Iterated exponential integral

Prove that

$\int_0^{2\pi} \exp \left ( \exp \left ( \exp (it) \right ) \right )\, \mathrm{d}t = 2\pi e$

Solution

We begin by stating a lemma:

Lemma: Let $f$ be an analytic function on some closed disk $\mathbb{D}$ which has center $a$ and radius $r$ . Let $\mathcal{C}$ denote the the boundary of the disk. It holds that

$\int_{0}^{2\pi} f \left ( a + re^{i \theta} \right )\, \mathrm{d}\theta = 2 \pi f(a)$

Proof: By the Cauchy integral formula we have that

$f(a) = \frac{1}{2\pi i }\oint \limits_{\mathcal{C}} \frac{f(z)}{z-a} \, \mathrm{d}z$

The equation of a circle of radius $r$ and centre $a$ is given by $z=a + re^{i \theta}$ . Hence,

$\begin{align*} f(a) &= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(a + re^{i\theta})ire^{i\theta}}{re^{i\theta}}\,\mathrm{d}\theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,\mathrm{d}\theta \end{align*}$

and the proof is complete.

Something quickie: Given the assumptions in Gauss’ MVT, we have

$\left|f(a)|\right\leq \frac{1}{2\pi}\int_{0}^{2\pi}\left|f(a+re^{i\theta})\right|\,\mathrm{d} \theta$

The proof of the result is pretty straight forward by using the fact that

$\left | \int_{a}^{b} f(x) \, \mathrm{d}x \right | \leq \int_{a}^{b} \left|f(x) \right| \, \mathrm{d}x$

Back to the problem the result now follows by the lemma.

Iterated exponential integral

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