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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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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A curious logarithmic and trigonometric integral

Let \gamma denote the Euler – Mascheroni constant. Prove that

    \[\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\ln x {\rm d}x=\frac{12\gamma^2-\pi^2}{ 2(4n)^2}\]


Let us consider the following

    \[\mathcal{J}\left ( \alpha \right )=\int_{0}^{\infty }x^{\alpha-1}\left ( \cos x-\cos x^{2} \right )\mathrm{d}x\]

It is known that \mathcal{J}(0) = -\frac{\gamma}{2} . Also, the Mellin transform of \cos x^b is known to be

\displaystyle\int_{0}^{\infty}x^{a-1}\cos\left(x^{b}\right)\,\mathrm{d}x=\frac{1}{b}\cos\left(\frac{\pi a}{2b}\right)\Gamma \left(\frac{a}{b} \right), \;\ b>a>1


\displaystyle \mathcal{J}\left ( \alpha \right )=\Gamma \left ( a \right )\cos\left ( \frac{a\pi }{2} \right )-\frac{1}{2}\Gamma \left ( \frac{a}{2} \right )\cos\left ( \frac{a\pi }{4} \right )

On the other hand

\displaystyle \Gamma \left ( x \right )\sim \frac{1}{x}-\gamma +\frac{6\gamma ^{2}+\pi ^{2}}{12}x+o(x)

What we are seeking is \mathcal{J}'(0).

\begin{aligned} \mathcal{J}'(0) &= \lim_{a\rightarrow 0} \frac{\mathcal{J}(a) - \mathcal{J}(0)}{a -0} \\ &= \lim_{a\rightarrow 0} \frac{-\frac{1}{2} \cos \left ( \frac{a \pi}{4} \right ) \Gamma \left ( \frac{a}{2} \right )+ \cos \left ( \frac{a \pi}{2} \right ) \Gamma(a) + \frac{\gamma}{2}}{a} \\ &=\frac{3 \gamma^2}{8} - \frac{\pi^2}{32} \end{aligned}

and the result follows.

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On a classic logarithmic computation

Let \mathbb{R} \ni a>1 . Find a closed form for the integral

    \[\mathcal{J} = \int_1^{a^2} \frac{\ln x}{\sqrt{x} (x+a)} \, {\rm d}x\]


We are invoking a powerful theorem stating that:

Theorem: Let {\rm P}(x) = {\rm A}x^2 + {\rm B} x + {\rm C} be a second degree polynomial (this implies {\rm A} \neq 0 ) . Then it holds that

    \[\int_a^b \frac{\log (dx + c)}{{\rm P}(x)} \, {\rm d}x = \frac{\log K}{2} \int_a^b \frac{{\rm d}x}{{\rm P}(x)}\]

where K= (ad +c) (db +c).

Making use of the above theorem we have that

\begin{aligned} \int_{1}^{a^2}\frac{\ln x}{\sqrt{x}\left ( x+a \right )}\, {\rm d}x&\overset{u=\sqrt{x}}{=\! =\! =\! =\!} \int_{1}^{a}\frac{2u \ln u^2}{u(u^2+a)} \, {\rm d}u \\ &=4\int_{1}^{a}\frac{\ln u}{u^2+a}\, {\rm d}u \\ &=2 \ln a \int_{1}^{a}\frac{{\rm d}u}{u^2 +a} \\ &=2\ln a \left [ \frac{\arctan \frac{u}{\sqrt{a}}}{\sqrt{a}} \right ]_1^a \\ &= 2\ln a \left [ \frac{\arctan \sqrt{a}-\arctan \frac{1}{\sqrt{a}}}{\sqrt{a}} \right ] \\ &= -\ln a \left [ \frac{\pi -4 \arctan \sqrt{a}}{\sqrt{a}} \right ] \end{aligned}

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