A log – trigonometric integral

Prove that

    \[\int_0^\infty \frac{\ln \left(1+x^2 \right) \mathrm{arccot}x}{x} \; \mathrm{d}x = \frac{\pi^3}{12}\]

Solution

We state two lemmata:

Lemma 1: \displaystyle \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x =\frac{\pi^2}{24}

Proof:

Successively we have:

    \begin{align*} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{1}{x} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n} \, \mathrm{d}x \\ &=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^{2n-1} \, \mathrm{d}x \\ &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \\ &= \frac{\eta(2)}{2}\\ &= \frac{\left ( 1-2^{1-2} \right ) \zeta(2)}{2} \\ &=\frac{\left ( 1-\frac{1}{2} \right ) \zeta(2)}{2} \\ &=\frac{\zeta(2)}{4} \\ &= \frac{\pi^2}{24} \end{align*}

where \eta is the Dirichlet eta function and \zeta the Riemann zeta function.

Lemma 2: It holds that \displaystyle \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x = - \frac{\pi^3}{32}.

Proof:

Successively we have:

    \begin{align*} \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln x}{x} \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{2n+1} \, \mathrm{d}x \\ &=\int_{0}^{1} \ln x \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{2n+1} \, \mathrm{d}x \\ &=\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1}x^{2n} \ln x \, \mathrm{d}x \\ &=-\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\ &= -\beta(3) \\ &=-\frac{\pi^3}{32} \end{align*}

where \beta is the Dirichlet Beta function.

Hence,

    \begin{align*} \int_{0}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{1}^{\infty} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot} x}{x} \, \mathrm{d}x \\ &=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\ln \left ( 1+\frac{1}{x^2} \right ) \arctan x}{x} \, \mathrm{d}x \\ &=\int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \mathrm{arccot}x }{x} \, \mathrm{d}x + \\ &\quad \quad \quad + \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \arctan x}{x}\, \mathrm{d}x - \\ &\quad \quad \quad -\int_0^1 \frac{2 \ln x \arctan x}{x} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \mathrm{arccot} x + \arctan x \right )}{x} \, \mathrm{d}x - \\ & \quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right ) \left ( \arctan \frac{1}{x} + \arctan x \right )}{x} \, \mathrm{d}x - \\ &\quad \quad \quad - 2 \int_{0}^{1} \frac{\ln x \arctan x}{x} \, \mathrm{d}x \\ &=\frac{\pi}{2} \int_{0}^{1} \frac{\ln \left ( 1+x^2 \right )}{x} \, \mathrm{d}x + \frac{\pi^3}{16} \\ &= \frac{\pi^2}{24} \cdot \frac{\pi}{2} + \frac{\pi^3}{16} \\ &= \frac{\pi^3}{12} \end{align*}

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MacLaurin of exp(exp(x))

Let \mathcal{B}_n denote the n-th Bell number. Prove that

    \[\exp \left(\exp(x) \right) = e \sum_{n=0}^{\infty} \frac{\mathcal{B}_n}{n!} x^n\]

Solution

Taking derivatives we get that

    \[f^{(n)}(x)=\sum_{k=0}^n{n\brace k}\exp(e^x+kx)\]

where f(x)=\exp \left(\exp(x) \right) and n\brace k are the Stirling numbers of second kind. We also note that

    \[f^{(n)}(0)=e\sum_{k=0}^n{n\brace k}=e\mathcal{B}_n\]

where \mathcal{B}_n are the Bell numbers. Thus,

    \[exp(\exp(x))=e\sum_{n=0}^\infty\frac{\mathcal{B}_n}{n!}x^n\]

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A root limit

Let a_i \; , \; i =1, 2, \dots, k be positive real numbers such that a_1\geq a_2\geq \cdots \geq a_k. Prove that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

Solution

Without loss of generation , let a_1 =\max\{ a_1, a_2 , \dots , a_k \}. Then,

    \begin{align*} a_1 &=(a_1^n)^{1/n}\\ &\leq (a_1^n+\cdots+a_k^n)^{1/n}\\ &=\sqrt[n]{a_1^n \left (1+\left (\frac{a_2}{a_1} \right )^n + ... + \left (\frac{a_k}{a_1} \right )^n \right )} \\ &\leq \sqrt[n]{{a_1}^n \cdot k} \\ &= a_1\sqrt[n]{k} \\ &=a_1 k^{1/n} \end{align*}

since a_i \leq  a_1 forall i=1,2, \dots,k. Thus, by the squeeze theorem it follows that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} =a_1 =  \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

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A limit!

