2 Integrals

[jacks]

Evaluation of \(\displaystyle \int\frac{x^{5n-1}+2x^{4n-1}}{\left(x^{2n}+x^n+1\right)^3}dx\) and \(\displaystyle \int_{0}^{\pi}\frac{\sin^2(nx)}{\sin^2(x)}dx\,,\) where \(n\in \mathbb{N}\,.\)

[Tolaso J Kos]

Hi Jacks,
I give a solution for the second integral (which is rather famous and can be found almost everywhere)

Successively we have: \( \displaystyle \int_{0}^{\pi}\left ( \frac{\sin n x}{\sin x} \right )^2\,dx=\int_{0}^{\pi}\frac{1-\cos n x}{1-\cos x}\,dx=\left | n\right | \pi \).

The latter integral (as far as I am concerned has at least \(3 \) solutions that can be found here)

I translate Demetre's solution which is the fastest:

Let:
$$I_n=\int_{0}^{\pi}\frac{1-\cos n \theta}{1-\cos \theta}\,d\theta=\frac{1}{2}\int_{0}^{2\pi}\frac{1-\cos n\theta}{1-\cos \theta}\,d\theta=\frac{1}{2}\int_{0}^{2\pi}\frac{2-\left ( e^{in\theta}+e^{-in\theta} \right )}{2-\left ( e^{i\theta}+e^{-i\theta} \right )}\,d\theta$$
$$\frac{1}{2}\oint_\gamma \frac{2-\left ( z^n+z^{-n} \right )}{2-\left ( z+z^{-1} \right )}\frac{1}{iz}\,dz$$
whereas \(\gamma\) is the curve defined as: \( \gamma:[0, 2\pi] \longrightarrow \mathbb{R} , \;\;\gamma(\theta)=e^{i\theta}\).
However it holds:
$$\frac{2-\left ( z^n+z^{-n} \right )}{2-\left ( z+z^{-1} \right )}\frac{1}{z}=\frac{\left ( z^n-1 \right )^2}{z^n\left ( z-1 \right )^2}=\frac{\left ( 1+z+z^2+\cdots+z^{n-1} \right )^2}{z^n}$$
Moreover the coefficient of \(z^{n-1} \) in the expansion of \( \left ( 1+z+z^2+\cdots+z^{n-1} \right )^2 \) equals \(n \). (That is because \(z^{n-1} \) shows up with a multiplication of \(z^a , z^{(n-1)-a} \) and we have \(n \) choices for \(a \) ). So, by Cauchy Residue we have that \(I_n= |n| \pi \).

Futher results: Left as an exercise:
$$\int_{0}^{2\pi} \frac{\sin nx}{\sin x}\,dx=2\int_{0}^{\pi}\frac{\sin nx}{\sin x}\,dx=4\int_{0}^{\pi/2}\frac{\sin nx}{\sin x}\,dx=\left\{\begin{matrix}
4\left ( 2-\frac{2}{3}+\frac{2}{5}+\cdots+\frac{2\left ( -1 \right )^n}{2n+1} \right )&, n=2m \\
2\pi&,n=2m-1
\end{matrix}\right.$$
$$\int_{0}^{2\pi}\left ( \frac{\sin nx}{\sin x} \right )^2\,dx=2\left | n \right |\pi$$
This one can be done using Fourier expansion. Someone first works for \(n \geq 0 \) and then generalizes for \(n<0 \).
$$\int_{0}^{\pi}\left ( \frac{\sin nx}{\sin x} \right )^m=\pi\sum_{k=0}^{\left \lfloor \frac{m\left ( 1-\frac{1}{n} \right )}{2} \right \rfloor}\left ( -1 \right )^k\binom{m}{1}\binom{\frac{m}{2}(n+1)-kn-1}{m-1}$$

[Tolaso J Kos]

As for the first one If I have done the operations right then it should result in:
$$\int \frac{x^{5n-1}+2x^{4n-1}}{\left ( x^{2n}+x^n+1 \right )^3}dx=-\frac{3x^{2n}+2x^{3n}+2x^n+1}{2n\left ( x^{2n}+x^n+1 \right )^2}+c, \;\; c\in \mathbb{R}, \;\; n \in \mathbb{N}$$