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Math Help - Proof required for integration involving exponential, cos and sine.

  1. #1
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    Question Proof required for integration involving exponential, cos and sine.

    This question appears in my university postgraduate advanced engineering mathematics course:

    Show that:


    ∫ e ^(ρ.cosθ) . cos(ρ sinθ – nθ) dθ = 2πρⁿ/n!
    0

    where ρ>0, and n = 1,2,3...
    Hint: cos(ρ sinθ – nθ)=Re{cos(ρ sinθ – nθ) + i sin (ρ sinθ – nθ)}

    (10 marks)

    Thanks a lot in advance. ;-)
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  2. #2
    Moo
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    Hello,

    e ^(ρ.cosθ) << this is a real number

    cos(ρ sinθ – nθ) << as mentioned this is the real part of cos(ρ sinθ – nθ) + i sin (ρ sinθ – nθ)

    But we know that cos(ρ sinθ – nθ) + i sin (ρ sinθ – nθ) is equal to e^(ρ sinθ – nθ) !

    Can you simplify the thing ?
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  3. #3
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    Wink

    hmmm......
    well, not yet.

    can i be tutored with a little more information?
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  4. #4
    Moo
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    Ok :

    \int e^{\rho \cos \theta} \underbrace{\cos (\rho \sin \theta – n \theta)}_{Re(\cos (\rho \sin ( \theta – n \theta))+i \sin(\rho \sin ( \theta – n \theta)) } d \theta

    So this integral is the same as :

    \int Re(e^{\rho \cos \theta} (\cos (\rho \sin( \theta – n \theta))+i \sin (\rho \sin ( \theta – n \theta)) d \theta


    It's really awful with latex... i have a little bug.
    I'll try to write it elsewhere.

    Just remember a thing : cos(ρ sinθ – nθ) + i sin (ρ sinθ – nθ)=e^(ρ sinθ – nθ) (de Moivre's formula)
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  5. #5
    Moo
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    Actually, i've got a problem, your integral is really strange oO
    Attached Thumbnails Attached Thumbnails Proof required for integration involving exponential, cos and sine.-ex-001.jpg  
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  6. #6
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    This is a complex analysis problem. I believe this is a Cauchy integral theorem problem. I do not have time now. By try integrating f(z) = e^z/z around the circle |z|=\rho.
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  7. #7
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    From Moo's comment, \int_0^{2\pi}e^{\rho\cos\theta}\cos(\rho\sin\theta -n\theta)d\theta = \text{re}\!\!\int_0^{2\pi}e^{\rho\cos\theta+i\rho\  sin\theta-in\theta}d\theta. This is equal to \text{re}\!\!\int_0^{2\pi}e^{\rho e^{i\theta}}e^{-in\theta}d\theta. Now (more or less following ThePerfectHacker's suggestion) make the substitution z = e^{i\theta}, so that as θ goes from 0 to 2π, z goes round the unit circle. Also, \textstyle d\theta = \frac{dz}{iz}. The integral becomes \text{re}\!\oint e^{\rho z}z^{-n}{\textstyle \frac{dz}{iz}} = \text{re}\Bigl(-i\oint e^{\rho z}z^{-(n+1)}dz\Bigr). You can evaluate this contour integral by the residue theorem, noting that the integrand has a single pole (of order n+1) at the origin. (Calculate the residue there by writing down the power series expansion of e^{\rho z} and picking out the coefficient of z^n.)
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    Ok :

    \int e^{ \rho \cos \theta} \underbrace{\ cos ( \rho \sin \theta – n \theta) }_{Re(\cos (\rho \sin ( \theta – n \theta)))+i \sin(\rho \sin ( \theta – n \theta)) } d \theta

    So this integral is the same as :

    \int Re(e^{\rho \cos \theta} (\cos (\rho \sin( \theta – n \theta))+i \sin (\rho \sin ( \theta – n \theta)))~ d \theta


    It's really awful with latex... i have a little bug.
    I'll try to write it elsewhere.

    Just remember a thing : cos(ρ sinθ – nθ) + i sin (ρ sinθ – nθ)=e^(ρ sinθ – nθ) (de Moivre's formula)
    here you go Moo
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  9. #9
    Moo
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    Hi Sir

    Actually, no... It still removes automatically the - sign...
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  10. #10
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    Quote Originally Posted by Moo View Post
    Hi Sir

    Actually, no... It still removes automatically the - sign...
    I think you're using the wrong symbol for the minus sign. Replace the en-dash by a short hyphen, and it comes out right:

    \color{blue}\int e^{\rho \cos \theta} \underbrace{\cos (\rho \sin \theta - n \theta)}_{\text{Re}(\cos (\rho \sin \theta - n \theta)+i \sin(\rho \sin \theta - n \theta))} d\theta

    So this integral is the same as:

    \color{blue}\int \text{Re}(e^{\rho \cos \theta} (\cos (\rho \sin \theta - n \theta)+i \sin (\rho \sin  \theta - n \theta)) d \theta
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