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 somethingĀ  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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Integral with floor and ceiling function

Let \left \lfloor \cdot \right \rfloor denote the floor function and \left \{ \cdot \right \} denote the ceiling function. Evaluate the integral

    \[\mathcal{J}= \int_{0}^{1}x \left \lfloor \frac{1}{x} \right \rfloor\left \{ \frac{1}{x} \right \} \, {\rm d}x\]


We have successively

    \begin{align*} \mathcal{J} &= \int_{0}^{1}x \left \lfloor \frac{1}{x} \right \rfloor\left \{ \frac{1}{x} \right \} \, {\rm d}x \\ &\!\!\!\! \overset{x \mapsto 1/x}{=\! =\! =\! =\!} \int_{1}^{\infty} \frac{\left \lfloor x \right \rfloor \left \{ x \right \}}{x^3} \, {\rm d}x\\ &= \sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n(x-n)}{x^3} \, {\rm d}x\\ &= \sum_{n=1}^{\infty} \left [ \int_{n}^{n+1} \frac{n}{x^2} \, {\rm d}x - \int_{n}^{n+1} \frac{n^2}{x^3} \, {\rm d}x \right ] \\ &= \sum_{n=1}^{\infty} \left ( \frac{1}{n+1} - \frac{2n+1}{2(n+1)^2} \right ) \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+1)^2}\\ &= \frac{1}{2} \left ( \zeta(2) - 1 \right ) \\ &= \frac{\pi^2}{12} - \frac{1}{2} \end{align*}

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Multiple integral

Let \langle \cdot, \cdot \rangle denote the usual inner product of \mathbb{R}^m. Evaluate the integral

    \[\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left( - ( \langle x, \mathcal{S} ^{-1} x \rangle )^a \right) \, {\rm d}x\]

where \mathcal{S} is a positive symmetric m \times m matrix and a>0.


Since \mathcal{S} is a positive symmetric matrix , so is \mathcal{S}^{-1}. For a positive symmetric matrix \mathcal{A} there exists an \mathcal{R} positive symmetric matrix such that \mathcal{A} = \mathcal{R}^2. Applying this to \mathcal{S}^{-1} our integral becomes

    \[\mathcal{M} = \int \limits_{\mathbb{R}^m} \exp \left ( - \left \| \mathcal{R} x \right \|^{2a} \right ) \, {\rm d}x\]

where \left \| \cdot \right \| is the Euclidean norm. Applying a change of variables we have that

    \[\mathcal{M} = \det \left ( \mathcal{R}^{-1} \right ) \int \limits_{\mathbb{R}^m} e^{-\left \| y \right \|^{2a}} \, {\rm d}y\]

Since \det \left ( \mathcal{R}^{-1} \right ) = \sqrt{\det \left ( \mathcal{S} \right )} then by converting to polar coordinates we have that

    \begin{align*} \mathcal{M} &=  \omega_m \sqrt{\det \left ( \mathcal{S} \right )} \int_{0}^{\infty} r^{m-1} e^{-r^{2a}} \, {\rm d}r \\ &= \frac{\omega_m}{m} \sqrt{\det \left ( \mathcal{S} \right )}\Gamma \left ( \frac{m}{2a} + 1 \right ) \end{align*}

Here \omega_m denotes the surface area measure of the unit sphere and it is known to be

    \[\omega_m = \frac{2 \pi^{m/2}}{\Gamma \left ( \frac{m}{2} \right )}\]


    \[\mathcal{M}=\frac{\sqrt{\det (\mathcal{S}) }\pi^{m/2}\Gamma\left(\frac{m}{2a}\right)}{2^{a-1}\Gamma\left(\frac{m}{2}\right)}\]

where \Gamma denotes the Gamma Euler function for which it holds that

    \[\Gamma(x+1) = x \Gamma(x) \quad \text{forall} \quad x>0\]

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