Like this? ##2\int_0^{\infty}\cos(kx) dk##maverick_76 said:okay so I have this integral:
2∫cos(kx)dk
The bounds are from zero to infinity, this doesn't converge but is there any other way to describe this?
The limit doesn't exist because sin(kx) oscillates endlessly.maverick_76 said:okay so the limit would be lim_k->∞ sin(kx)
xHow would I go about evaluating it? Is there a substitution trick?
The value of cosine evaluated at infinity is undefined or "does not exist".
No, cosine does not approach a specific value as the input approaches infinity. It oscillates between -1 and 1 infinitely.
No, most calculators do not have the capability to evaluate functions at infinity. However, you can use the limit definition of cosine to approximate its value at large input values.
As the input approaches infinity, cosine oscillates between -1 and 1 infinitely. It does not approach a specific value or "converge".
No, the concept of cosine evaluated at infinity is mostly used in theoretical mathematics and has limited practical applications.