Calc Question

[Nalcon]

<p>I usually don't have to ask questions but this one really bugs me and it doesn't explain it in the book.</p>

<p>The improper integral, from 0 to infinity of sin(x)</p>

<p>Our teacher told us it diverges but I don't understand why it's not zero because there is an equal amount of space above and below the x axis as x approaches infinity.</p>

<p>How do you get diverges?</p>

<p>Thanks</p>

[Gospy]

<p>Take it as the limit as t approaches infinity of the integral from 0 to t of sin(x). Then you get the limit as t approaches infinity of -cos(x) evaluated from 0 to t. Which simplifies to limit as t approaches infintiy of -cos(t) + 1. That limit doesn't exist. Thus, the integral diverges.</p>