of b)

These boundaries do not include infinity, right? Plato seems to have said as much by saying the value for infinity diverges, but I just wanted to clarify.

Sorry to make you keep coming back here. I am a little confused since my TA said that $\int_{-\infty}^{\infty}sin{x}\,dx$ = 0, but Plato in his post said that this integral diverges.
• June 11th 2013, 01:11 PM
Plato
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by blaisem

Now in Plato's post is written $\int_{-b}^{b}sinx\,dx$= 0 (HallsofIvy used a instead of b)

These boundaries do not include infinity, right? Plato seems to have said as much by saying the value for infinity diverges, but I just wanted to clarify.

Sorry to make you keep coming back here. I am a little confused since my TA said that $\int_{-\infty}^{\infty}sin{x}\,dx$ = 0, but Plato in his post said that this integral diverges.

First, this post is in the caculus sub-forum therefore it never occurred to me too consider the "Cauchy principal value".

In basic calculus most of us use this definition:
If $\int_{-\infty}^{c}f(x)\,dx$ and $\int_{c}^{\infty}f(x)\,dx$ both exist the we define $\int_{-\infty}^{\infty}f(x)\,dx$= $\int_{-\infty}^{c}f(x)\,dx+\int_{c}^{\infty}f(x)\,dx$.

Because neither $\int_{c}^{\infty}\sin(x)\,dx$ nor $\int_{c}^{\infty}\cos(x)\,dx$ exists then using that definition neither $\int_{-\infty}^{\infty}\sin(x)\,dx$ nor $\int_{-\infty}^{\infty}\cos(x)\,dx$ exists.

Again, definitions do differ. So your TA needs to clarify the exact definition used here.

Finally it is true that $\int_{-c}^{c}\sin(x)\,dx=0$ and $\int_{-c}^{c}\cos(x)\,dx=2\int_{0}^{c}\cos(x)\,dx$. exists
• June 19th 2013, 11:40 AM
bkarpuz
Re: Why does the integral of cos(x) from negative infinity to positive infinity diver
Quote:

Originally Posted by blaisem
Hello everyone.

I understand that sin(x) is an odd function and therefore its integral over a symmetrical domain is equal to zero. In class today, I was told that this is true even for boundaries from negative infinity to positive infinity. This is because for every positive area there is a corresponding negative area traced out by the function, so the integral cancels.

As an even function, cos(x) is not symmetrical about the origin, so this does not apply. However, when I look at the graph of cos(x) piecewise from negative infinity to negative pi/2, and pi/2 to infinity, it appears to be repeatedly symmetrical about various points on the x-axis. Another way of looking at it is expressing cos(x) = sin(x + pi/2). It appears that I can *convert* an even function into an odd function?

Why can't I use this reasoning to find a value for the integral of cos(x) from negative to positive infinity? Namely, I'd like to assume that areas in negative x cancel with each other, and areas in positive x cancel with each other, so that I can rearrange the integral:

$\int_{-\infty}^{\infty}cos{x}\,dx =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}cos{x}\,dx$

Thank you for reading my question and any advice you might be able to give! :)

A short note to first post. $\int_{-\infty}^{\infty}\sin(u) \mathrm{d}u$ turns out to be $\int_{-\infty}^{\infty}\cos(v) \mathrm{d}v$ with the substitution $v=u-\frac{\pi}{2}$. Hence both either diverge (for sure) or converge.