Asymptotic behaviour of an integral of trigonometric functions Consider the integral
$$
\int_{0}^1 \frac{\sin^2(\pi n x)}{(\sin{\pi x})^{1+a}} dx,
$$
with $a \in (0,2)$.
The integral is certainly well defined (possibly infinite) for $n>0$. My interest is to show that is integral is finite for every $n$ and find a lower bound asymptotic behavior on as $n \to \infty$ in function of $a$.
I believe that I managed to get
$\int_{0}^1 \frac{\sin^2(\pi n x)}{(\sin{\pi x})^{1+a}} dx \ge c_a n^{a}$ by taking a lower bound of $\sin(\pi n x)$ by step functions. But I am really not sure that this is the best method.
 A: Note first that your integral is equal (up to a factor, I leave the computation to you) to the integral $\displaystyle \int_0^{\pi/2}\frac{\sin(n x))^2}{\sin(x)^{1+a}}dx$ (first change of variable $\pi x=u$, and then $v=\pi-u$ on $[\pi/2,\pi]$).
Now let $x,y$ such that $0<y<x$. We have $x^{1+a}-y^{1+a}=(x-y)a c^{a}$ with $y<c<x$, hence $x^{1+a}-y^{1+a}\leq (x-y)a x^{a}$. We apply this with $x=t\in )0,\pi/2]$, $y=\sin(t)$, we get that $t^{1+a}-( \sin(t))^{1+a}\leq (t-\sin(t))at^a$. Now $t-\sin (t)\leq t^3/6$, and we have $t^{1+a}-( \sin(t))^{1+a}\leq \frac{a}{6}t^{a+3}$.
It is now easy to show that the function $F(t)=\frac{1}{(\sin(t)^{1+a}}-\frac{1}{t^{1+a}}$ is $\geq 0$ and integrable on $[0,\pi/2]$.
Now
$$\int_0^{\pi/2}\frac{\sin(n x))^2}{\sin(x)^{1+a}}dx=\int_0^{\pi/2}\sin(n x))^2F(x)dx+\int_0^{\pi/2}\frac{\sin(n x))^2}{x^{1+a}}dx=A_n+B_n$$
By the above, $A_n$ is bounded by $\int_0^{\pi/2}F(x)dx$. Using the change of variable $nx=t$, we get that
$$ B_n=n^a\int_0^{n\pi/2}\frac{(\sin(t))^2}{t^{1+a}}dt=n^a w_n $$
Now $w_n$ as $n\to +\infty$, goes to the value of the convergent integral $\int_0^{+\infty}\frac{(\sin(t))^2}{t^{1+a}}dt >0$, and we have the asymptotic behaviour. 
A: The only points that could induce non-finiteness are the zeroes of the denominator, $x=0$ and, symmetrically, $x=\pi$.
For small $x$, we can approximate the integrand as
$$\frac{(\pi nx)^2}{(\pi x)^{1+a}}=n^2(\pi x)^{1-a},$$ with the antiderivative
$$\frac{n^2\pi^{1-a} x^{2-a}}{2-a}, $$ which remains finite.
