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Integral of Jacobi Theta function

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

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

Solution

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*}

Hence,

    \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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Logarithmic Gaussian integral

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

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

Solution

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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Series of zeta sum

Let \zeta denote the Riemann zeta function. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( n - \sum_{k=2}^{n} \zeta(k) \right)\]

Solution

Let a \in \mathbb{R} \mid \left| a \right|<2. We are proving the more general result.

    \[\sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) = a \left ( \frac{\psi^{(0)}\left ( 2-a \right ) + \gamma}{1-a} + 1 \right )\]

where \psi^{(0)} denotes the digamma function.

First of all, we note that:

    \begin{align*} \sum_{k=2}^{n} \zeta(k) &= \sum_{k=2}^{n} \sum_{m=1}^{\infty} \frac{1}{m^k} \\ &= \sum_{m=1}^{\infty} \sum_{k=2}^{n} \frac{1}{m^k}\\ &=\sum_{m=1}^{\infty} \sum_{k=0}^{n-2} \frac{1}{m^{k+2}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} \\ &=\sum_{m=1}^{1} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} + \sum_{m=2}^{\infty} \frac{1}{m^2}\sum_{k=0}^{n-2} \frac{1}{m^k} \\ &= n-1 + \sum_{m=2}^{\infty} \frac{1}{m^2} \left ( \frac{1-\frac{1}{m^{n-1}}}{1-\frac{1}{m}} \right ) \\ &= n-1 +\sum_{m=2}^{\infty} \frac{m^{n-1}-1}{\left ( m-1 \right ) m^n} \\ & = n -1 + \sum_{m=1}^{\infty} \frac{1}{m(m+1)} - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \\ &= n - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \end{align*}

Thus,

    \begin{align*} \sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) &= \sum_{m=2}^{\infty} \frac{1}{m-1}\sum_{n=2}^{\infty} \left ( \frac{a}{m} \right )^n \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1) m \left ( m-a \right )} \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{m-a} \left ( \frac{1}{m-1} - \frac{1}{m} \right ) \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1)(m-a)} -a^2 \sum_{m=2}^{\infty} \frac{1}{m(m-a)} \\ &=a^2 \sum_{m=1}^{\infty} \frac{1}{m(m+1-a)} - a^2 \sum_{m=1}^{\infty} \frac{1}{m(m-a)} + \\ & \quad \quad \quad \quad \quad + \frac{a^2}{1-a} \\ &=\frac{a^2}{1-a} \left ( \psi^{(0)} (2-a) + \gamma \right ) +\\ & \quad \quad \quad \quad \quad + a \left ( \psi^{(0)} (1-a) + \gamma \right )+\frac{a^2}{1-a}\\ &= a \left ( \frac{\psi^{(0)}(2-a) + \gamma}{1-a} + 1 \right ) \end{align*}

due to the reflection formula \displaystyle \psi^{(0)} (z+1) = \psi^{(0)}(z) + \frac{1}{z}.

Side note: If a=1 then the sum equals \zeta(2) - 1.

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On a log Gamma integral using Riemann sums

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \log \Gamma (x) \, \mathrm{d}x\]

using Riemann sums.

Solution

Partition the interval [0, 1] into n subintervals of length \frac{1}{n}. This produces,

(1)   \begin{equation*}  \int_{0}^{1} \log \Gamma(x) \, \mathrm{d}x = \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) \end{equation*}

On the other hand, assuming n is even:

    \begin{align*} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) &= \frac{1}{n} \log \prod_{k=1}^{n} \Gamma \left ( \frac{k}{n} \right ) \\ &= \frac{1}{n} \log \prod_{k=1}^{n/2} \Gamma \left ( \frac{k}{n} \right ) \Gamma \left ( 1 - \frac{k}{n} \right )\\ &=\frac{1}{n} \log \prod_{k=1}^{n/2} \frac{\pi}{\sin \frac{\pi k}{n}} \\ &= \log \sqrt{\pi} - \log \left ( \prod_{k=1}^{n} \sin \frac{\pi k}{n} \right )^{1/n}\\ &= \log \frac{\sqrt{2 \pi}}{\left ( 2n \right )^{1/2n}} \end{align*}

since it holds that

    \[\prod_{k=1}^{n} \sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}\]

Euler’s Gamma reflection formula was used at line (3). Letting n \rightarrow +\infty we get that

    \[\int_0^1 \log \Gamma(x) \, \mathrm{d}x = \log \sqrt{2\pi}\]

If n is odd we work similarly.

 

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A beautiful limit with harmonic number

Let \mathcal{H}_n denote the n – th harmonic number. Prove that

    \[\lim_{n \rightarrow  +\infty} \left(\mathcal{H}_n - \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} \mathcal{H}_k \right) = \log 2\]

Solution

We note that

    \begin{align*} \frac{1}{2^n} \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} \frac{1}{t} \right] &= \frac{1}{2^n} \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} \int_{0}^1 x^{t-1} \mathrm{d}x \right] \\ &=\frac{1}{2^n} \int_{0}^1 \sum\limits_{k=0}^n \left[ \binom{n}{k} \sum\limits_{t=1}^{k} x^{t-1} \right] \;\mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \sum\limits_{k=0}^n \left[ \binom{n}{k} \cdot \frac{x^k-1}{x-1} \right]\; \mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \frac{\sum\limits_{k=0}^n \binom{n}{k} x^k- \sum\limits_{k=0}^n \binom{n}{k}}{x-1} \; \mathrm{d}x \\ &=\frac{1}{2^n} \int_{0}^1 \frac{(x+1)^n- 2^n}{x-1} \; \mathrm{d}x \end{align*}

At the last integral we apply the substitution y = \frac{x+1}{2}. Thus,

    \begin{align*} \frac{1}{2^n} \int_{0}^1 \frac{(x+1)^n- 2^n}{x-1}\; \mathrm{d}x &=\int_{1/2}^{1} \frac{y^n-1}{y-1} \, \mathrm{d}y \\ &= \int_{1/2}^{1} \left ( 1+ y +y^2 + \cdots + y^{n-1} \right ) \; \mathrm{d}y \\ &= \mathcal{H}_n - \sum_{i=1}^{n} \frac{1}{i} \left ( \frac{1}{2} \right )^i \end{align*}

After all these we have that

    \begin{align*} \lim_{n \rightarrow +\infty} \left(\mathcal{H}_n - \frac{1}{2^n} \sum_{k=1}^n \binom{n}{k} \mathcal{H}_k \right) &= \lim_{n \rightarrow +\infty} \left [\cancel{ \mathcal{H}_n - \mathcal{H}_n} + \sum_{i=1}^{n} \frac{1}{i} \left ( \frac{1}{2} \right )^i \right ] \\ &= \sum_{i=1}^{\infty} \frac{1}{i} \left ( \frac{1}{2} \right )^i\\ &= \log 2 \end{align*}

proving the result.

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