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Math Help - Question of change of variables in definite integral.

  1. #1
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    Question of change of variables in definite integral.

    I want to proof:
    \int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} e^{jz\cos \theta}\cos(m\theta) d\theta

    I did it in two different way as shown below:

    (1) Let \theta=-\theta\;\Rightarrow\; d\theta=-d\theta,\;\cos(-\theta)=\cos\theta,\;\cos[m(-\theta)]=\cos(m\theta)

    \int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=-\int_{\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi}e^{jz\cos \theta}\cos(m\theta) d\theta


    BUT

    (2) Let \theta=\pi+\theta\;\Rightarrow\; d\theta=d\theta,\;\cos(\pi+\theta)=-\cos\theta

    \int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} e^{-jz\cos \theta}\cos(m\pi+m\theta) d\theta

    \cos(m\pi+m\theta)=\cos(m\pi)\cos(m\theta)-\sin(m\pi)\sin(m\theta)=(-1)^m \cos(m\theta)

    \Rightarrow\;\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m \cos(m\theta) d\theta


    Obviously the two ways give different result. What happened?

    Thanks
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  2. #2
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    Re: Question of change of variables in definite integral.

    Going from -\pi to \pi is adding 2\pi NOT \pi.
    Cos(m(\theta+ 2\pi)= cos(m\theta+ 2m\pi)= cos(m\theta)cos(2m\pi)- sin(m\theta)sin(2m\pi)= cos(m\theta)
    because cos(2m\pi)= 1 and sin(2m\pi)= 0 for all m.
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  3. #3
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    Re: Question of change of variables in definite integral.

    Quote Originally Posted by HallsofIvy View Post
    Going from -\pi to \pi is adding 2\pi NOT \pi.
    Cos(m(\theta+ 2\pi)= cos(m\theta+ 2m\pi)= cos(m\theta)cos(2m\pi)- sin(m\theta)sin(2m\pi)= cos(m\theta)
    because cos(2m\pi)= 1 and sin(2m\pi)= 0 for all m.
    This is not the case in (2). I am going from \;\int_{-\pi}^{0}\; to \;\int_{0}^{\pi}\;. Nothing about 2 \pi. This is done by \;(\theta=\theta + \pi)\;, not \;(\theta=\theta + 2\pi)\;
    Last edited by Alan0354; July 25th 2013 at 09:55 AM.
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  4. #4
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    Re: Question of change of variables in definite integral.

    The limits for the second transformation are wrong aren't they ?

    If \theta \rightarrow \pi + \theta,

    then

    (-\pi,0) \rightarrow (-2\pi,-\pi).

    Your problem, as I see it, is caused by your use of the same symbol for both the original and the transformed variable. What you have done, is to use the transformation \theta \rightarrow \pi + \theta, for the integrand, but the transformation \theta + \pi \rightarrow \theta for the limits.
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  5. #5
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    Re: Question of change of variables in definite integral.

    Quote Originally Posted by BobP View Post
    The limits for the second transformation are wrong aren't they ?

    If \theta \rightarrow \pi + \theta,

    then

    (-\pi,0) \rightarrow (-2\pi,-\pi).

    Your problem, as I see it, is caused by your use of the same symbol for both the original and the transformed variable. What you have done, is to use the transformation \theta \rightarrow \pi + \theta, for the integrand, but the transformation \theta + \pi \rightarrow \theta for the limits.
    You can use \;u=\pi+\theta\; to make it less confusing if you like.

    It is not right to say If \theta \rightarrow \pi + \theta, (-\pi,0) \rightarrow (-2\pi,-\pi).

    Lets make it clearer, Let \; (u=\pi+\theta),\;[-\pi,0]\;\rightarrow\;[0,\pi]

    Then

    (2) Let u=\pi+\theta\;\Rightarrow\; du=d\theta,\;\cos(\pi+\theta)=-\cos\theta

    \int_{-\pi}^0 e^{jz\cos u}\cos(mu) du=\int_0^{\pi} e^{-jz\cos \theta}\cos(m\pi+m\theta) d\theta

    \cos(m\pi+m\theta)=\cos(m\pi)\cos(m\theta)-\sin(m\pi)\sin(m\theta)=(-1)^m \cos(m\theta)

    \Rightarrow\;\int_{-\pi}^0 e^{jz\cos \theta}\cos(m\theta) d\theta=\int_0^{\pi} \frac{1}{e^{jz\cos \theta}}(-1)^m \cos(m\theta) d\theta

    Sorry about the confusion

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    Last edited by Alan0354; July 25th 2013 at 02:20 PM.
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  6. #6
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    Re: Question of change of variables in definite integral.

    You seem to be going out of your way to confuse yourself, and you're still making the same mistake.

    For your last calculation, you are now using u as the 'original' variable and \theta as the 'transformed' variable and the relationship between them is u=\pi  + \theta.

    For the stuff to the right of the integral sign u is being replaced by \pi + \theta.

    If you use the same replacement method for the limits, as you must, then you have to say that,

    when u = -\pi, then \theta is given by -\pi = \pi + \theta. That is, \theta = -2\pi.

    Similarly when u = 0,\text{   } \theta = -\pi.

    The limits (-\pi, 0) \text{ transform to } (-2\pi, -\pi).
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