Hi, I need to find this integral:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt

[/tex]

Here's the working out I did:

[tex]

G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}]

[/tex]

Therefore:

[tex]

G(f)=2TAsinc(fT) -3TAsinc(3fT)

[/tex]

But when I used Mathematica I typed this:

Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, (-3*T)/2}] +

Integrate[A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, T/2}] +

Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, T/2, (3*T)/2}]

and it gave me this:

[tex]\frac{ASin(f\pi T)}{f\pi}-\frac{Ae^{-2j\pi ft}Sin(f\pi T)}{f\pi}+\frac{Ae^{2j\pi ft}Sin(f\pi t)}{f\pi}[/tex]

which equals:

ATsinc(fT) + 2jATsinc(πFT)Sin(2πfT)

Is the answer from Mathematica the correct one?

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# Mathematica Integration Question

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