Show that the Fourier transform of $g(t)$ may be expressed as
$$
G(f)=\int_{-\infty}^{\infty} g(t) \cos 2 \pi f t d t-j \int_{-\infty}^{\infty} g(t) \sin 2 \pi f t d t
$$
Hence, show that if $g(t)$ is an even function of $t$, then
$$
G(f)=2 \int_0^{\infty} g(n) \cos 2 \pi f t d t
$$
and if $g(t)$ is an odd function of $t$, then
$$
G(f)=-2 i \int_0^{\infty} g(t) \sin 2 \pi f t d t
$$
Hence, prove that:
\begin{tabular}{ll}
If $g(t)$ is: & Then $G(f)$ is: \\
a real and even function of $t$ & a real and even function of $f$ \\
a real and odd function of $t$ & an imaginary and odd function of $f$ \\
an imaginary and even function of $t$ & an imaginary and even function of $f$ \\
a complex and even function of $t$ & a complex and even function of $f$ \\
a complex and odd function of $t$ & a complex and odd function of $f$
\end{tabular}