Joy of Mathematics

A logarithmic integral

May 14, 2022 / Leave a comment

Let $\alpha \in \mathbb{R}$ . Prove that:

$\int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2}\, \mathrm{d}x = \frac{\pi}{\sqrt{2}} \cdot \frac{\alpha}{\sqrt{1+ \sqrt{1+\alpha^2}}}$

Solution

Using the known Fourier series $\displaystyle \sum_{n=-\infty}^{\infty} \frac{1}{\left( x+2\pi n \right)^2}=\frac{1}{4} \csc^2 \frac{x}{2}$ as well as the known generating function of the Catalan numbers

$\sum_{n=0}^{\infty} \mathcal{C}_n x^n = \frac{1- \sqrt{1-4x}}{2x} \Rightarrow \sum_{n=0}^{\infty} \mathcal{C}_{2n} x^{2n} = \frac{\sqrt{1+4x} - \sqrt{1-4x}}{4x}$

Then, we have successively:

$\begin{align*} \int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2} \, \mathrm{d}x &=\frac{1}{2} \int_{-\infty}^{\infty}\frac{\arctan \left(\alpha \sin^{2} x\right)}{x^{2}}\, \mathrm{d}x\\ &=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan \left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}} \, \mathrm{d}x\\ &=\frac{1}{8}\int_{0}^{2\pi} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}\frac{x}{2} \, \mathrm{d}x\\ &=\frac{1}{4}\int_{0}^{\pi/2}\arctan \left(\alpha \sin^{2} x \right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right) \, \mathrm{d} x\\ &=\int_{0}^{\pi/2} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}x\, \mathrm{d} x\\ &=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{2n+1} \alpha^{2n+1} \int_{0}^{\pi/2}\sin^{4n}x \, \mathrm{d}x\\ &=\pi \alpha \sum_{n=0}^{\infty} \mathcal{C}_{2n}\left(-\frac{\alpha^{2}}{16}\right)^{n} \\ &=\frac{\pi}{i}\left(\sqrt{1+i \alpha}-\sqrt{1-i\alpha}\right)\\ &=\frac{\pi}{\sqrt{2}}\frac{\alpha}{\sqrt{1+\sqrt{1+\alpha^{2}}}} \end{align*}$

A Gamma summation

May 11, 2022 / Leave a comment

Let $a \notin \mathbb{Z}$ and $a > \frac{1}{2}$ . Prove that

$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$

Solution

Consider the function

$f(x) = \left\{\begin{matrix} \cos^{2a-2} \pi x & , & \left | x \right |< \frac{1}{2} \\ 0 & , & \text{elsewhere} \end{matrix}\right.$

We easily evaluate that

$\hat{f}(y)= \int_{-\infty}^{\infty} f(x) e^{-2i \pi x y} \, \mathrm{d}x = \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+y) \Gamma(a-y)}$

Hence it follows from Poisson summation formula that

$\begin{align*} 1 &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) \\ &= \sum_{n=-\infty}^{\infty} \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+n) \Gamma(a-n)} \end{align*}$

Thus,

$\begin{align*} \frac{1}{2^{2-2a} \Gamma(2a-1)} &= \sum_{n=-\infty}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &=\sum_{n=-\infty}^{-1} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(\alpha)} + \\ &\quad \quad \quad + \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &= 2 \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(a)} \\ &= 2 \sum_{n=0}^{\infty}\frac{1}{\Gamma(a+n) \Gamma(a-n)} - \frac{1}{\Gamma^2(a)} \end{align*}$

The result follows.

A Gudermannian integral

May 8, 2022 / 1 Comment on A Gudermannian integral

Let $n \in \mathbb{N}$ . Prove that

$\int_{0}^{\infty} \int_{0}^{\infty} \frac{\mathrm{gd}^n \left ( xy \right )}{\cosh xy} \sin y \, \mathrm{d} \left ( x, y \right ) = \frac{1}{n+1} \cdot \left ( \frac{\pi}{2} \right )^{n+2}$

where $\mathrm{gd}$ is the Gudermannian function.

Solution

First of all we note that $\displaystyle{\dfrac{\partial}{\partial x} \mathrm{gd}(xy)=\frac{y}{\cosh xy}}$ and $\displaystyle{\mathrm{gd}(x)=2\arctan e^x-\frac{\pi}{2}}$ . Hence,

$\begin{align*} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\mathrm{gd}^n \left ( xy \right )}{\cosh xy} \sin y \, \mathrm{d} \left ( x, y \right ) &= \int_{0}^{\infty} \frac{\sin y}{y} \int_{0}^{\infty} \mathrm{gd}^n (xy) \left ( \mathrm{gd}(xy) \right )'\, \mathrm{d}(x, y) \\ &=\frac{\pi^{n+1}}{\left ( n+1 \right ) 2^{n+1}} \int_{0}^{\infty} \frac{\sin y}{y}\, \mathrm{d}y \\ &=\frac{1}{n+1} \left ( \frac{\pi}{2} \right )^{n+2} \end{align*}$

An odd trigonometric integral

May 8, 2022 / Leave a comment

Let $n \in \mathbb{N}$ . Prove that

$\int_{0}^{\infty} \frac{\sin^{2n+1} y}{y}\, \mathrm{d}y = \frac{\pi}{2^{2n+1}} \binom{2n}{n}$

Solution

Since $\displaystyle \sin x =\frac{e^{ix}-e^{-ix}}{2i}$ we deduce that

$\sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^{n} (-1)^r \binom{2n+1}{r}\sin(2r+1)x$

Hence,

$\begin{align*} \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right) \end{align*}$

since $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x} \, \mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x} \, \mathrm{d}x=\frac{\pi}{2}$ and $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ . Finally, we note that the sum telescopes hence the result.

A rational harmonic series

April 26, 2022 / Leave a comment

Can the rational numbers in the interval $[0,1]$ be enumerated as a sequence $\{q_n\}_{n \geq 1}$ in such a way that the series $\displaystyle \sum_{n=1}^{\infty} \frac{q_n}{n}$ is convergent?

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