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''Integration by Parts''
 
''Integration by Parts''
  
<math>\int udv = uv - \int vdu</math>
+
<math>\int u \; dv = uv - \int v \; du</math>
  
 
<math>u=t^2 \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t}</math>
 
<math>u=t^2 \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t}</math>
  
 
<math>du=2t dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math>
 
<math>du=2t dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math>

Revision as of 08:44, 3 October 2008

Fourier Transform

$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $

$ x(t)=t^2 u(t) $

$ X(\omega)=\int_{-\infty}^{\infty}t^2 u(t) e^{-j\omega t}dt \; = \int_{0}^{\infty}t^2 e^{-j\omega t}dt $


Integration by Parts

$ \int u \; dv = uv - \int v \; du $

$ u=t^2 \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t} $

$ du=2t dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett