Joy of Mathematics

Trigonometry with golden ratio

March 4, 2023 Tolaso

Let $\varphi$ denote the golden ratio. Prove that

$4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} = \frac{\pi}{2}$

Solution

Since $\varphi^2 = \varphi + 1$ we deduce that $\varphi^3 = \varphi^2 + \varphi = 2 \varphi + 1$ . Furthermore,

$\varphi^6 - 1 = \left ( \varphi^3 \right )^2 -1 = \left ( 2 \varphi + 1 \right )^2 -1 = 4\varphi^2 + 4 \varphi = 4 \varphi^3$

Also, taking into consideration that $\displaystyle 2 \arctan x = \arctan \frac{2x}{1-x^2} \; ,\; x^2<1$ we deduce that:

$\begin{align*} 4 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{\varphi^6 - 1}} &= 2 \cdot 2 \arctan \frac{1}{\sqrt{\varphi^3}} - \arctan \frac{1}{\sqrt{4 \varphi^3}} \\ &= 2 \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \sqrt{\varphi} - \arctan \frac{1}{2 \sqrt{\varphi^3}} \\ &= \arctan \sqrt{\varphi} + \arctan \frac{1}{\sqrt{\varphi}} \\ &= \frac{\pi}{2} \end{align*}$

In the last step we made use of the identity $\displaystyle \arctan \alpha + \arctan \beta = \arctan \frac{x+y}{1-xy} \;,\; xy<1$ .

Arctan integral

July 25, 2020July 25, 2020 Tolaso

Prove that

$\int_0^{\infty}\frac{\arctan x -x e^{-x}}{x^2}\,\mathrm{d}x=1+\gamma$

where $\gamma$ denotes the Euler – Mascheroni constant.

Solution

Beginning by parts we have,

$\begin{align*} \int_{0}^{\infty} \frac{\arctan x- xe^{-x}}{x^2} \, \mathrm{d}x &= \left [ -\frac{\arctan x - xe^{-x}}{x} \right ]_0^\infty + \\ &\quad \quad \quad + \int_{0}^{\infty} \frac{1}{x} \left ( \frac{1}{x^2+1} +e^{-x} (x-1) \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x\left ( x^2+1 \right )} + \frac{e^{-x}(x-1)}{x} \right ) \, \mathrm{d}x \\ &=\int_{0}^{\infty} \left ( \frac{1}{x} - \frac{x}{x^2+1} + e^{-x} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x \end{align*}$

However,

$\begin{aligned} \int_{\epsilon}^{M} \left ( \frac{1}{x} - \frac{x}{x^2+1} - \frac{e^{-x}}{x} \right ) \, \mathrm{d}x &= \left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \int_{\epsilon}^{M} \frac{e^{-x}}{x} \, \mathrm{d}x \\ &=\left ( \ln M - \ln \epsilon - \frac{\ln \left ( M^2+1 \right ) - \ln \left ( \epsilon^2+1 \right ) }{2} \right ) - \\ &\quad \quad \quad - \left [ \ln x e^{-x} \right ]_\epsilon^{M} - \int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \\ &=\left ( \ln M - \frac{\ln \left ( M^2+1 \right )}{2} \right ) + \frac{\ln \left ( \epsilon^2+1 \right )}{2} - \\ &\quad \quad \quad \quad -\ln M e^{-M} + \left (\ln \epsilon \; e^{-\epsilon} - \ln \epsilon \right ) - \\ &\quad \quad \quad \quad -\int_{\epsilon}^{M} e^{-x} \ln x \, \mathrm{d}x \end{aligned}$

The result now follows taking $\epsilon \rightarrow 0^+$ and $M \rightarrow +\infty$ .

