Computing $\lim\limits_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\mathrm{d}x} $

Question

$\def\d{\mathrm{d}}$I would like to compute the following limit, $$\displaystyle{\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x} .$$

I am looking for a high school answer.

I tried writing $$\lim_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x = \lim_{n \to \infty} \lim_{ε \to \frac{\pi}{2}}\int_0^ε{\frac{(\sin(x))^n}{1-\sin(x)}}\,\d x},$$

but it doesn't help me, since $1 - \sin(x) \leq 1, \forall x \in \left[0, \dfrac{\pi}{2}\right]$.

Answer 1

Your integral does event convergence, for each $n$ we have $$ \int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}=\infty$$

In fact Since see here $$\frac2πx≤\sin x≤x,~~~~~~\forall x \in \left[0, \displaystyle \frac{\pi}{2}\right]$$ we have

$$\frac{(\frac2πx)^n}{1-\frac2πx}≤\frac{(\sin x)^n}{1-\sin x}≤\frac{x^n}{1-x}\implies \int_0^{\fracπ2}\frac{(\frac2πx)^n}{1-\frac2πx}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{\fracπ2}\frac{x^n}{1-x}dx$$ then let $u= \frac2πx$ the we get

$$\infty=\int_0^{1}\frac{x^n}{1-x}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{1}\frac{x^n}{1-x}dx+\int_1^{\fracπ2}\frac{x^n}{1-x}dx=\infty$$

Answer 2

just a hint

Write the integral as $$I_1+I_2=$$

$$\int_0^{\frac {\pi}{2}-\epsilon}+\int_{\frac {\pi}{2}-\epsilon}^\frac\pi 2$$ with $$I_1\le \frac {(\cos (\epsilon))^n}{1-\cos (\epsilon) }$$ goes to zero. and

$$I_2\le \epsilon \frac {1}{\cos (\epsilon)} $$

Answer 3

The integral fails to converge for all $n$ as $1-\sin x \sim (\pi/2-x)^2$ near $\pi/2.$