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Integrate 2exp(2+jωt)

  1. Sep 1, 2013 #1
    I do not understand the following integral:

    [itex]\int^{\infty}_{0}2e^{2+jωt}dt[/itex] = [itex]\frac{j2e^{2}}{\omega}[/itex]

    Why is it not ∞? Here are my steps:

    Let u = 2+jωt, du = jωdt, dt = [itex]\frac{1}{jω}[/itex]du = -[itex]\frac{j}{ω}[/itex]du


    = -[itex]\frac{2j}{ω}[/itex][itex]\int^{\infty}_{2}2e^{u}du[/itex]

    = -[itex]\frac{2j}{ω}[/itex][itex]\stackrel{lim}{h\rightarrow∞}[/itex][itex]\int^{h}_{2}2e^{u}du[/itex]

    = -[itex]\frac{2j}{ω}[/itex][itex]\stackrel{lim}{h\rightarrow∞}[/itex]([itex]e^{h}-e^{2}[/itex])

    To me, this limit does not exist, so why is the answer [itex]\frac{j2e^{2}}{\omega}[/itex]?
  2. jcsd
  3. Sep 2, 2013 #2


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    You also need to change your integration limits, if ##t \to \infty## then ##u \to 2 + j \infty##.

    First thing I would do is separate ##e^{2 + j \omega t} = e^2 e^{j \omega t}## and pull the e² in front of the integration sign. The remaining integral can be solved using for example contour integration.
  4. Sep 2, 2013 #3
    That's cus you don't know what [itex]\omega[/itex] is. Let's just look at:
    [tex]\int_0^{\infty} e^{iwt}dt=\frac{1}{iw}e^{iwt}\biggr|_0^{\infty}[/tex]

    Let [itex]\omega=a+bi[/itex] then we have

    Now, for what values of a and b will that expression converge?
    Last edited: Sep 2, 2013
  5. Sep 2, 2013 #4
    a, b < 0 which means ω < 0. So:

    [itex]\frac{1}{iw}e^{iwt}\biggr|_0^{\infty} = -\frac{1}{iω}[/itex] for ω < 0.

    From that, the complete answer to the original integral is:

    [itex]-\frac{2e^{2}}{jω} = \frac{j2e^{2}}{ω}[/itex] for ω < 0.

    Thanks for the help! If I made any mistakes above, let me know.
    Last edited: Sep 2, 2013
  6. Sep 2, 2013 #5
    Your analysis of a and b is not correct. for w=a+bi, in order for the integral to converge, b has to be less than zero. a can be any real number. Go through that to make sure you understand it.
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