# evaluating sine (or cos) at infinity

• October 12th 2009, 07:39 PM
walleye
evaluating sine (or cos) at infinity
so ive hit a snag in a problem (the topic is actually fourier integral transformations)
i was wondering whether there was some approach to evaluating
$\int_0^{\infty}sin(x) dx$

in my mind $sin(\infty)$ could be either -1, 0 or 1. is there a orthodox way of approaching this?
• October 12th 2009, 08:42 PM
tonio
Quote:

Originally Posted by walleye
so ive hit a snag in a problem (the topic is actually fourier integral transformations)
i was wondering whether there was some approach to evaluating
$\int_0^{\infty}sin(x) dx$

in my mind $sin(\infty)$ could be either -1, 0 or 1. is there a orthodox way of approaching this?

There is a very orthodox solution as well: your integral diverges, since it involves evaluating the limit of the primitive function of sin x, i.e. -cos x, when x --> oo, and this limit just doesn't exist.

Tonio
• October 12th 2009, 11:58 PM
chisigma
The limit...

$\lim_{t \rightarrow \infty} \int_{0}^{t} \sin \tau\cdot d\tau$ (1)

... doesn't exist but exists the limit...

$\lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} \sin \tau\cdot d\tau = \langle \sin t \rangle = 0$ (2)

In general if $f(*)$ is the sum of a finite number of periodic functions not necessarly with the same period the limit...

$\lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} f(\tau)\cdot d\tau = \langle f(t) \rangle$ (3)

... exists and $\langle f(*) \rangle$ is called [improperly] 'mean value of $f(*)$'. It is...

$\langle f(t) \rangle = \sum_{n} \langle f_{n}(t) \rangle$ (4)

Kind regards

$\chi$ $\sigma$
• October 13th 2009, 02:05 AM
tonio
Quote:

Originally Posted by chisigma
The limit...

$\lim_{t \rightarrow \infty} \int_{0}^{t} \sin \tau\cdot d\tau$ (1)

... doesn't exist but exists the limit...

$\lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} \sin \tau\cdot d\tau = \langle \sin t \rangle = 0$ (2)

Welll...yes, in a rather trivial way: since INT{0,t} sin(s) ds = -cos(t) + 1,
then lim (1/t)[1 - cos(t)] = 0 when t --> oo since 1 - cos t is bounded and (1/t) --> 0 .
This though doesn't seem to appropiately address the OP's question.

Tonio

In general if $f(*)$ is the sum of a finite number of periodic functions not necessarly with the same period the limit...

$\lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} f(\tau)\cdot d\tau = \langle f(t) \rangle$ (3)

... exists and $\langle f(*) \rangle$ is called [improperly] 'mean value of $f(*)$'. It is...

$\langle f(t) \rangle = \sum_{n} \langle f_{n}(t) \rangle$ (4)

Kind regards

$\chi$ $\sigma$

..............