Evaluate the limit:

    \[\ell = \lim_{m\rightarrow +\infty}\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\]

Solution

Let a_m=\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}. Then,

    \[\log a_m = \log \sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}} = \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k}\]

It follows by Stolz–Cesàro that

    \begin{align*} \lim_{m \rightarrow +\infty} \log a_m &= \lim_{m \rightarrow +\infty} \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k} \\ &=\lim_{m \rightarrow +\infty} \frac{\sum \limits_{k=1}^{m+1} \log \binom{m+1}{k} - \sum \limits_{k=1}^{m} \log \binom{m}{k}}{(m+1)^2 + (m+1) -(m^2+m)} \\ &=\lim_{m \rightarrow +\infty} \frac{1}{2m+2} \sum_{k=1}^{m} \log \frac{m+1}{m+1-k} \\ &= \lim_{m \rightarrow +\infty} \frac{\log \frac{(m+1)^m}{m!}}{2m+2}\\ &=\lim_{m \rightarrow +\infty} \frac{1}{2} \cdot \frac{m}{m+1} \cdot \log \frac{m+1}{\sqrt[m]{m!}} \\ &= \frac{1}{2} \cdot 1 \cdot \log e \\ &= \frac{1}{2} \end{align*}

In addition,

    \begin{align*} \lim_{m \rightarrow +\infty} \sqrt[m]{m+1} &= \lim_{m \rightarrow +\infty} \left ( m+1 \right )^{1/m} \\ &=\lim_{m \rightarrow +\infty} e^{\frac{\log m}{m+1}} \\ &=1 \end{align*}

Hence \ell = \sqrt{e}.

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Floor series

Let \left \lfloor \cdot \right \rfloor denote the floor function. Evaluate the series

    \[\mathcal{S} = \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)}\]

Solution

First of all we note that 4k+3 and 4k+2 are never squares. Thus, there exists a positive integer m such that

    \[m^2 \leq \sqrt{4k+1} < \sqrt{4k+2}< \sqrt{4k+3} < \left ( m+1 \right )^2\]

It is easy to see that \sqrt{4x+1} \leq\sqrt{x} + \sqrt{x+1} < \sqrt{4x+3} and thus we conclude that

    \[\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor = \left \lfloor 4k+1 \right \rfloor\]

Now \left \lfloor 4k+1 \right \rfloor is equal to the even number 2n if-f

    \[\left ( 2n \right )^2 \leq 4k+1 < \left ( 2n+1 \right )^2 \Leftrightarrow k \in \left [ n^2, n^2+n \right )\]

Hence, since the series is absolutely convergent we can rearrange the terms and by noting that the finite sums are telescopic , we get that:

    \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{\left \lfloor \sqrt{k} + \sqrt{k+1} \right \rfloor}}{k(k+1)} &= \sum_{n=1}^{\infty} \bigg ( \sum_{k=n^2}^{n^2+n-1} \left ( \frac{1}{k} - \frac{1}{k+1} \right )- \\ &\quad \quad \quad  \quad \quad - \sum_{k=n^2+n}^{\left ( n+1 \right )^2-1}\left ( \frac{1}{k} - \frac{1}{k+1} \right ) \bigg )\\ &=\sum_{n=1}^{\infty} \left ( \frac{1}{n^2} -\frac{1}{n^2+n}-\frac{1}{n^2+n} +\frac{1}{(n+1)^2} \right ) \\ &=\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} - 2 \sum_{n=1}^{\infty} \frac{1}{n^2+n} \\ &=\frac{\pi^2}{6} + \frac{\pi^2}{6}-1 -2 \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \\ &=\frac{\pi^2}{3} -1-2 \\ &= \frac{\pi^2}{3}-3 \end{align*}

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