Fourier coefficients of $\mathrm{e}^{\mathrm{e}^{\mathrm{i}x}}$?

Question

The question is to find the complex Fourier coeff. of $ f(x)=\mathrm{e}^{\mathrm{e}^{\mathrm{i}x}}$

This leads to an integral $\int_{0}^{2\pi} \mathrm{e}^{\mathrm{e}^{\mathrm{i}x}-\mathrm{i}kx} \,dx$ , which I have no idea how to solve.

Answer 1

(incomplete answer, we have the result up to a constant coefficient)

We can do some work with integration by parts.

Let

$$u_k=\int_0^{2\pi} \exp(\exp(ix)-kix)\,\mathrm dx=\int_0^{2\pi} \exp(\exp(ix))\exp(-kix)\,\mathrm dx$$

Then, integrating by parts for $k>0$, (differentiating the first factor in the integrand), we get:

$$u_k=\left[\exp(\exp ix)\frac{\exp(-ikx)}{-ik}\right]_0^{2\pi}+\frac1{ki}\int_0^{2\pi}i\exp(ix)\exp(\exp ix)\exp(-ikx)\,\mathrm dx$$

The first term is zero and the second one simplifies to:

$$u_{k}=\frac1{k}\int_0^{2\pi}\exp(\exp ix)\exp(-i(k-1)x)\,\mathrm dx=\frac 1ku_{k-1}$$

Then by recurrence, for all $k\ge0$,

$$u_k=\frac{u_0}{k!}$$

It remains to find out the value of $u_0$. Maple tells me it's $2\pi$, but it's not a proof.

Actually, the final step is already here on math.se: Integral $\int_0^{2π} e^{e^{ix}} dx$

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For the negative coefficients, let, for $k>0$, $v_k=u_{-k}$, and note that the recurrence formula above is still valid (it's not for $k=0$ only), i.e.

$$v_{k+1}=kv_k$$

But now, as $v_k$ is obviously bounded, we must have $v_k=0$ for all $k>0$.

That is, for all $k<0$, $u_k=0$.

Answer 2

Here is an answer that requires no integration. Consider that

$$e^{e^{ix}} = \sum_{n=0}^\infty \frac{e^{inx}}{n!}$$

via Taylor series. Coincidentally, this is also already a Fourier series - thus the coefficients are

$$c_n = \begin{cases} \frac{1}{n!} & n\geq 0\\ 0 & n<0 \end{cases}$$

Answer 3

While no effort was shown on OP's behalf, others had already answered, so I'll post my answer.

A complex analysis approach (which is really the same as the series approach, but it's still nice to see it through this lens):

Let $z = e^{ix}$, then the image of $[0, 2\pi)$ after this change of variables is the unit circle. Further, $dz = ie^{ix}\,dx$ so that $dx = \frac{1}{i} z^{-1}\,dz$, giving

$$ \int_0^{2\pi} e^{e^{ix}} e^{-ikx}\,dx = \frac{1}{i}\int_{|z|=1} e^z z^{-k-1}\,dz = \frac{1}{i}\int_{|z|=1} \frac{e^z}{z^{k+1}}\,dz.$$

From the generalized Cauchy integral formula, we have that

$$ f^{(n)}(0) = \frac{n!}{2\pi i}\int_{|z|=1} \frac{f(z)}{z^{n+1}}\,dz.$$

Thus,

$$\int_0^{2\pi} e^{e^{ix}} e^{-ikx}\,dx = \frac{2\pi}{k!} \frac{d^k}{dz^k}e^z \bigg|_{z=0} = \frac{2\pi}{k!}.$$