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Zeta logarithmic series

September 15, 2019 / Leave a comment

Let \zeta denote the zeta function. Prove that

    \[\sum_{k=1}^{\infty} \left( \log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1} \right)=0\]

Solution

    \begin{align*} \sum_{k=1}^{\infty} \left ( \frac{\zeta(k+1)}{k+1} - \frac{1}{k+1} \right ) &=\sum_{k= 2}^{\infty} \frac{\zeta(k)-1}{k}\\ &=\sum_{k =2}^{\infty} \sum_{n =2}^{\infty} \frac{1}{kn^k}\\ &=\sum_{n=2}^{\infty} \sum_{k= 2}^{\infty} \frac{1}{kn^k}\\ &=-\sum_{n=2}^{\infty} \left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)\\ &=-\sum_{n= 1}^{\infty} \left(\ln\left(1-\frac{1}{n+1}\right)+\frac{1}{n+1}\right) \\ &-\sum_{n =1}^{\infty} \left(\ln\left(\frac{n}{n+1}\right)+\frac{1}{n+1}\right)\\ &=\sum_{n= 1}^{\infty} \left(\ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right) \\ &=\sum_{n= 1}^{\infty} \left(\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right) \end{align*}

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Mellin transform integral

September 14, 2019 / Leave a comment

Evaluate the integral:

    \[\mathcal{J} = \int_0^\infty x \log x e^{-\sqrt{x}} \, \mathrm{d}x\]

Solution

We are evaluating the Mellin transform of the function f(x)=e^{-\sqrt{x}}.

    \begin{align*} \mathcal{M}\left ( f \right ) &= \int_{0}^{\infty} x^{s-1} e^{-\sqrt{x}} \, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\sqrt{x}}{=\! =\! =\! =\! =\!} 2\int_{0}^{\infty} u^{2s-1} e^{-u} \, \mathrm{d}u\\ &= 2 \Gamma\left ( 2s \right ) \end{align*}

where \Gamma is the Euler’s Gamma function. Hence,

    \begin{align*} \int_{0}^{\infty} x \log x e^{-\sqrt{x}} \, \mathrm{d}x &= \mathcal{M}'(f) \bigg|_{s=2} \\ &= \left ( 2 \Gamma(2s) \right )'\bigg|_{s=2}\\ &= 4 \Gamma(2s) \psi^{(0)}(2s) \bigg|_{s=2}\\ &=4 \Gamma(4) \psi^{(0)}(4) \\ &= 4 \cdot 6 \cdot \left ( \frac{11}{6} - \gamma \right ) \\ &=44 - 24 \gamma \end{align*}

where \gamma is the Euler – Mascheroni constant and \psi^{(0)} is the digamma.

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Contour integral

September 14, 2019 / Leave a comment

Let f be analytic in the disk |z|<2. Prove that:

    \[\frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\overline{f(z)}}{z-\alpha} \, \mathrm{d}z = \left\{\begin{matrix} \overline{f(0)} & , & \left | \alpha \right |<1 \\\\ \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} & , & \left | \alpha \right |>1 \end{matrix}\right.\]

Solution

It follows from Taylor that f(z)=\sum \limits_{n=0}^{\infty} c_n z^n and the convergence is uniform. Hence,

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\overline{c_n} \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &= \sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{ \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{\left | z \right |=1\Rightarrow \bar{z}^n = \frac{1}{z^n}}{=\! =\! =\! =\! =\! =\! =\!=\! =\!=\!}\sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{1}{z^n(z-\alpha)} \,\mathrm{d}z \end{align*}

We have that \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};\alpha\right) = \frac{1}{\alpha^n} and for n \geq 1 we also have that \displaystyle \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};0\right) = -\frac{1}{\alpha^n} due to

    \[\frac{1}{z^n(z-\alpha)} = -\frac{1}{\alpha z^n} \frac{1}{1-z/\alpha} = -\frac{1}{\alpha z^n}\left(1 + \frac{z}{\alpha} + \frac{z^2}{\alpha^2} + \cdots \right)\]

So if |\alpha|<1 then \alpha lies within the disk |z|=1; hence the integral equals \overline{c_0} = \overline{f(0)} whereas if |\alpha|>1 then \alpha lies outside the disk |z|=1; hence

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &= -\sum_{n=1}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{c_n} - \sum_{n=0}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{f(0)} - \overline{\left ( \sum_{n=0}^{\infty} \frac{c_n}{\bar{\alpha}^n} \right )}\\ &= \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} \end{align*}

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Functions that preserve convergent serieses

September 13, 2019 / Leave a comment

Find all functions f:\mathbb{R} \rightarrow \mathbb{R} that preserve convergent serieses.

Root inequality

September 12, 2019 / Leave a comment

Let a, b, c be positive real numbers such that a+b+c=3. Prove that

    \[\sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} \leq \frac{3}{2} \sqrt[4]{\frac{1}{abc}}\]

Solution

Due to the AM – GM we have that

(1)   \begin{equation*} a^2+3 \geq 4\sqrt{a} \end{equation*}

and

(2)   \begin{equation*} 3b + c \geq 4 \sqrt[4]{b^3c} \end{equation*}

Thus,

    \begin{align*} \sum \sqrt{\frac{b}{a^2+3}} &\leq \sqrt{\frac{b}{4\sqrt{a}}} + \sqrt{\frac{c}{4\sqrt{b}}} + \sqrt{\frac{a}{4\sqrt{c}}} \\ &=\frac{1}{2\sqrt[4]{abc}} \left ( \sqrt[4]{b^3c} + \sqrt[4]{c^3a} + \sqrt[4]{a^3b} \right ) \\ &\leq \frac{1}{2\sqrt[4]{abc}} \left ( \frac{3b+c}{4} + \frac{3c+a}{4} + \frac{3a+b}{4} \right ) \\ &=\frac{3}{2} \sqrt[4]{\frac{1}{abc}} \end{align*}

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