ASN6 a,b- Prove given Fourier Transform property

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Problem Statement

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

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

Answer

a)Remember that dummy variable $\lambda \!$ was used in substitution such that $\lambda =t-t_{0}\!$

Then $s(\lambda )=s(t-t_{0})={\mathcal {F}}\left[S(f)-S(f_{0})\right]\!$

and $\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)-S(f_{0})\right]\,d\lambda \!$

The problem statement says to make $S(f_{0})=0\!$ that makes the above equation simplify to

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}\left[S(f)\right]\,dt\!$

Taking the inverse Fourier Transform and changing the order of intgration

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}e^{j2\pi ft}\,dt\int _{-\infty }^{\infty }S(f)\,df={\frac {e^{j2\pi ft}}{j2\pi f}}\int _{-\infty }^{\infty }S(f)\,df=\!$

Then

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{\infty }^{\infty }S(f){\frac {e^{j2\pi ft}}{j2\pi f}}\,df={\mathcal {F}}^{-1}\left[{\frac {S(f)}{j2\pi f}}\right]\!$

Therefore it is demonstrated that ${\mathcal {F}}\left[\int _{-\infty }^{t}s(\lambda )\,d\lambda \right]={\frac {S(f)}{j2\pi f}}\!$

b) If $S(f_{0})\neq 0$

Then $\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}{\mathcal {F}}^{-1}\left[S(f)-S(f_{0})\right]\,d\lambda =\int _{-\infty }^{t}\int _{-\infty }^{\infty }e^{j2\pi ft}[S(f)-S(f_{0})]\,d\lambda \,d\lambda \!$

$\int _{-\infty }^{t}s(\lambda )\,d\lambda =\int _{-\infty }^{t}e^{j2\pi ft}S(f)\,d\lambda -\int _{-\infty }^{\infty }e^{j2\pi ft}S(f_{0})\,d\lambda \!$