ASN6 a,b- Prove given Fourier Transform property

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(a) Show ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}{\mbox{ if }}S(0)=0$. HINT: $S(0)=S(f)\vert _{_{f=0}}=\int _{-\infty }^{\infty }s(t)e^{-j2\pi (f\rightarrow 0)t}\,dt=\int _{-\infty }^{\infty }s(t)\,dt$

(b) If $S(0)\neq 0$ can you find ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]$ in terms of $\displaystyle S(0)$?

${\mathcal {F}}\left[g(t-t_{0})e^{j2\pi f_{0}t}\right]=\int _{-\infty }^{\infty }\left[g(t-t_{0})e^{j2\pi f_{0}t}\right]e^{-j2\pi ft}\,dt$

$\int _{-\infty }^{\infty }g(t-t_{0})e^{-j2\pi (f-f_{0})t}\,dt=\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})(\lambda +t_{0})}\,dt$

And

$\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})(\lambda +t_{0})}\,dt=\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }e^{-j2\pi (f-f_{0})t_{0}}\,dt=$

And $\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }e^{-j2\pi (f-f_{0})t_{0}}\,dt=e^{-j2\pi (f-f_{0})t_{0}}\left[\int _{-\infty }^{\infty }g(\lambda )e^{-j2\pi (f-f_{0})\lambda }\,dt\right]$

Result

${\mathcal {F}}\left[g(t)e^{j2\pi f_{0}t}\right]=G(f-f_{0})e^{-j2\pi (f-f_{0})t_{0}}$