A limit

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,

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

A limit

They might interest you:

Hyperbolic series

On zeroes and poles of a meromorphic function

A zeta series

Rational function and polynomial

Coincidences from ... hell

A convergent series

Peculiar alternating series

A product

On idempotent invertible matrices

Leave a Reply Cancel reply