Finding value of $\lim_{r\rightarrow \infty}r\cdot (2/\pi)^{r+1}\cdot f(r)$ is

Question

If $\displaystyle f(r)=\int^{\frac{\pi}{2}}_{0}x^r\cdot \sin xdx$. Then $\lim_{r\rightarrow \infty}r\cdot \bigg(\frac{2}{\pi}\bigg)^{r+1}\cdot f(r)$ is

Try: Given $f(r)=\int^{\frac{\pi}{2}}_{0}x^r\cdot \sin xdx$. Applying by parts , We get

$$f(r)=\frac{1}{(r+1)}\cdot \bigg(\frac{\pi}{2}\bigg)^{r+1}-\int^{\frac{\pi}{2}}_{0}\frac{x^r}{(r+1)}\cdot \cos xdx$$

Now i did not understand how to find $f(r)$ so that i can find limit, could some help me, Thanks

Answer 1

Note that for $x\in[0,\pi/2]$

$$x^r\cdot \frac2\pi x\le x^r\cdot \sin x\le x^r \cdot 1$$

then

$$\int^{\frac{\pi}{2}}_{0}x^{r+1} \frac2\pi dx=\frac{(\frac{\pi}2)^{r+1}}{r+2}\le f(r)=\int^{\frac{\pi}{2}}_{0}x^r\cdot \sin xdx\le \int^{\frac{\pi}{2}}_{0}x^rdx=\frac{(\frac{\pi}2)^{r+1}}{r+1}$$

thus

$$\frac r{r+2}\le r\cdot \left(\frac2\pi\right)^{r+1}\cdot f(r)\le \frac r{r+1}$$

Answer 2

First notice that: \begin{align} f(r)=\int^{\pi/2}_0 e^{r\log(x)}\sin(x)\,dx \end{align} By applying Laplace's method, then one has as $r\to\infty$ the following asymptotics for $f(r)$, namely: \begin{align} f(r) \sim \pi\frac{ e^{r \log(\pi/2)}}{2r}=\frac{1}{r}\left(\frac\pi 2\right)^{r+1} \end{align} Hence: \begin{align} \lim_{r\to\infty} r \left( \frac 2 \pi\right)^{r+1} f(r) = \lim_{r\to\infty} r \left( \frac 2 \pi\right)^{r+1}\frac{1}{r}\left(\frac\pi 2\right)^{r+1} = 1 \end{align}