Joy of Mathematics

An odd trigonometric integral

May 8, 2022August 22, 2022 Tolaso

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,

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

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.

An integral

November 27, 2021 Tolaso

Let $n \in \mathbb{Z}_{\geq 0}$ . Evaluate the integral

$\mathcal{J}_n = \int_0^\pi \frac{1- \cos nx}{1-\cos x}\, \mathrm{d}x$

Solution

We note that

$\begin{align*} \frac{\mathcal{J}_{n+1}+\mathcal{J}_{n-1}}{2} &=\int_{0}^{\pi}\frac{2-\cos (n+1)x-\cos (n-1)x}{2(1-\cos x)}\, \mathrm{d}x \\ &= \int_{0}^{\pi}\frac{1-\cos n x \cos x}{1-\cos x} \, \mathrm{d}x \\ &=\int_{0}^{\pi}\frac{\left ( 1-\cos n x \right )+\cos n x(1-\cos x)}{1-\cos x}\, \mathrm{d}x \\ &=\mathcal{J}_n+\int_{0}^{\pi} \cos n x\, \mathrm{d}x \\ &=\mathcal{J}_n \end{align*}$

This shows that we have a geometric progression. Since $\mathcal{J}_0 = 0$ , $\mathcal{J}_1 = \pi$ it follows that $\mathcal{J}_n = n \pi$ .

Symmetry

July 24, 2021November 18, 2021 Tolaso

Let $f, g : \mathbb{R} \rightarrow \mathbb{R}$ be continuous functions. If $\mathcal{C}_f$ is symmetric around the line $x=\frac{\alpha + \beta}{2}$ then prove that:

$\int_{\alpha}^{\beta} x g \left ( f(x) \right )\, \mathrm{d}x = \frac{\alpha+\beta}{2} \int_{\alpha}^{\beta}g \left ( f(x) \right )\, \mathrm{d}x$

Solution

Since $\mathcal{C}_f$ is symmetric around the line $x=\frac{\alpha + \beta}{2}$ this means that

$f \left ( 2 \cdot \frac{\alpha + \beta}{2} - x \right ) = f(x) \Leftrightarrow f \left ( \alpha + \beta - x \right ) = f(x)$

Hence,

$\begin{align*} \int_{\alpha}^{\beta} x g \left ( f(x) \right )\, \mathrm{d}x &\overset{u = \alpha + \beta - x}{=\! =\! =\! =\! =\! =\! =\!} \int_{\alpha}^{\beta} \left ( \alpha + \beta - x \right ) g \left ( f \left ( \alpha + \beta - x \right ) \right )\, \mathrm{d}x \\ &= \int_{\alpha}^{\beta} \left ( \alpha + \beta - x \right ) g \left ( f(x) \right )\, \mathrm{d}x \\ &= \left ( \alpha + \beta \right ) \int_{\alpha}^{\beta} g \left ( f(x) \right )\, \mathrm{d}x - \int_{\alpha}^{\beta} x g \left ( f(x) \right )\, \mathrm{d}x \end{align*}$

The result follows.

Sophomore’s dream constant

May 14, 2021 Tolaso

Evaluate the integral

$\mathcal{J} = \int_0^1 \int_0^1 (xy)^{xy} \,\mathrm{d}(x, y)$

Solution

Let $t=xy$ and $s=y$ . The Jacobian is

$\frac{\partial (s, t)}{\partial (x, y)} = y \Rightarrow \left (\frac{\partial (s, t)}{\partial (x, y)} \right )^{-1} = \frac{1}{y} = \frac{1}{s}$

Hence,

$\begin{align*} \mathcal{J} &= \iint \limits_{[0, 1]^2} \left ( xy \right )^{xy} \, \mathrm{d}(x, y) \\ &=\int_{0}^{1} \int_{0}^{s} \frac{t^t}{s} \, \mathrm{d}(t, s) \\ &= \int_{0}^{1} \int_{t}^{1} \frac{t^t}{s} \, \mathrm{d} ( s, t)\\ &= -\int_{0}^{1} t^t \log t \, \mathrm{d}t \end{align*}$