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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]


We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}


    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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Complex inequality

Prove that forall z \in \mathbb{C} and n \in \mathbb{N} it holds that

    \[\left | \left ( 1+z \right )^n -1 \right | \leq \left ( 1+\left | z \right | \right )^n -1\]



    \begin{align*} \left | \left ( 1+z \right )^n -1 \right | &= \left | \sum_{k=0}^{n} \binom{n}{k} z^k -1 \right | \\ &= \left | \sum_{k=1}^{n} \binom{n}{k} z^k\right |\\ &\leq \sum_{k=1}^{n} \binom{n}{k} \left | z^k \right | \\ &=\sum_{k=0}^{n} \binom{n}{k} \left | z \right |^n -1 \\ &= \left ( 1+\left | z \right | \right )^n -1 \end{align*}


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Value of parameter

This is a very classic exercise and can be dealt with various ways. We know the result in advance. Why? Because it is the Taylor Polynomial of the exponential function. Let us see however how we gonna deal with it with High School Methods.

Find the positive real number \alpha such that

    \[e^x \geq 1+x+\frac{x^2}{2} + \alpha x^3 \quad  \quad \text{forall} \;\; x \in \mathbb{R}\]


Define the function

    \[f(x) = \frac{1+x+\frac{x^2}{2}+\alpha x^3}{e^x} \quad , \quad x \in \mathbb{R}\]

and note that f(x) \leq 1 forall x \in \mathbb{R}. Clearly , f is differentiable and its derivative is given by

    \[f'(x)= \frac{x^2\left ( 6\alpha-1-2\alpha x \right )}{e^x} \quad , \quad x \in \mathbb{R}\]

It follows that \displaystyle f'(x) =0 \Leftrightarrow \left\{\begin{matrix} x &= &0 \\\\ x & = & \dfrac{6\alpha-1}{2\alpha} \end{matrix}\right.. Suppose that \frac{6\alpha-1}{2\alpha}>0. Then the monotony of f as well as the sign of f' is seen at the following table.

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It follows then that \displaystyle f\left ( \frac{6\alpha-1}{2\alpha} \right )> f(0)=1. This is an obscurity due to the fact that f(x) \leq 1. Similarly, if we suppose that \frac{6\alpha-1}{2\alpha}<0. Hence

    \[\frac{6\alpha-1}{2\alpha}=0 \Leftrightarrow \alpha=\frac{1}{6}\]

For \alpha=\frac{1}{6} we easily see that the given inequality holds.

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Inequality of a convex function

Let f:[0, 2] \rightarrow \mathbb{R} be a twice differentiable function such that f''(x)>0 forall x \in [0, 2]. Prove that:

    \[\int_0^2 f(x) \, \mathrm{d}x > 2 f(1)\]


The equation of the tangent of \mathcal{C}_f at x_0=1 is

    \[\left ( \varepsilon \right ): y - f(1) = f'(1) \left ( x-1 \right ) \Leftrightarrow y = f'(x) \left ( x-1 \right ) + f(1)\]

Since f''(x)>0 forall x \in [0, 2] we deduce that f is strictly convex. Thus, the tangent lies below the graph of f except at x_0=1. Hence,

    \begin{align*} f(x) \geq f'(1)\left ( x-1 \right ) + f(1) &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > \int_{0}^{2} f'(1) \left ( x-1 \right ) +f(1) \int_{0}^{2} \, \mathrm{d}x \\ &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > \cancelto{0}{f'(1) \left [ \frac{x^2}{2} - x \right ]_0^2} + 2f(1) \\ &\Rightarrow \int_{0}^{2} f(x) \, \mathrm{d}x > 2f(1) \end{align*}

and the conclusion follows.

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Logarithmic Gaussian integral

Let \gamma denote the Euler’s constant. Prove that:

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = -\frac{\sqrt{\pi}}{4} \left ( 2 \log 2 + \gamma \right )\]


First of all by making the substitution x \mapsto \sqrt{x} we get that

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = \frac{1}{4} \int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x\]

For t>-1 let us consider the function

    \[F(t) = \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}x = \Gamma\left ( t+1 \right )\]

where \Gamma is the Euler’s Gamma function. Differentiating once we get:

    \begin{align*} \Gamma'(t+1) &=\frac{\mathrm{d} }{\mathrm{d} t} \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}t \\ &=\int_{0}^{\infty} \frac{\partial }{\partial t} x^t e^{-x}\; \mathrm{d}t \\ &=\int_{0}^{\infty} x^t \log x e^{-x} \; \mathrm{d}x \end{align*}

Thus, the desired integral is obtained by setting t=-\frac{1}{2}. Hence,

    \begin{align*} \frac{1}{4}\int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x &= \frac{1}{4} \cdot \Gamma' \left ( 1-\frac{1}{2} \right )\\ &= \frac{1}{4} \cdot \Gamma' \left ( \frac{1}{2} \right )\\ &=\frac{1}{4} \cdot \psi^{(0)} \left ( \frac{1}{2} \right ) \Gamma \left ( \frac{1}{2} \right ) \\ &= \frac{1}{4} \cdot \left ( -2\ln 2 -\gamma \right ) \cdot \sqrt{\pi} \\ &= -\frac{\sqrt{\pi}}{4} \left ( 2\ln 2+\gamma \right ) \end{align*}

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