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how do we find the value of the following definite integral?

$$I=\int_{-1}^{\infty}\cos(\pi t)dt$$

$-1 \le \cos(\pi t) \le 1$

$$I=\left[\frac{\sin(\pi t)}{\pi}\right]_{-1}^{\infty}$$

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3 Answers 3

If the upper limit of an integral is $\infty$ (along with some other cases), it is an improper integral. In this case it should be considered to be $$\lim_{x \to \infty}\int_{-1}^x \cos(\pi t)\; dt=\lim_{x \to \infty} \frac {\sin \pi x}\pi - \frac {\sin (-\pi)}\pi$$ which does not converge

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it does not diverge either. does it ? –  Rajesh K Singh Feb 22 '13 at 5:13
@RajeshKSingh: It does not go off to infinity, but oscillation is another form of divergence. –  Ross Millikan Feb 22 '13 at 14:04

This diverges. There is no definite integral. Recall that $I = \lim_{t\to\infty}\frac{\sin(\pi t)}{\pi} - \frac{\sin(-\pi)}{\pi}$

Its obvious that the sine does not converge. It oscillates forever.

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+1 Nicely done! –  amWhy Feb 22 '13 at 4:44

To prove that the sine function does not converge we can use the sequential characterization of limit so we take for example the sequence $(u_n)_n=\left((2n+1)\frac{\pi}{2}\right)_n$ that diverges to $+\infty$ and $\sin u_n=(-1)^n$ also diverge.

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