Question

convergence of the integral ln(cos(1/t)) de ]2/pi,+oo[

Answer 1

This? \(\int\limits_{2\pi}^{\infty}\ln(\cos(1/t))\;dt\)

Answer 2

First things first, replace \[\infty\] with 'm', then set up the integral as
\[lim_{m \to \infty} \int_{2 \pi}^{m} ln(cos(\frac{1}{t})) \;\;dt\]

Answer 3

from \[\int\limits_{2/\pi}^{+oo} \ln(\cos(1/t))dt\] exsist or no i found that it doesn't converge

Answer 4

How did you prove that?

Answer 5

we have lim(t->2/pi) ln(cos(1/t)) -> -oo which means |f(t)|>= 1 \[\int\limits_{2/\pi}^{+oo}f(t)dt = +oo \] and i'm not sure

Answer 6

You do have to cut it up and consider both ends separately.

On the far end, as t increases without bound, cos(1/t) approaches 1 from <1, and ln(cos(1/t)) approaches 0 from the negative side.

It's zero. That's at least encouraging.

Answer 7

Yeah i figured that mush but the problem is ln is not compatible with equivalence and i was trying to prove that lim t-> +oo ln(cos(1/t)) equivalent to ln(1+1/t) and it diverges in +oo which mean the whole integral divergese but i'm stuck at proving that ln(cos(1/t))/ln(1+1/t))=1 t->+oo

Answer 8

1 + 1/t is no good. That approaches 1 from >1

Try 1 - 1/t

Answer 9

but it's a bit hard to prove ln(cos(1/t)) equivalente to ln(1-1/t))

Answer 10

It's also no good. If we're going to find a rational expression to substitute for cosine, it might be:

\(1 - \dfrac{1}{2t^{2}} + \dfrac{1}{24t^{4}}\)

This is strictly greater than \(\cos(1/t)\).

Answer 11

but logarithme is not compatible with equivalnte

Answer 12

No, the logarithm is fine for large enough 't'. This is a natural consequence of examining the two ends in different pieces.