EE102 Systems and Signals (Discussion)


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Examples

  1. $ \int t e^{at} dt$
    Solution: In the usual way, we identify

    $\displaystyle u = t, dv = e^{at} dt \Longrightarrow du = dt, v = \frac{e^{at}}{a} $

    Therefore, integrating by parts,
    $\displaystyle \int t e^{at} dt $ $\displaystyle =$ $\displaystyle t \frac{e^{at}}{a} - \int \frac{e^{at}}{a} dt$  
      $\displaystyle =$ $\displaystyle \frac{t e^{at}}{a} - \frac{e^{at}}{a^2} + C$  

    where $ C$ is an arbitrary constant, usually included in indefinite integrals.

  2. $ \int_0^\infty t^n e^{-t} dt$
    Solution: Let

    $\displaystyle I_n = \int_0^\infty t^n e^{-t} dt $

       and let $\displaystyle u = t^n, dv = e^{-t} dt \Longrightarrow du = n t^{n-1} dt, v = -e^{-t} $

    Therefore, integrating by parts, we get
    $\displaystyle I_n$ $\displaystyle =$ $\displaystyle \left[ t^n(-e^{-t}) \right]_0^\infty - \int_0^\infty (-e^{-t}) n t^{n-1} dt$  
      $\displaystyle =$ $\displaystyle 0 + n \int_0^\infty t^{n-1} e^{-t} dt$  
      $\displaystyle =$ $\displaystyle n I_{n-1}, ~n \geq 1.$  

    So,
    $\displaystyle I_1$ $\displaystyle =$ $\displaystyle 1 \cdot I_{0} = I_0$  
    $\displaystyle I_2$ $\displaystyle =$ $\displaystyle 2\cdot I_1 = 2\cdot 1 \cdot I_0$  
    $\displaystyle I_3$ $\displaystyle =$ $\displaystyle 3\cdot I_1 = 3 \cdot 2\cdot 1 \cdot I_0$  

    and in general,

    $\displaystyle I_n = n(n-1)...1 \cdot I_0 = (n!) \cdot I_0$

    Finally, since

    $\displaystyle I_0 = \int_0^\infty t^0 e^{-t} dt = 1$

    we get

    $\displaystyle I_n =(n!),~n \geq 1$

    Actually $ I_n$ can be defined for non-integral values of $ n$, and is a generalization of the factorial to non-integral $ n$. $ \Gamma(n) = I_{n-1}$ is called the gamma function.


next up previous contents
Next: Using Laplace Transform to Up: Integrals Previous: Integration by Parts

visitors since January 7., 2002. -

Sankaran Panchapagesan
2002-06-05