I think this should be the logic.
Let f(x) = $\frac{\cos ^{m}x}{x^{n}}$
Value of x |
Value of n |
<1 |
<1 |
<1 |
>=1 |
>=1 |
<1 |
>=1 |
>=1 |
Now, we focus for first 2 cases (i.e. x<1 since x has more values less than 1 rather than greter than 1)
Now, if we consider n>=1, then consider what will be instantaneous value of f(x).
e.g n=100 and x=0.02
$\therefore$ cos^{m}x<1 but x^{n}$\rightarrow$0
So, instantaneous value is very high and thus, integration of f(x) will never converge.
Now, if we consider n<1, then consider what will be instantaneous value of f(x).
e.g n=0.9 x=0.02
$\therefore$ instantaneous value of f(x) will be finite. (just try taking an example)
So, this will converge.
So, my answer was just a calculated guess. I dont know if there is any solution by using the series expansion of cos(x).