An integral with Gamma and digamma

Let \Gamma denote the Euler’s Gamma function and \psi denote the digamma function. Evaluate

\displaystyle \bigintsss_{0}^{2}\frac{\Gamma^2\left ( \frac{1}{t} \right )}{2t^3 \Gamma \left ( \frac{2}{t} \right )} \bigg[ t + 2\psi\left ( \frac{1}{t} \right) - 2 \psi \left ( \frac{2}{t} \right ) \bigg] \, {\rm d}t

Solution

We have successively

\begin{aligned} \int_{0}^{2}\frac{\Gamma^2\left ( \frac{1}{t} \right )}{2t^3 \Gamma \left ( \frac{2}{t} \right )} \bigg[ t + 2\psi\left ( \frac{1}{t} \right) - 2 \psi \left ( \frac{2}{t} \right ) \bigg]\, {\rm d}t & =\int_{0}^{2} \frac{\Gamma^2\left ( \frac{1}{t} \right )}{t^3 \Gamma\left ( \frac{2}{t} \right )} \bigg[ \frac{t}{2}+ \psi\left ( \frac{1}{t} \right ) - \psi\left ( \frac{2}{t} \right ) \bigg] \, {\rm d}t \\ &\!\!\overset{u=2/t}{=\! =\! =\!} \frac{1}{4}\int_{1}^{\infty} \frac{\Gamma \left ( \frac{u}{2} \right ) \Gamma \left ( \frac{u}{2} \right )}{\Gamma \left ( \frac{u}{2}+ \frac{u}{2} \right )} \bigg[ 1+ u \psi \left ( \frac{u}{2} \right ) - u \psi (u) \bigg] \, {\rm d}u\\ &= \frac{1}{4}\int_{1}^{\infty} {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \left ( 1+ u \psi \left ( \frac{u}{2} \right )- u \psi (u) \right ) \, {\rm d}u\\ &=\frac{1}{4}\int_{1}^{\infty} {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \bigg( 1+ u \left ( \psi \left ( \frac{u}{2} \right ) - \psi\left ( \frac{u}{2}+ \frac{u}{2} \right ) \right ) \bigg )\, {\rm d}u \\ &=\frac{1}{4}\int_{1}^{\infty } \bigg[{\rm B} \left ( \frac{u}{2}, \frac{u}{2} \right ) + u {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \left ( \psi\left ( \frac{u}{2} \right ) - \psi \left ( \frac{u}{2}+ \frac{u}{2} \right ) \right )\bigg]\, {\rm d}u \\ &=-\frac{1}{4}\int_{1}^{\infty} \left ( u {\rm B}\left ( \frac{u}{2}, \frac{u}{2} \right ) \right )'\, {\rm d}u \\ &= \frac{\pi}{4} \end{aligned}

The exercise can also be found at Aops.com .

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An inequality involving harmonic number

Let \mathcal{H}_n denote the n – th harmonic number and let  n \in \mathbb{N}. Prove that

\displaystyle \int_{0}^{1} \frac{{\rm d}x}{x^{n-1} + x^{n-2} + \cdots+x+1} \geq \frac{1}{\mathcal{H}_n}

Solution

We might begin with the integral representation of the harmonic number, namely the equation:

\displaystyle \mathcal{H}_n = \int_{0}^{1} \left ( 1+x+\cdots+x^{n-1} \right ) \, {\rm d}x = \sum_{k=1}^{n} \frac{1}{k}

So we have to prove the equivelant inequality

\displaystyle \int_{0}^{1}\left ( x^{n-1} + \cdots + x +1 \right ) \int_{0}^{1} \frac{{\rm d}x}{x^{n-1} + \cdots + x +1} \geq 1

and this is obvious using the Cauchy – Schwarz inequality.

The exercise can also be found here .

