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!
The correct integral is .
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, for all a so that the "Cauchy principle value" is 0 while the integral does not exist, as topsquark said.
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.
In basic calculus most of us use this definition:
If and both exist the we define = .
Because neither nor exists then using that definition neither nor exists.
Again, definitions do differ. So your TA needs to clarify the exact definition used here.
Finally it is true that and . exists