HW 06

Problem

Figure out why $\int_{0}^{\infty} \cos (2 π f u) d u$ seems to equal an imaginary odd function of frequency, but there is no j.

$F [\frac{sgn (t)}{2}]$	$= \int_{- \infty}^{\infty} \frac{sgn (t)}{2} e^{- j 2 π f t} d t$
	$= \frac{1}{2} [\int_{- \infty}^{0} - 1 \cdot e^{- j 2 π f t} d t + \int_{0}^{\infty} 1 \cdot e^{- j 2 π f t} d t]$
	$= \frac{1}{2} [\int_{0}^{- \infty} 1 \cdot e^{- j 2 π f t} d t + \int_{0}^{\infty} 1 \cdot e^{- j 2 π f t} d t]$
	$= \underset{\begin{matrix} u = - t \\ d u = - d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{\infty} - e^{j 2 π f u} d u}} + \underset{\begin{matrix} u = t \\ d u = d t \end{matrix}}{\underset{⏟}{\frac{1}{2} \int_{0}^{\infty} e^{- j 2 π f u} d u}}$
	$= \int_{0}^{\infty} \frac{- e^{j 2 π f u} + e^{- j 2 π f u}}{2} d u$
	$= \int_{0}^{\infty} - j \frac{e^{j 2 π f u} - e^{- j 2 π f u}}{2 j} d u$
	$= \int_{0}^{\infty} - j \sin (2 π f u) d u$
	$\neq \int_{0}^{\infty} \cos (2 π f u) d u$