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A logarithmic integral

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \left(x \ln x \right)^{2020} \, \mathrm{d}x\]

Solution

Recall the idenity

    \[\mathbf{\ln x = \lim_{n \rightarrow +\infty} n \left ( x^{1/n} - 1 \right )}\]

thus,

    \begin{align*} \int_{0}^{1} \left ( x \ln x \right )^{2020}\, \mathrm{d}x &= \lim_{n \rightarrow +\infty} n^{2020} \int_{0}^{1} \left ( x^{2020} \left ( x^{1/n} -1 \right )^{2020} \right )\, \mathrm{d}x \\ &\!\!\!\!\!\!\overset{u=x^{1/n}}{=\! =\! =\! =\!} \lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2020} \, \mathrm{d}u \\ &=\lim_{n \rightarrow +\infty} n^{2021} \int_{0}^{1} u^{2021n-1} \left ( 1-u \right )^{2021-1} \, \mathrm{d}u \\ &=\lim_{n \rightarrow +\infty} n^{2021} \mathrm{B} \left ( 2021n, 2021 \right ) \\ &=\lim_{n \rightarrow +\infty} n^{2021} \; \frac{\Gamma \left ( 2021 n \right ) \Gamma \left ( 2021 \right )}{\Gamma \left ( 2021 n + 2021 \right )} \\ &= \Gamma \left ( 2021 \right ) \lim_{n \rightarrow +\infty} n^{2021} \frac{\Gamma \left ( 2021 n \right )}{\Gamma \left ( 2021 n + 2021 \right )} \end{align*}

Using Gautschi’s Inequality it follows that

\displaystyle n^{2021}\left ( 2021n -1 \right )^{1-2022}<\frac{ n^{2021} \Gamma\left ( 2021n -1 + 1 \right )}{\Gamma\left ( 2021n-1 + 2022 \right )}< n^{2021}\left ( 2021n\right )^{1-2022}

and hence the integral equals

    \[\mathcal{J} = \frac{\Gamma(2021)}{2021^{2021}}\]

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On the factorial

Let \mu denote the Möbius function and \left \lfloor \cdot \right \rfloor denote the floor function. Prove that:

    \[n! = \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)}\]

Solution

The RHS equals

    \begin{align*} \prod_{j=1}^{\infty} \prod_{i=1}^{\infty} \left ( \left \lfloor \frac{n}{ij} \right \rfloor! \right )^{\mu(i)} &= \prod_{k=1}^n \prod_{i|k} \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\mu(i)} \\ &= \prod_{k=1}^n \left( \left \lfloor \frac{n}{k}\right\rfloor!\right)^{\sum_{i|k}\mu(i)} \\ &= n! \end{align*}

since \sum \limits_{i | k} \mu(i) = 0 for k>1.

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Sum over all positive rationals

For a rational number x that equals \frac{a}{b} in lowest terms , let f(x)=ab. Prove that:

    \[\sum_{x \in \mathbb{Q}^+} \frac{1}{f^2(x)} = \frac{5}{2}\]

Solution

First of all we note that

    \[F(s) = \sum_{x \in \mathbb{Q}^+} \frac{1}{f^s(x)} = \sum_{\substack{a,b=1 \\ \gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s}\]

Moreover for s>1 we have that

    \begin{align*} \zeta^2(s) &= \left ( \sum_{a=1}^{\infty} \frac{1}{a^s} \right )^2 \\ &=\sum_{a, b=1}^{\infty} \frac{1}{(ab)^s} \\ &=\sum_{d=1}^{\infty} \sum_{\substack{a, b=1 \\\gcd(a, b)=d}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^{2s}} \sum_{\substack{a, b=1 \\\gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \zeta(2s) F(s) \end{align*}

Hence for s=2 we have that

    \[F(2) = \frac{5}{2}\]

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Bessel series

Let J_n denote the Bessel function of the first kind. Prove that:

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]

Solution

The Jacobi – Anger expansion tells us that

    \[e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}\]

Hence by Parseval’s Theorem it follows that:

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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On the dubious function

The dubious function \Delta:\mathbb{N} \rightarrow \mathbb{N} is defined as follows : \Delta(1)=1 and

    \[\mathbf{\Delta(n) = \sum_{\substack{\mathbf{d \mid n } \\ \mathbf{d \neq n} }} \Delta (d)} \quad \text{for} \;\; n>1\]

Evaluate the sum

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\Delta\left ( 15^n \right )}{15^n}\]

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