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.

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.

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On a nested sin sequence

Consider the sequence x_n defined recursively as

    \[x_1=1 \quad, \quad x_{n+1}=\sin x_n\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt{n} x_n = \sqrt{3}.

Solution

Lemma: If a_n is a sequence for which \displaystyle \lim_{n\to+\infty}(a_{n+1}-a_n)=a then

    \[\lim_{n\to + \infty}\frac{a_n}n=a.\]

Proof: In Stolz theorem we set x_{n}=a_{n+1} and y_n=n.

It is easy to see that x_n is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives

    \[\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}\]

Therefore

    \[\lim_{n\to+\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)=\frac{1}{3}\]

Now, due to the lemma we have \lim\limits_{n\to+\infty} na_n^2 = 3 and the result follows.

Remark : The asymptotic now follows to be \displaystyle x_n \sim \sqrt{\frac{3}{n}}.

Problem: Find what inequality should \beta satisfy such that the series

    \[\mathcal{S}=\sum_{n=1}^{\infty} x_n^\beta\]

converges.

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An interesting limit

Let \displaystyle s_n=\sum_{k=1}^{\infty} \frac{1}{k(k+1)^n}. Prove that

    \[\lim_{n \rightarrow +\infty}  2^n s_n =1\]

Solution

We note that

    \[s_{n} - s_{n-1} = -\sum_{k=1}^{\infty} \frac{k}{k(k+1)^n}= 1- \zeta(n)\]

Thus,

    \begin{align*} 1 & = 2^n \cdot \frac{1}{2^n} \\ &\leq 2^n \left ( \zeta(n) -1 \right )\\ &=2^n \left ( \frac{1}{2^n} + \sum_{k=3}^{\infty} \frac{1}{k^n} \right ) \\ &\leq 2^n \left ( \frac{1}{2^n} + \int_{2}^{\infty} \frac{{\rm d}x}{x^n} \right ) \\ &= 1+ \frac{2}{n-1} \longrightarrow 1 \end{align*}

and the result follows from the sandwich theorem.

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Nested radical inequality

Let n \in \mathbb{N}. Prove that

    \[\sqrt{2\sqrt[3]{3\sqrt[4]{4\cdots \sqrt[n]{n}}}}<2\]

Solution

The LHS is equal to 2^{1/2}3^{1/6} \cdots n^{1/n!} which by AM – GM is less or equal to

    \[\left( \frac{\sum_{k=2}^n (k/k!)}{\sum_{k=2}^n (1/k!)}\right)^{\sum_{k=2}^n (1/k!)} = \left(1 + \frac{1}{a_n} \right)^{a_n}\]

where a_n=\sum \limits_{k=2}^{n} \frac{1}{k!}. Since a_n \nearrow e-2 <2 it follows from Bernoulli inequality that \displaystyle \left(1 + \frac{1}{a_n} \right)^{a_n} <2.

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No invertible matrices

Show that there do not exist invertible matrices A, B \in \mathcal{M}_n \left( \mathbb{C} \right) such that A^2+B^2 = ( A + B )^2 and A^3+B^3 = ( A + B)^3.

Solution

Suppose, on the contrary, that such matrices do exist. Then

    \[(A + B) ^2 = A^2 +B^2 \implies AB + BA = \mathbb{O} \implies AB = - BA\]

and also

    \[(A+B)^3 = A^3 + B^3 \implies A^2 B = - B^2 A\]

Using the fact that AB =-BA we deduce that

    \begin{align*} -BA^2 &= A^2B \\ &= A \cdot A B\\ &= -A B \cdot A\\ &= BA \cdot A \\ &= BA^2 \end{align*}

The last means that BA^2 =\mathbb{O} which is impossible because both A and B are invertible ( and so must be the product ). Hence, the conclusion follows.

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