HW 06

Problem

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

Solution

$F\left[{\frac {\operatorname {sgn}(t)}{2}}\right]$	$=\int _{-\infty }^{\infty }{\frac {\operatorname {sgn}(t)}{2}}e^{-j\,2\,\pi \,f\,t}\,dt$
	$={\frac {1}{2}}\left[\int _{-\infty }^{0}-1\cdot e^{-j\,2\,\pi \,f\,t}\,dt+\int _{0}^{\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt\right]$
	$={\frac {1}{2}}\left[\int _{0}^{-\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt+\int _{0}^{\infty }1\cdot e^{-j\,2\,\pi \,f\,t}\,dt\right]$
	$=\underbrace {{\frac {1}{2}}\int _{0}^{\infty }-e^{j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=-t\\du=-dt\end{matrix}}+\underbrace {{\frac {1}{2}}\int _{0}^{\infty }e^{-j\,2\,\pi \,f\,u}\,du} _{\begin{matrix}u=t\\du=dt\end{matrix}}$
	$=\int _{0}^{\infty }{\frac {-e^{j\,2\,\pi \,f\,u}+e^{-j\,2\,\pi \,f\,u}}{2}}\,du$
	$=\int _{0}^{\infty }-j\,{\frac {e^{j\,2\,\pi \,f\,u}-e^{-j\,2\,\pi \,f\,u}}{2j}}\,du$
	$=\int _{0}^{\infty }-j\,\sin(2\,\pi \,f\,u)\,du$
	$\neq \int _{0}^{\infty }\cos(2\,\pi \,f\,u)\,du$