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Iterated exponential integral

Prove that

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


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.

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On a log Gamma integral using Riemann sums

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \log \Gamma (x) \, \mathrm{d}x\]

using Riemann sums.


Partition the interval [0, 1] into n subintervals of length \frac{1}{n}. This produces,

(1)   \begin{equation*}  \int_{0}^{1} \log \Gamma(x) \, \mathrm{d}x = \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) \end{equation*}

On the other hand, assuming n is even:

    \begin{align*} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) &= \frac{1}{n} \log \prod_{k=1}^{n} \Gamma \left ( \frac{k}{n} \right ) \\ &= \frac{1}{n} \log \prod_{k=1}^{n/2} \Gamma \left ( \frac{k}{n} \right ) \Gamma \left ( 1 - \frac{k}{n} \right )\\ &=\frac{1}{n} \log \prod_{k=1}^{n/2} \frac{\pi}{\sin \frac{\pi k}{n}} \\ &= \log \sqrt{\pi} - \log \left ( \prod_{k=1}^{n} \sin \frac{\pi k}{n} \right )^{1/n}\\ &= \log \frac{\sqrt{2 \pi}}{\left ( 2n \right )^{1/2n}} \end{align*}

since it holds that

    \[\prod_{k=1}^{n} \sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}\]

Euler’s Gamma reflection formula was used at line (3). Letting n \rightarrow +\infty we get that

    \[\int_0^1 \log \Gamma(x) \, \mathrm{d}x = \log \sqrt{2\pi}\]

If n is odd we work similarly.


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Multiple logarithmic integral

Prove that

    \[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y,z)}{ \ln x + \ln y + \ln z} = - \frac{1}{2}\]


First of all we observe that the integral \bigintsss_0^1 \frac{\mathrm{d}x}{\ln x} as well as the integral \bigintsss_0^1 \bigintsss_0^1 \frac{\mathrm{d}(x, y)}{\ln x + \ln y} diverge whereas the proposed integral converges which is an interesting fact. Now,

    \begin{align*} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln x + \ln y + \ln z} &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\mathrm{d}(x, y, z)}{\ln xyz} \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{u=xyz \; , \; v=y \; , \; w=z}{=\! =\! =\! =\! =\! =\! =\! =\! =\!=\!=\!} \iiint \limits_{\mathbb{D}} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=\int_{0}^{1} \int_{u}^{1} \int_{u/w}^{1} \frac{\mathrm{d}(v, w, u)}{v w \ln u} \\ &=-\frac{1}{2} \end{align*}

since \mathbb{D} is determined by the inequalities

    \[\frac{u}{w}\leq v\leq 1 \quad , \quad u\leq w\leq 1\quad ,\quad 0\leq u\leq 1\]

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Limit of a multiple integral

Prove that

    \[\lim_{n \to +\infty} \idotsint \limits_{[0, 1]^n}\frac{x_1^2+x_2^2+ \cdots +x_n^2}{x_1+x_2+ \cdots +x_n} \, \mathrm{d} (x_1, x_2, \dots, x_n) = \frac{2}{3}\]


Let X_1, X_2,\dots be independent and uniform (0,1) random variables. By the law of large numbers we have

    \[\begin{matrix} \dfrac{X_1+\cdots + X_n}{ n}&\rightarrow & \mathbb{E}(X)= \dfrac{1}{2} \\\\ \dfrac{X_1^2+\cdots + X^2_n}{n}&\rightarrow & \mathbb{E}(X^2)={1\over 3} \end{matrix}\]

in probability as n \rightarrow +\infty. Therefore ,

    \[\frac{X_1^2+\cdots +X^2_n}{ X_1+\cdots +X_n}=\frac{X_1^2+\cdots +X^2_n}{ n}\cdot{n\over X_1+\cdots +X_n}\to \frac{2}{3}\]

in probability as n \rightarrow +\infty.

The ratio random variables {X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n} are bounded below by zero and above by one. This guarantees convergence of the expectations, as well.


    \[\mathbb{E}\left({X_1^2+\cdots +X^2_n\over X_1+\cdots +X_n}\right)\rightarrow{2\over 3}\]

which is the required result.

Remark: In general it holds that

\displaystyle \lim_{n\rightarrow+\infty} \int_0^1 \cdots \int_0^1 f\left(\frac{x_1 + \cdots + x_n}{n}\right) \, \mathrm{d}(x_1, \dots, x_n) = \mathbb{E}\left [ f \left ( \frac{X_1+X_2+\cdots+X_n}{n} \right ) \right ]

because the distribution of (X_1,\dots,X_n) is the Lebesgue measure on [0,1]^n hence for every measurable function g,

    \[\mathbb E(g(X_1,\ldots,X_n))=\iint \limits_{[0,1]^n}g(x_1,\ldots,x_n)\mathrm dx_1\ldots\mathrm dx_n\]

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An extraordinary sin integral

I was surfing the net today and I fell on this cute integral

    \[\mathcal{J}= \int_{-\infty}^{\infty} \sin \left( x^2 + \frac{1}{x^2} \right) \, {\rm d}x\]

I have seen integrals of such kind before like for instance this \displaystyle \int_{0}^{\infty} \sin \left( x^2 - \frac{1}{x^2} \right) \, {\rm d}x .In fact something  more general holds

    \[\int_0^{\infty}\sin \left( ax^2 - \frac{b}{x^2} \right) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{2a}} e^{-2 \sqrt{ab}}\]

where a, b>0.


The original integral does not fall into this category which is a real shame. Yet it does have a closed form and it does not contain an e in its final answer. Strange, huh? So similar but so different at the same time these two integrals. A sign changes everything.

We begin by exploring the integral

    \[\mathcal{J}'  = \int_0^\infty \exp \left( \frac{ia}{x^2} + ib x^2 \right) \, {\rm d}x\]

Manipulating the integral ( substitutions and known Gaussian results) reveals that

    \[\int_{0}^{\infty} \exp \left (\frac{ia}{x^2} + i b x^2 \right ) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{b}}\exp\left ( \frac{i \pi}{4} \right ) e^{2i \sqrt{ab}}\]

where a, b>0. Taking the imaginary part of the last expression we get that

    \[\int_{-\infty}^{\infty} \sin \left ( x^2 + \frac{1}{x^2} \right ) \, {\rm d}x = \sqrt{\frac{\pi}{2}} \left ( \sin 2 + \cos 2 \right )\]

and this is the final answer. See, no e!. Of course we can also extract the real part and calculate the corresponding integral involving \cos.

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