Joy of Mathematics

An arcosine integral

January 20, 2021January 20, 2021 Tolaso

Evaluate the integral

$\mathcal{J} = \int_{-1}^{1} \frac{\arccos x}{\sqrt{3x^4+2x^2+3}} \, \mathrm{d}x$

A limit

January 20, 2021 Tolaso

For any nonnegative integer $n$ , define

$\mathcal{J}_n = \int_{0}^{\pi/2} \frac{\sin^2 nt}{\sin t}\, \mathrm{d}t$

Evaluate the limit $\ell = \lim \limits_{n \rightarrow +\infty} \left( 2 \mathcal{J}_n - \ln n \right)$ .

Solution

Lemma: Let $n \in \mathbb{Z}$ and $x \neq \kappa \pi$ . It holds that

$\sin x + \sin 3x + \cdots + \sin(2n-1)x = \frac{\sin^2 nx}{\sin x}$

Proof: The LHS is just the imaginary part of

$\mathcal{S}=e^{ix}+e^{3ix}+\cdots+e^{(2n-1)ix}$

This is a geometric progression , hence:

$\begin{align*} \mathcal{S} &=e^{ix}\frac{e^{2in x}-1}{e^{2ix}-1}\\ &=e^{inx}\frac{e^{inx}-e^{-inx}}{e^{i x}-e^{-ix}} \\ &=e^{inx}\frac{\sin nx}{\sin x}\\ &=(\cos nx+i\sin nx) \\ &=\frac{\sin nx}{\sin x} \end{align*}$

The result follows. $\blacksquare$

Hence,

$\begin{align*} \mathcal{J}_n &= \int_{0}^{\pi/2} \frac{\sin^2 nt}{\sin t}\, \mathrm{d}t \\ &= \int_{0}^{\pi/2} \sum_{k=1}^{n} \sin \left ( 2k-1 \right )t \, \mathrm{d}t \\ &=\sum_{k=1}^{n} \int_{0}^{\pi/2} \sin \left ( 2k-1 \right )t \, \mathrm{d}t \\ &=\sum_{k=1}^{n} \frac{1}{2k-1} \\ &=\sum_{k=1}^{2n}\frac{1}{k}-\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k} \end{align*}$

Now,

$\begin{align*} \ell &= \lim_{n \rightarrow +\infty} \left ( 2 \mathcal{J}_n - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} - \ln n \right ) \\ &=\lim_{n \rightarrow +\infty} \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - 2 \ln 2n + 2 \ln 2n - \sum_{k=1}^{n} \frac{1}{k} - \ln n \right ) \\ &= \lim_{n \rightarrow +\infty} \left [ \left ( 2 \sum_{k=1}^{2n} \frac{1}{k} - 2 \ln 2n \right ) + 2 \ln 2 - \left ( \sum_{k=1}^{n} \frac{1}{k} -\ln n \right ) \right ] \\ &= 2 \gamma + 2 \ln 2 - \gamma \\ &= \gamma + 2 \ln 2 \end{align*}$

Fractional integral

January 17, 2021January 17, 2021 Tolaso

Let $\{ \cdot \}$ denote the fractional part. Prove that

$\int_{0}^{\pi/2}\sin 2x \{\ln^{2n-1} \tan x \} \, \mathrm{d}x$

for the different values of the integer number $n$ .

Solution

Let $\mathcal{J}$ denote the integral,

$\begin{align*} \mathcal{J} &= \int_{0}^{\pi/2} \sin 2x \left \{ \ln^{2n-1} \tan x \right \}\, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \sin 2 \left ( \frac{\pi}{2} - u\right ) \left \{ \ln^{2n-1} \tan \left ( \frac{\pi}{2} - u \right ) \right \}\, \mathrm{d}u \\ &= \int_{0}^{\pi/2} \sin 2u \left \{ \ln^{2n-1} \frac{1}{\tan u} \right \} \, \mathrm{d}u\\ &=\int_{0}^{\pi/2} \sin 2u \left \{ - \ln^{2n-1} \tan u \right \}\, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \left ( \left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} \right ) \, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \, \mathrm{d}u \\ &= \frac{1}{2} \end{align*}$

since $\left \{ x \right \} + \left \{ -x \right \} = 1$ if $x \notin \mathbb{Z}$ whereas $\left \{ x \right \} + \left \{ -x \right \} = 0$ if $x \in \mathbb{Z}$ . Therefore,

$\left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} =1$

except of a countable set $\displaystyle A = \left\{ x_k \in \left ( 0, \frac{\pi}{2} \right ) \bigg| \ln^{2n-1} \tan x_k \in \mathbb{Z} \right\}$ whose measure is $0$ .

