ASN6 a,b- Prove given Fourier Transform property

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(a) Show ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]=\frac{S(f)}{j2\pi f}\text{ if }S(0)=0$. HINT: $S(0)=S(f){|}_{{f=0}_{}}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}s(t)e^{-j2\pi (f\to 0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}s(t)dt$

(b) If $S(0)\ne 0$ can you find ${ℱ}[\underset{-\mathrm{\infty }}{\overset{t}{\int }}s(\lambda )d\lambda ]$ in terms of $S(0)$?

Find ${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]$

${ℱ}[g(t-t_0)e^{j2\pi f_{0}t}]=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[g(t-t_0)e^{j2\pi f_{0}t}]e^{-j2\pi ft}dt$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(t-t_0)e^{-j2\pi (f-f_0)t}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt$

After some simplification and rearranging terms, we get:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)(\lambda +t_0)}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }{e}^{-j2\pi (f-f_0)t_0}dt=e^{-j2\pi (f-f_0)t_0}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}g(\lambda )e^{-j2\pi (f-f_0)\lambda }dt]$

Result

${ℱ}[g(t)e^{j2\pi f_{0}t}]=G(f-f_0)e^{-j2\pi (f-f_0)t_0}$