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Exponential fractional part

Let \{ \cdot \} denote the fractional part of x. Evaluate the integral

    \[\mathcal{J} = \int_0^1 \int_0^1 \left\{ \frac{e^x}{e^y} \right\} \, \mathrm{d}(x, y)\]

Solution

By definition it holds that

    \[\left \{ x \right \} = x - \left \lfloor x \right \rfloor\]

hence

    \[\iint \limits_{\left [ 0, 1 \right ]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y)= \iint \limits_{\left [ 0, 1 \right ]^2} e^x e^{-y} \, \mathrm{d}(x, y)-\iint \limits_{\left [ 0, 1 \right ]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d}(x, y)\]

We note however that

    \[\left \lfloor e^x e^{-y} \right \rfloor = \left\{\begin{matrix} 0 & , & x< y \\ 2 & , & y< x - \ln 2 \\ 1 & , & \text{otherwise} \end{matrix}\right.\]

Hence,

    \begin{align*} \iint \limits_{[0, 1]^2} \left \lfloor e^x e^{-y} \right \rfloor \, \mathrm{d} (x, y) &= \int_{0}^{1} \int_{x}^{1} 0 \, \mathrm{d}(y, x) + \int_{0}^{1} \int_{0}^{x} \, \mathrm{d} (y, x) + \\ & \quad \quad + \int_{\ln 2}^{1} \int_{0}^{x-\ln 2} 1 \, \mathrm{d} \left ( y, x \right ) \\ &= 1 - \ln 2 + \frac{\ln^2 2}{2} \end{align*}

Summing up,

    \[\iint \limits_{[0, 1]^2} \left \{ e^x e^{-y} \right \} \, \mathrm{d}(x, y) = \frac{1}{e} +e + \ln 2 - \frac{\ln^2 2}{2} - 3\]

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Arctan squared integral

Evaluate the integral

    \[\mathcal{J} =\int_0^\pi \arctan^2 \left( \frac{\sin x}{2+\cos x} \right) \, \mathrm{d}x\]

Solution

First of all we note that for x \in (0, \pi)

    \begin{align*} \arctan \left ( \frac{\sin x}{2 + \cos x} \right ) &= \mathfrak{Im} \log \left ( 2 + e^{ix} \right ) \\ &= \mathfrak{Im} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n 2^n} e^{inx} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} }{n 2^n} \sin nx \end{align*}

Hence by Parseval we get that

    \[\int_{0}^{\pi} \arctan^2 \left(\frac{\sin x}{2+\cos x}\right)\,\mathrm{d} x=\frac{\pi}{2} \cdot \sum_{n=1}^{\infty} \frac{1}{n^2 4^n}=\frac{\pi}{2} \cdot \mathrm{Li}_2 \left ( \frac{1}{4} \right )\]

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Contour radical integral

Consider the branch of \displaystyle f(z) =\sqrt{z^2-1} which is defined outside the segment [-1, 1] and which coincides with the positive square root \sqrt{x^2-1} for x>1. Let R>1 then evaluate the contour integral:

    \[\ointctrclockwise \limits_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}\]

Solution

It is a classic case of residue at infinity. Subbing z \mapsto \frac{1}{z} the counterclockwise contour integral rotates the northern pole of the Riemannian sphere to the southern one and the contour integral is transformed to a clockwise one. Hence:

    \begin{align*} \oint \limits_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\ &=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\ &=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\ &= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\ &=2\pi i \end{align*}

The equality w\sqrt{w^{-2}-1}=\sqrt{1-w^2}does hold for all |w|<1 if we take the standard branch \sqrt{1-w^2}=\exp \left ( \frac{1}{2}\mathrm{Log} \left ( 1-w^2 \right ) \right ) , otherwise it is not that obvious why this holds, since we are dealing with a multi-valued function.

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Square logarithmic integral

Prove that

    \[\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x = \frac{\pi^3}{24} + \frac{\pi \ln^2 2}{2}\]

Solution

Lemma 1: Let n, m \in \mathbb{N}. It holds that

    \[\int_{0}^{\pi/2}\cos nx \cos mx \, \mathrm{d}x = \left\{\begin{matrix} 0 & , & n \neq m \\ \frac{\pi}{2} & , & n=m \end{matrix}\right.\]

Lemma 2: Let x \in (0, 2\pi). It holds that

    \[\ln 2\sin \frac{x}{2} = - \sum_{n=1}^{\infty} \frac{\cos nx}{n}\]

We begin by squaring the identity of lemma 2. Hence,

    \[\ln^2 \left ( 2 \sin \frac{x}{2} \right ) = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\cos nx \cos mx}{nm}\]

