Evaluate this integral - HW1

Max Woesner

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Homework #1 - Evaluate this integral

Problem Statement

Evaluate the integral $\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt\!$

Solution
For $n=m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T{\Bigg |}_{\frac {-T}{2}}^{\frac {T}{2}}={\frac {T}{2}}-{\frac {-T}{2}}=T\!$
For $n\neq m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt={\frac {e^{j2\pi (n-m)t/T}}{\frac {j2\pi (n-m)}{T}}}{\Bigg |}_{\frac {-T}{2}}^{\frac {T}{2}}={\frac {e^{j\pi (n-m)}-e^{-j\pi (n-m)}}{\frac {j2\pi (n-m)}{T}}}=0\!$
So, $\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt={\begin{cases}T,&{\mbox{for }}n=m\\0,&{\mbox{for }}n\neq m\end{cases}}\!$

Alternate method:
$\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt\!$
For $n=m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt=\int _{-{\frac {T}{2}}}^{\frac {T}{2}}1dt=T{\Bigg |}_{\frac {-T}{2}}^{\frac {T}{2}}={\frac {T}{2}}-{\frac {-T}{2}}=T\!$
For $n\neq m,\int _{-{\frac {T}{2}}}^{\frac {T}{2}}[cos(2\pi (n-m)t/T)+jsin(2\pi (n-m)t/T)]dt=0\!$
So, $\int _{-{\frac {T}{2}}}^{\frac {T}{2}}e^{j2\pi (n-m)t/T}dt={\begin{cases}T,&{\mbox{for }}n=m\\0,&{\mbox{for }}n\neq m\end{cases}}\!$