# Thread: Proof required for integration involving exponential, cos and sine.

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

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. ;-)

2. 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 ?

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

4. Ok :

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

So this integral is the same as :

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

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)

5. Actually, i've got a problem, your integral is really strange oO

6. This is a complex analysis problem. I believe this is a Cauchy integral theorem problem. I do not have time now. By try integrating $\displaystyle f(z) = e^z/z$ around the circle $\displaystyle |z|=\rho$.

7. From Moo's comment, $\displaystyle \int_0^{2\pi}e^{\rho\cos\theta}\cos(\rho\sin\theta -n\theta)d\theta = \text{re}\!\!\int_0^{2\pi}e^{\rho\cos\theta+i\rho\ sin\theta-in\theta}d\theta$. This is equal to $\displaystyle \text{re}\!\!\int_0^{2\pi}e^{\rho e^{i\theta}}e^{-in\theta}d\theta$. Now (more or less following ThePerfectHacker's suggestion) make the substitution $\displaystyle z = e^{i\theta}$, so that as θ goes from 0 to 2π, z goes round the unit circle. Also, $\displaystyle \textstyle d\theta = \frac{dz}{iz}$. The integral becomes $\displaystyle \text{re}\!\oint e^{\rho z}z^{-n}{\textstyle \frac{dz}{iz}} = \text{re}\Bigl(-i\oint e^{\rho z}z^{-(n+1)}dz\Bigr)$. 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 $\displaystyle e^{\rho z}$ and picking out the coefficient of z^n.)

8. Originally Posted by Moo
Ok :

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

So this integral is the same as :

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

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

9. Hi Sir

Actually, no... It still removes automatically the - sign...

10. Originally Posted by Moo
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:

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

So this integral is the same as:

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