Multiple logarithmic integral

January 17, 2021 Tolaso

Let $\zeta$ denote the Riemann zeta function. Evaluate the integral

$\mathcal{J} = \int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln \prod_{k=1}^{n} x_k \ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )$

Solution

Based on symmetries,

$\begin{align*} \mathcal{J} &=n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1} \ln x_{1}\ln \left ( 1 - \prod_{k=1}^{n} x_k \right ) \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &=-n\int_{0}^{1} \int_{0}^{1} \cdots \int_{0}^{1}\ln x_{1}\sum_{i=1}^{ \infty}\frac{(x_{1}\dots x_{n})^i}{i}\, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )\\ &=-n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n-1}}\int_{0}^{1}x_{1}^i\ln x_{1}dx_{1} \\ &=n \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}} \end{align*}$

Let $\displaystyle \mathcal{S}_n = \sum_{i=1}^{\infty}\frac{1}{i(i+1)^{n+1}}$ . It follows that

$\begin{align*} \mathcal{S}_n &= \sum_{i=1}^{\infty} \frac{1}{(i+1)^n} \left ( \frac{1}{i} - \frac{1}{i+1} \right ) \\ &= \mathcal{S}_{n-1} - \sum_{i=2}^{\infty} \frac{1}{i^{n+1}} \\ &= \mathcal{S}_{n-1} + 1 - \zeta (n +1) \end{align*}$

Using the recursion we get that

$\mathcal{S}_n = n + 1 - \sum_{k=2}^{n+1} \zeta(k)$

Thus,

$\mathcal{J} = n \left( n +1 - \sum_{k=2}^{n+1} \zeta(k) \right)$

Fourier transformation

November 25, 2020November 25, 2020 Tolaso

Let $\xi \in \mathbb{R}$ and $\alpha \in (0, 1)$ . Show that

$\int_{- \infty}^{\infty} {e}^{-2 \pi i x \xi} \frac{ \sin \pi \alpha }{ \cosh \pi x + \cos \pi \alpha } \mathrm{d} x = \frac{2 \sinh 2 \pi \alpha \xi }{ \sinh 2 \pi \xi }$

Solution

We note that

$\begin{align*} \frac{\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha} &= \frac{1}{1+e^{-\pi(x+i\alpha)}}-\frac{1}{1+e^{-\pi(x - i\alpha)}}\\ &= 2\sum\limits_{n=1}^{\infty} (-1)^{n}e^{-n\pi x}\sin n\pi\alpha \end{align*}$

Thus,

$\begin{align*} \int_{-\infty}^{\infty} \frac{e^{-2\pi i \xi x}\sin \pi \alpha}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x &= \int_{0}^{\infty} \frac{2\sin \pi \alpha \cos 2\pi\xi x}{\cosh \pi x + \cos \pi \alpha}\,\mathrm{d}x\\ &= 4\int_0^{\infty}\sum\limits_{n=1}^{\infty}(-1)^n e^{-n\pi x}\cos 2\pi\xi x \sin n\pi \alpha\,\mathrm{d}x\\ &= \frac{4}{\pi}\sum\limits_{n=1}^{\infty}(-1)^n \frac{n\sin n\pi\alpha}{n^2+4\xi^2}\\ &= \frac{2}{\pi}\sum\limits_{n=1}^{\infty} (-1)^n\left(\frac{1}{n+2i\xi}+\frac{1}{n-2i\xi}\right)\sin n\pi \alpha\\ &= \frac{2}{\pi}\sum\limits_{n = -\infty}^{\infty} \frac{(-1)^n\sin n\pi\alpha}{n+2i\xi}\\ &= \frac{2 \sin 2\pi i \alpha\xi}{\sin 2\pi i\xi} \qquad \textrm{ (by residue theorem) }\\ &= \frac{2 \sinh 2\pi \alpha\xi}{\sinh 2\pi \xi} \end{align*}$