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:

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:

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

does not exist because 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:

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:

also doesn't converge?

Yes, it does mean that.

Because does not converge neither does .

Now .

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

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

does not converge neither does

.

Now

.

Thank you all for your helpful replies.

Now in Plato's post is written = 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 = 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

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:

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

A short note to first post. turns out to be with the substitution . Hence both either **diverge** (for sure) or converge.