Integrating the last equation we get,

    \[\int_{0}^{\pi}\ln^2 \left ( 2 \sin \frac{x}{2} \right ) \, \mathrm{d}x = \int_{0}^{\pi}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\cos nx \cos mx}{nm} \, \mathrm{d}x = \frac{\pi^3}{12}\]

Expanding the LHS we get that

    \begin{align*} \frac{\pi^3}{12} &= \int_{0}^{\pi} \ln^2 \left ( 2 \sin \frac{x}{2} \right ) \, \mathrm{d}x \\ &=\int_0^\pi \left ( \ln 2 + \ln \sin \frac{x}{2} \right )^2 \, \mathrm{d}x \\ &= \int_{0}^{\pi} \ln^2 2 \, \mathrm{d}x + 2 \ln 2 \int_{0}^{\pi} \ln \sin \frac{x}{2} \, \mathrm{d}x + \\ &\quad \quad + \int_{0}^{\pi} \ln^2 \sin \frac{x}{2} \, \mathrm{d}x \\ &= \pi \ln^2 2 + 4 \ln 2 \int_{0}^{\pi/2} \ln \sin x \, \mathrm{d}x + 2\int_{0}^{\pi/2} \ln^2 \sin x \,\mathrm{d}x \\ &= \pi \ln^2 2 - 2 \pi \ln^2 2 + 2\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x \\ &= -\pi \ln^2 2 + 2\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x \end{align*}

Finally,

    \[\mathbf{\int_{0}^{\pi/2} \ln^2 \sin x \, \mathrm{d}x = \frac{\pi \ln^2 2}{2} + \frac{\pi^3}{24}}\]

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A parametric logarithmic integral

Let n \in \mathbb{N} \mid n>1. Prove that

    \[\int_{0}^{\infty} \frac{n^2 x^n \ln x}{1+x^{2n}} \, \mathrm{d}x = \frac{\pi^2}{4} \frac{\sin \frac{\pi}{2n}}{\cos^2 \frac{\pi}{2n}}\]

Solution

We are basing the whole solution on the Beta function and its derivative. We recall that

(1)   \begin{equation*} \mathrm{B}(x, y) = \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}} \, \mathrm{d}t \end{equation*}

Setting x=\frac{1}{2} + \frac{1}{2n} and y=\frac{1}{2} - \frac{1}{2n} back at (1) we get that

    \begin{align*} \mathrm{B}\left ( \frac{1}{2}+ \frac{1}{2n}, \frac{1}{2}- \frac{1}{2n} \right ) &= \Gamma \left ( \frac{1}{2}+ \frac{1}{2n} \right ) \Gamma \left ( \frac{1}{2}- \frac{1}{2n} \right )\\ &= \pi \sec \left ( \frac{\pi}{2n} \right ) \end{align*}

Differentiating (1) with respect to x we get that

    \begin{align*} \mathrm{B} '\left ( \frac{1}{2n}+\frac{1}{2}, \frac{1}{2} - \frac{1}{2n} \right ) & = \Gamma \left ( \frac{1}{2}+ \frac{1}{2n} \right ) \Gamma \left ( \frac{1}{2}- \frac{1}{2n} \right ) \bigg [ \psi^{(0)} \left ( \frac{1}{2}+ \frac{1}{2n} \right )- \\ &  \quad \quad \quad - \psi^{(0)} \left ( \frac{1}{2} - \frac{1}{2n} \right ) \bigg ]\\ &=\pi^2 \sec \left ( \frac{\pi}{2n} \right) \tan \left ( \frac{\pi}{2n} \right ) \end{align*}

where we made use of the reflection formulae of both the Gamma and the digamma function; \Gamma (x) \Gamma(1-x) = \pi \csc \pi x and \psi^{(0)} (1-x)- \psi^{(0)}(x) = \pi \cot \pi x.

Now for our integral we have successively:

    \begin{align*} \int_{0}^{\infty} \frac{ n^2 x^n \log x}{1+x^{2n}}\, {\rm d}x &\overset{u=x^n}{=\! =\! =\!} \int_{0}^{\infty} \frac{u^{1/n} \log u}{1+u^2} \, {\rm d}u \\ &\overset{y=u^2}{=\! =\! =\! =\!} \frac{1}{4}\int_{0}^{\infty} \frac{y^{1/2n -1/2} \log y}{1+y} \, {\rm d}y \\ &=\frac{1}{4}\pi^2 \sec \left ( \frac{\pi}{2n} \right ) \tan \left ( \frac{\pi}{2n} \right ) \\ &= \frac{\pi^2}{4} \frac{\sin \left ( \frac{\pi}{2n} \right )}{\cos^2 \left ( \frac{\pi}{2n} \right )} \end{align*}

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