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A limit

Let \mathcal{H}_n denote the n -th harmonic number. Evaluate the limit

\displaystyle \lim_{n\rightarrow +\infty} \left [ e^{\mathcal{H}_n-\gamma} - n \right ]

Solution

We begin by the simple observation that

\displaystyle \gamma = \lim_{n \rightarrow +\infty} \left ( \mathcal{H}_n - \ln n \right )

and of course we have the series represantation of the \gamma constant, namely this:

\displaystyle \gamma = \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \ln \left ( 1 + \frac{1}{n} \right ) \right ]

Hence:

    \begin{align*} \mathcal{H}_n-\gamma&=\sum_{k=1}^n\ln\left(1+\frac{1}{k} \right)+ \\ & \; \quad \quad \quad +\sum_{k=n+1}^{\infty}\left[\ln\left(1+\frac{1}{k} \right)-\frac{1}{k} \right]\\ &=\ln(n+1)+\sum_{k=n+1}^{\infty}\left[\ln\left(1+\frac{1}{k} \right)-\frac{1}{k} \right]\\ &=\ln(n+1)-\frac{1}{2n}+\mathcal{O}(n^{-2}) \end{align*}

Thus

    \begin{align*} e^{H_n-\gamma}-n&=e^{\ln(n+1)-\frac{1}{2n}+\mathcal{O}(n^{-2})}-n\\ &=(n+1)e^{-\frac{1}{2n}+\mathcal{O}(n^{-2})}-n\\ &=(n+1)\left(1-\frac{1}{2n}+\mathcal{O}(n^{-2})\right)-n\\ &=n+\frac{1}{2}+\mathcal{O}(n^{-1})-n \end{align*}

We conclude that the desired limit is just \frac{1}{2}.

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A differential equation

Let f:\mathbb{R} \rightarrow \mathbb{R} be a differential equation such that

f'(x)=f^2(x)f(-x)

Find an explicit formula of f.

Solution

We consider the function g(x)=f(x)f(-x) which is clearly constant because

    \begin{align*} g'(x) &= f(x) f(-x) \\ &= f'(x) f(-x) - f(x) f'(-x)\\ &= \cancel{f^2(x) f^2(-x) - f^2(x) f^2(-x)}\\ &= 0 \end{align*}

Making use of the initial condition we get that g(x)=1. Thus f(x)f(-x)=1. This also means that f has no roots in the domain given. Hence the initial condition gives us f'(x)=f(x) and the function follows to be f(x)=e^x.

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Does the trigonometric series converge?

Examine if the series

\displaystyle \sum_{n=1}^{\infty} \sin \left[ \pi \left( 2+\sqrt{3} \right)^n \right]

converges.

Solution

The key lies in the fact that the number

\alpha = \left ( 2 + \sqrt{3} \right )^n +\left ( 2-\sqrt{3} \right )^n

is an integer. Let  x \in \mathbb{R} and let d(x) be the distance from x to the nearest integer. That is

d(x) = \left\{\begin{matrix} x & , & x \in \left [ 0, \frac{1}{2} \right ] \\\\ 1-x& , & x \in \left [ \frac{1}{2}, 1 \right ] \end{matrix}\right.

It is immediate that d can be expanded periodically with period T=1. Since \alpha is an integer we can get that

\displaystyle d\left [ \left ( 2 + \sqrt{3} \right )^n \right ] = \left ( 2-\sqrt{3} \right )^n = \frac{1}{\left ( 2+\sqrt{3} \right )^n}

Thus:

    \begin{align*} \left |\sum_{n=1}^{\infty} \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | &\leq \sum_{n=1}^{\infty} \left | \sin \left [ \pi \left ( 2+\sqrt{3} \right )^n \right ] \right | \\ &=\sum_{n=1}^{\infty} \sin \left [ \pi \; d\left ( 2+\sqrt{3} \right )^n \right ] \\ &\leq \pi \sum_{n=1}^{\infty} \frac{1}{\left ( 2+\sqrt{3} \right )^n} \end{align*}

and the conclusion follows.

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