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Math Help - Difficult Integration

  1. #1
    Forum Admin topsquark's Avatar
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    Difficult Integration

    (sigh) I know I've seen how to do this before, but I can't find the notes!

    The problem is this:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} where a is real.

    It is simple enough to show that the integral is equal to a real number, so
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} = \int_0^{\pi} d \theta \, sin \theta \, cos(a \, sin \theta)
    as an alternate formula.

    Comparing two lines in the book I got this from the integral should be equal to
    2 \frac{sin \, a}{a}
    (For the case a = 0, the integral is simple and comes out to be equal to 2. Obviously I'm not worried about this case.)

    Any takers? Thanks in advance.

    -Dan
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  2. #2
    MHF Contributor Quick's Avatar
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    You should know that the answer is:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}=2







    You should also know never to listen to me when you get past algebra
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    You should know that the answer is:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta}=2







    You should also know never to listen to me when you get past algebra
    No no no! When it's Calculus the answer is always 8.

    -Dan
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    No no no! When it's Calculus the answer is always 8.

    -Dan
    Oh, sorry you're right (I really need to review my notes)
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    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    (sigh) I know I've seen how to do this before, but I can't find the notes!

    The problem is this:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} where a is real.

    It is simple enough to show that the integral is equal to a real number, so
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, sin \theta} = \int_0^{\pi} d \theta \, sin \theta \, cos(a \, sin \theta)
    as an alternate formula.
    Why do you think the integral is real?

    See attachment.

    RonL
    Attached Thumbnails Attached Thumbnails Difficult Integration-gash.jpg  
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Why do you think the integral is real?

    See attachment.

    RonL
    Heavens. I had transformed the integral by substituting \theta' = \theta - \pi/2 and took a look at sin(acos(\theta')) and decided it was an odd function over  [-\pi/2, \pi/2] . It isn't, of course.

    Now I've got a "real" problem (pun intended) because my book clearly indicates the real.

    Thanks! Back to the books to find out what's going on.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Okay, I fixed my boo-boo. I wish I could say I used the wrong coordinate system or something face-saving, but I simply used the wrong trig function.

    The correct integral is:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, cos \theta} = \int_0^{ \pi} d \theta \, sin \theta \, cos(a \, cos \theta) = 2 \frac{sin(a)}{a} where a > 0.

    Anyone know how to do this one? Thanks!

    -Dan
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    Okay, I fixed my boo-boo. I wish I could say I used the wrong coordinate system or something face-saving, but I simply used the wrong trig function.

    The correct integral is:
    \int_0^{\pi} d \theta \, sin \theta \, e^{ia \, cos \theta} = \int_0^{ \pi} d \theta \, sin \theta \, cos(a \, cos \theta) = 2 \frac{sin(a)}{a} where a > 0.

    Anyone know how to do this one? Thanks!

    -Dan
    Perhaps the following might help:

    <br />
\sin (\theta)\ e^{ia \cos( \theta)} =-\frac{1}{ia}\ \frac{d}{d\theta} e^{ia \cos( \theta)}<br />

    RonL
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Perhaps the following might help:

    <br />
\sin (\theta)\ e^{ia \cos( \theta)} =-\frac{1}{ia}\ \frac{d}{d\theta} e^{ia \cos( \theta)}<br />

    RonL
    I don't mind that you pointed out something I should have seen almost immediately. I don't mind that I figured this out late last night and didn't get to the post early enough to "save face" by mentioning I figured it out.

    What I mind is that I spent hours working on a problem that I should have been able to answer in minutes.

    Ah well. The learning process! Thanks CaptainBlack.

    -Dan
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