How do you find the integral of #int sin(6x) dx# from negative infinity to infinity?

Question

Luke Alvarez · Accepted Answer

That integral does not converge (it diverges).

#int_-oo^oo sin(6x) dx = int_-oo^0 sin(6x) dx + int_0^oo sin(6x) dx# Provided that both integrals on the right converge. Let's look first at the integral on the positives. #int_0^oo sin(6x) dx = lim_(brarroo)int_0^b sin(6x) dx # # = lim_(brarroo) [-1/6cos(6x)]_0^b# # = lim_(brarroo) [-1/6cos(6b)+1/6cos(0)]# However #lim_(brarroo) [-1/6cos(6b)]# does not exist, so the integral on the positives diverges and the big integral (on #(-oo,oo)#) also diverges.

Emily Ammerman · Answer

undefined To find the integral of $ \int \sin(6x) \, dx $ from negative infinity to infinity, you can apply the properties of sine functions and limits. Since the sine function is periodic with a period of $ 2\pi $, the integral from negative infinity to positive infinity can be split into intervals of $ 2\pi $ where the function repeats itself. The integral of sine over one period from $ -\pi $ to $ \pi $ is zero. Therefore, the integral of $ \sin(6x) $ from negative infinity to positive infinity is zero.