An Euler’s gamma constant limit

June 30, 2020July 11, 2020 Tolaso

Let $\gamma_n = \mathcal{H}_n - \ln n$ . Evaluate the limit

$\ell = \lim_{n \rightarrow +\infty} \left( \sin \gamma_n - \sin \gamma \right) \sqrt[n]{n!}$

Solution

Multiplying up and bottom with $\gamma_n-\gamma$ and using the facts that

(1) $\begin{equation*} \gamma_n -\gamma \sim \frac{1}{2n} \end{equation*}$

(2) $\begin{equation*} \sqrt[n]{n!} \sim \frac{n\sqrt[2n]{2\pi n}}{e} \end{equation*}$

(3) $\begin{equation*} \frac{\sin \gamma_n - \sin \gamma}{\gamma_n - \gamma} \rightarrow \cos \gamma \end{equation*}$

Hence,

$\lim_{n \rightarrow +\infty} \left( \sin \gamma_n - \sin \gamma \right) \sqrt[n]{n!} = \frac{\cos \gamma}{2e}$

Asymptotic expansion

June 28, 2020 Tolaso

Prove that

$\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i^2+j^2} = \frac{\pi \log n}{2} + \frac{\pi+4}{4n} + c + \mathcal{O} \left ( \frac{1}{n^2} \right )$

for some constant $c$ .

Logarithmic – Trigonometric integral

April 16, 2020April 16, 2020 Tolaso

Let $\gamma$ denote Euler – Mascheroni’s constant. Prove that

$\int_{0}^{\infty} \frac{\ln x \left ( 1 - \cos t \right )}{t^2} \, \mathrm{d}t = \frac{\pi}{2} \left ( 1 - \gamma \right )$

Solution

We are applying a powerful technique; using Laplace and Inverse Laplace transformation. Hence,

$\begin{align*} \int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t^2} \, {\rm d}t &= \int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t \cdot t} \, {\rm d}t \\ &= -\frac{1}{2}\int_{0}^{\infty} \left ( \log x + \gamma \right ) \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x \\ &=- \frac{1}{2} \int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ & \quad \quad \quad -\frac{1}{2} \gamma \int_{0}^{\infty} \log \left ( \frac{1}{x^2} +1 \right ) \, {\rm d}x \\ &= \frac{\pi}{2} - \frac{\pi \gamma}{2}\\ &= \frac{\pi}{2} \left ( 1-\gamma \right ) \end{align*}$

since

$\begin{align*} \int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x + \\ &\quad \quad \quad + \int_{1}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x\\ &\!\!\!\!\!\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{0}^{1}\log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ &\quad \quad \quad - \int_{1}^{0} \log \left ( \frac{1}{x} \right ) \log \left ( x^2+1 \right ) \frac{1}{x^2} \, {\rm d}x \\ &=\int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \\ &\quad \quad \quad - \int_{0}^{1} \frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x \\ &=-2\mathcal{G} - \frac{\pi}{2} - \log 2 - \left ( -2\mathcal{G} + \frac{\pi}{2} - \log 2\right )\\ &=- \pi \cancel{- 2\mathcal{G} - \log 2 + 2\mathcal{G} + \log 2} \\ &= -\pi \end{align*}$

The integrals $\displaystyle \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x$ , $\displaystyle \int_{0}^{1}\frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x$ are quite easy to calculate. For instance:

$\begin{align*} \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{x^2+1}{x^2} \right ) \, {\rm d}x \\ &= \int_{0}^{1} \log x \left ( \log \left ( x^2+1 \right ) - 2 \log x \right ) \, {\rm d}x\\ &=\int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x - 2 \int_{0}^{1} \log^2 x \, {\rm d}x \\ &= \int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x -4 \\ &= -2\mathcal{G} - \frac{\pi}{2} -\log 2 + 4 -4 \\ &= -2\mathcal{G} - \frac{\pi}{2} - \log 2 \end{align*}$

where $\mathcal{G}$ denotes the Catalan’s constant. The other integral is done similarly. Throughout the solution it was used the fact that

$\mathcal{L}\left \{ \frac{1-\cos t}{t} \right \} = \int_{0}^{\infty} e^{-st} \left ( \frac{1-\cos t}{t} \right ) \, {\rm d}t= \frac{1}{2} \log \left ( \frac{1}{s^2}+1 \right )$