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:

$\displaystyle \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! :)

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:

$\displaystyle \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! :)

Your comments apply only to finite integration limits. Please be aware that

$\displaystyle \int_{-\infty}^{\infty} cos(x)~dx = \lim_{a \to \infty} \int_{-a}^a cos(x)~dx$

does not exist because $\displaystyle \lim_{x \to \infty}cos(x)$ does not exist.

A similar argument is true for sin(x).

-Dan

Re: Why does the integral of cos(x) from negative infinity to positive infinity diver

Thanks for your reply, Dan.

Does this mean that:

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

also doesn't converge?

Re: Why does the integral of cos(x) from negative infinity to positive infinity diver

Quote:

Originally Posted by

**blaisem** Thanks for your reply, Dan.

Does this mean that:

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

also doesn't converge?

Yes, it does mean that.

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

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

Re: Why does the integral of cos(x) from negative infinity to positive infinity diver

Quote:

Originally Posted by

**topsquark** Your comments apply only to finite integration limits. Please be aware that

$\displaystyle \int_{-\infty}^{\infty} cos(x)~dx = \lim_{a \to \infty} \int_{-a}^a cos(x)~dx$

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 $\displaystyle \int_{-\infty}^\infty cos(x) dx= \lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b cos(x) dx$.

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, $\displaystyle \int_{-a}^a sin(x) dx= \left[- cos(x)\right]_{-a}^a= 0$ for all a so that the "Cauchy principle value" is 0 while the integral $\displaystyle \int_{-\infty}^\infty sin(x) dx= \lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b cos(x) dx$ does not exist, as topsquark said.

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

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

Thank you all for your helpful replies.

Now in Plato's post is written $\displaystyle \int_{-b}^{b}sinx\,dx$= 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 $\displaystyle \int_{-\infty}^{\infty}sin{x}\,dx$ = 0, but Plato in his post said that this integral diverges.

Re: Why does the integral of cos(x) from negative infinity to positive infinity diver

Quote:

Originally Posted by

**blaisem** Thank you all for your helpful replies.

Now in Plato's post is written $\displaystyle \int_{-b}^{b}sinx\,dx$= 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 $\displaystyle \int_{-\infty}^{\infty}sin{x}\,dx$ = 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 $\displaystyle \int_{-\infty}^{c}f(x)\,dx$ and $\displaystyle \int_{c}^{\infty}f(x)\,dx$ **both** exist the we define $\displaystyle \int_{-\infty}^{\infty}f(x)\,dx$=$\displaystyle \int_{-\infty}^{c}f(x)\,dx+\int_{c}^{\infty}f(x)\,dx$.

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

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

Finally it is true that $\displaystyle \int_{-c}^{c}\sin(x)\,dx=0$ and $\displaystyle \int_{-c}^{c}\cos(x)\,dx=2\int_{0}^{c}\cos(x)\,dx$. exists

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:

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