Multiple logarithmic integral

Prove that

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

Solution

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

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

Read more

Limit of a multiple integral

Prove that

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

Solution

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.

So,

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

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,

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

Read more

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.

Solution

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.

Read more

 

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

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

Solution

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*}

Read more

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.

Solution

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

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

hence

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

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

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

Read more