Integrating along the unit circle

Prove that

$$\int^{2\pi}_0e^{\cos \theta}\cos(n\theta -\sin \theta)\,d \theta=\frac{2\pi}{n!}$$

$$\textit{proof}$$

 

Consider the following function

$$f(z)=e^{z^{-1}}z^{n-1}$$

Now we integrate the function along a circle of radius 1

The contour encloses a pole at \(z = 0\)

$$\oint_{|z|=1}e^{z^{-1}}z^{n-1} dz=2\pi i\mathrm{Res}(f(z),0) $$

Now we need to find the residue which is the coefficient of \( 1/z \)

 

$$e^{z^{-1}}z^{n-1} = \sum_{k\geq 0}\frac{1}{k!}z^{n-k-1} $$

The coefficient of \( z^{-1} \) implies that \( k=n \), hence

$$ \mathrm{Res}(f(z),0) =\frac{1}{n!}$$

So we have

$$\oint_{|z|=1}e^{z^{-1}}z^{n-1} dz= 2\pi i \frac{1}{n!} $$

Using a parametirzation of the circle \(z = e^{it}\) where \(t \in (0,2\pi)\)

$$i\int_{0}^{2\pi }e^{e^{-i\theta }}e^{i n\theta } d\theta = 2\pi i \frac{1}{n!} $$

Finally we have

$$\int_{0}^{2\pi }e^{e^{-i\theta }}e^{i n\theta } d\theta = 2\pi\frac{1}{n!} $$

 

Notice that

$$e^{e^{-i\theta }}e^{i n\theta } = e^{\cos(\theta)} (\cos(\sin \theta)-i\sin(\sin \theta ))(\cos(n\theta)+i\sin(n\theta))$$

Take the real part

$$\boxed{\int^{2\pi }_0 e^{\cos \theta} \cos(n\theta-\sin \theta)\, d\theta= \frac{2\pi}{n!}} $$

As a bonus we have

$$\boxed{\int^{2\pi }_0 e^{\cos \theta} \sin(n\theta-\sin \theta)\, d\theta= 0} $$

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