Factorial series

April 26, 2020 / 1 Comment on Factorial series

Evaluate the series

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{1}{(3n)!}\]

Solution

Let \delta_{n, k} denote Kronecker’s delta and \zeta_m = e^{2\pi m i /3}. We have successively,

    \begin{align*} \sum_{n=0}^{\infty} \frac{1}{(3n)!} &=\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\delta_{k, 3n}}{k!} \\ &=\frac{1}{2\pi i } \sum_{n=0}^{\infty} \sum_{k=0}^{\infty}\; \oint \limits_{\left | z \right |=R>1} z^{3n-k+1} \, \mathrm{d} z\\ &= \frac{1}{2\pi i }\oint \limits_{ \left | z \right |=R>1} \frac{1}{z} \sum_{n=0}^{\infty} \frac{1}{z^{3n}} \sum_{k=0}^{\infty} \frac{z^k}{k!} \, \mathrm{d} z\\ &= \frac{1}{2\pi i} \oint \limits_{\left | z \right |=R>1} \frac{e^z}{z \left ( 1-\frac{1}{z^3} \right )}\, \mathrm{d} z\\ &= \frac{1}{2\pi i } \oint \limits_{\left | z \right |=R>1} \frac{e^z z^2}{z^3-1} \, \mathrm{d}z \\ &=\sum_{m=-1}^{1}\frac{\zeta_m^2 e^{\zeta_m}}{3\zeta_m^2} \end{align*}

It follows that

    \[\mathcal{S} =\frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}\]

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Contour radical integral

April 26, 2020 / Leave a comment

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

April 17, 2020 / 2 Comments on 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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Otto Dunkel Memomorial

April 17, 2020 / Leave a comment

Prove that

    \[\sum_{n=1}^{\infty}\frac{1}{n}\int_{2n\pi}^{\infty}\frac{\sin z}{z}\,\mathrm{d}z=\pi-\frac{\pi \ln 2 \pi}{2}\]

Solution

We have successively:

    \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \int_{2\pi n}^{\infty} \frac{\sin z}{z} \,\mathrm{d}z &= \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{\sin nt}{t+2\pi} \, \mahrm{d}t \\ &=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \int_{0}^{2\pi} \frac{\sin nt}{t + 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \sum_{n=1}^{\infty} \frac{1}{n} \frac{\sin nt}{t+ 2 m \pi} \, \mathrm{d}t \\ &= \sum_{m=1}^{\infty} \int_{0}^{2\pi} \frac{\pi -t}{2\left (t + 2 m\pi \right )} \, \mathrm{d}t\\ &= \pi \sum_{n=1}^{\infty} \left [ \left ( 1 + \frac{1}{2} \right ) \log \left ( 1 + \frac{1}{n} \right ) -1 \right ] \\ &=\pi \log \left [ \lim_{N \rightarrow +\infty} e^{-N} \prod_{n=1}^{N} \left ( \frac{n+1}{n} \right )^{n+1/2} \right ] \\ &= \pi \log \left ( \lim_{N \rightarrow +\infty} \frac{\sqrt{N+1}\left ( N+1 \right )^N e^{-N}}{N!} \right ) \\ &= \pi \log \left ( \frac{e}{\sqrt{2\pi}} \right ) \\ &= \pi - \frac{ \pi \log 2\pi}{2} \end{align*}

since for x \in (0, 2\pi) it holds that

    \[\sum_{n=1}^{\infty} \frac{\sin nt}{n} = \frac{\pi-t}{2}\]

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A squared trigamma series

April 17, 2020 / Leave a comment

Prove that

    \[\sum_{n=1}^{\infty} \left( \psi^{(1)} (n) \right)^2 = 3 \zeta(3)\]

Solution

Since \displaystyle \psi^{(1)}(n) = \sum_{k=n}^{\infty} \frac{1}{k^2} we have successively:

    \begin{align*} \sum_{n=1}^\infty \left( \psi^{(1)}(n) \right)^2 &=\sum_{n=1}^\infty\sum_{j=n}^\infty\frac1{j^2}\sum_{k=n}^\infty\frac1{k^2} \\ &=\sum_{n=1}^\infty\left(\sum_{j=n}^\infty\frac1{j^4}+2\sum_{j=n}^\infty\sum_{m=1}^\infty\frac1{j^2}\frac1{(j+m)^2}\right) \\ &=\sum_{j=1}^\infty\sum_{n=1}^j\frac1{j^4}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\sum_{n=1}^j\frac1{j^2}\frac1{(j+m)^2} \\ &=\sum_{j=1}^\infty\frac1{j^3}+2\sum_{j=1}^\infty\sum_{m=1}^\infty\frac1{j(j+m)^2} \\ &=\zeta(3)+2\sum_{n=1}^\infty\frac{\mathcal{H}_{n-1}}{n^2} \\ &=\zeta(3)-2\zeta(3)+2\sum_{n=1}^\infty\frac{\mathcal{H}_n}{n^2} \\ &=\zeta(3)-2\zeta(3)+4\zeta(3) \\ &=3\zeta(3) \end{align*}

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