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I'm working on a school project about the aerodynamics of an helicopter blade. I'm trying to adapt my aerodynamics class ( centered around planes wings mostly ) to this case.

In our class, we used what we called the integrals of Glauert and more specifically, this integral : $$ \int_{0}^{\pi}\frac{\cos\left(nx\right)} {\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x = -\pi\,\frac{\sin\left(n\theta\right)}{\sin\left(\theta\right)} $$ In the helicopter case, I need to evaluate a similar integral but with $\sin\left(nx\right)$ instead of $\cos\left(nx\right)$.

I found a beautiful proof of the first result in the book Inside Interesting Integrals by Paul Nahin ( page $60$ ) which uses recursivity. I tried to adapt it to the other integral using the formula $$\sin\left(\left(n + 1\right)x\right) + \sin\left(\left(n - 1\right)x\right) = 2\sin\left(nx\right)\cos\left(x\right)$$ $$ \mbox{Calling}\quad I_{n} = \int_0^\pi\frac{\sin\left(nx\right)}{\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x, $$ I find that $$ I_{n+1} - 2\cos\left(\theta\right)I_{n} + I_{n - 1} = \int_{0}^{\pi}2\sin\left(nx\right)\,{\rm d}x $$ but I then have a problem to solve the recursive equation be the result is $0$ or $\frac{4}{n}$ depending on the parity of n. I decoupled the system to get two recursive equations for the different parities :

  • For $a_n=I_{2n}$, I get $a_{n+1}-2\cos(2\theta)a_n+a_{n-1}=8\cos\theta(\frac{1}{2n+1}+\frac{1}{2n-1}$)
  • And for $b_n=I_{2n+1}$, I get $b_n-2\cos(2\theta)b_{n-1}+b_{n-2}=\frac{8}{2n-1}\cos(\theta)$

I entered these formula on Wolfram Alpha and got some really ugly results. I was thus wondering if a more simple formula could exist for my integral.

I also quickly looked at the wikipedia proof which uses residual theorems but I got lost quite fast...

Do you know some other way to evaluate this integral or maybe if it's possible to solve the recursive equations to get to a simpler result than the one from Wolfram ?

Thank you very much !

Felix Marin
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Julie Ben Zenou
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    You go to a really cool school. – Vishu Dec 22 '20 at 13:31
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    I do in fact ! I'm in a French engineering school and I'm specialized in Aeronautics and Space physics :) – Julie Ben Zenou Dec 22 '20 at 13:52
  • For real $\theta$ the integral diverges (logarithmically). Mathematica gives the closed form result $\frac{\pi , _3\tilde{F}_2\left(\frac{1}{2},1,1;1-n,n+1;\sec ^2\left(\frac{\theta }{2}\right)\right)}{\cos (\theta )+1}$ provided $\Im(\cos (\theta ))\neq 0\lor \Re(\cos (\theta ))\leq -1$. – Dr. Wolfgang Hintze Dec 22 '20 at 13:57
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    Note $\int_0^\pi\frac{\cos(nx)-cos(n\theta)}{\cos(\theta)-\cos(x)}dx=-\pi\frac{\sin(n\theta)}{\sin(\theta)}$. $\int_0^\pi\frac{\cos(nx)}{\cos(\theta)-\cos(x)}dx$ diverges – Quanto Dec 22 '20 at 14:14
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    what you have is a Cauchy Principal Value http://arion.naval.ntua.gr/~banthy/Glauert_integrals.pdf – Will Jagy Dec 22 '20 at 16:41
  • When I try to solve with Mathematica for $n=2$ $$\int_0^{\pi } \frac{\sin (2 x)}{\cos t-\cos x} , dx$$ I get $$2 \cos t \log \left(\frac{\cos t+1}{\cos t-1}\right)-4$$ which implies $\cos t\ge 1$ which makes little to zero sense. In other words the integral diverges if I set $\cos \theta=1$. The same happens for any other value of $n$. – Raffaele Dec 22 '20 at 17:52
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    Those are principal value integrals, for $0 < t < \pi$ we get $$\operatorname {v. ! p.} \int_0^\pi \frac {\cos n x} {\cos t - \cos x} dx = -\pi U_{n - 1}(\cos t), \ \operatorname {v. ! p.} \int_0^\pi \frac {\sin n x} {\cos t - \cos x} dx = 2 U_{n - 1}(\cos t) \ln \cot \frac t 2 - 4 \sum_{k = 1}^{\lfloor n/2 \rfloor} \frac {U_{n - 2 k}(\cos t)} {2 k - 1} ,$$ where $U_n$ are the Chebyshev polynomials, $U_n(\cos t) = \csc t \sin , (n + 1) t$. (Since all $I_n$ depend on $I_1$, I don't see how you can split the recurrence relation into two.) – Maxim Dec 22 '20 at 22:14
  • @Maxim Would you please share the details of obtaining the integral for sin nx/(cost-cosx)? – Rajai Mar 15 '24 at 01:02

1 Answers1

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Probably, there is no a manageable general expression but we can get an exact answer for moderate values of n.

The idea is to get rid of singularity in the integrand.

So we replace $\theta$ with $i\theta$ ($i$ is the imaginary unit).

Since $\cos(i\theta)=\cosh(\theta)$ we get

$$J_n(\theta)=\int_0^\pi\frac{\sin(nx)}{\cosh(\theta)-\cos(x)}dx$$ Now, $J_1(\theta)=-2\ln(\tanh\frac{\theta}{2})$

To return to the original integral we replace $\theta$ with $i\theta$ again and equate the real parts on both sides of the equal sign:

$$I_1(\theta)=P\;V\int_0^\pi\frac{\sin(x)}{\cos(\theta)-\cos(x)}dx=-2\ln(\tan\frac{\theta}{2})$$

Completely analogously:

$$J_2(\theta)=\int_0^\pi\frac{\sin(2x)}{\cosh(\theta)-\cos(x)}dx$$

$$=2\cosh(\theta)\ln\frac{\cosh(\theta)+1}{\cosh(\theta)-1}-4$$

and $$I_2(\theta)=P\;V\int_0^\pi\frac{\sin(2x)}{\cos(\theta)-\cos(x)}dx$$

$$=4\cos(\theta)\ln(\cot\frac{\theta}{2})-4$$

For $n=3,4,...$ the procedure is the same.

REMARK:

In similar manner we can establish the initial integral.

Namely, consider

$$\int_0^\pi\frac{\cos(nx)}{\cosh(\theta)-\cos(x)}dx=\pi\frac{e^{-n\theta}}{\sinh(\theta)}$$

Replacing here $\theta$ with $i\theta$ ($i$ is the imaginary unit) and equating real parts gives

$$P\;V\int_0^\pi\frac{\cos(nx)}{\cos(\theta)-\cos(x)}dx=-\pi\frac{\sin(n\theta)}{\sin(\theta)}$$

Martin Gales
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    Thanks a lot for this answer ! Can you detail a bit more how you find the results for J1 and J2 ? I think I get J1 approximately since you have some u'/u type to integrate but I have no idea how to integrate J2 ? Couldn't we adapt this to any n ? – Julie Ben Zenou Dec 30 '20 at 13:36