Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$ 
Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$

My try:
I want to use Direct comparison test so: $$0\le \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}}=\frac{\sin(\pi-x)}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} \le \frac{\pi-x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}}=\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}$$
However in this moment I don't know what I can do with $\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}$ because for $x \in [1,\pi)$ I have $\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}\le \frac{1}{(\pi-x)^{\alpha-1}}$ but for $x\in (0,1]$ it is not right.Have you got any ideas?
 A: Consider $$f(x)=\frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}}$$ and look what happens when close to the bounds.
Using Taylor expansions, we have close to $x=0$
$$f(x)=\frac{\pi ^{-a}}{\sqrt{x}}+a \pi ^{-a-1} \sqrt{x}+O\left(x^{3/2}\right)$$ which does not seem to make any problem.
Similarly, close to $x=\pi$
$$f(x)=(\pi -x)^{-a} \left(\frac{\pi-x }{\pi ^{3/2}}+O\left((x-\pi )^2\right)\right)\sim \frac{(\pi-x)^{1-a} }{\pi ^{3/2}}$$
Now, it is your turn.
A: Divide the integral $$ \int_0^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx = \int_0^{\pi/2} \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx  + \int_{\pi/2}^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx $$
For $x\in[0,\pi/2]$ we have $$ \frac{2}{\pi} \le \frac{\sin x}{x} \le 1$$
$$ 0 < M_1(\alpha) = \min\{\frac{1}{\pi^\alpha},\frac{1}{(\pi/2)^\alpha}\} \le \frac{1}{(\pi-x)^\alpha} \le \max\{\frac{1}{\pi^\alpha},\frac{1}{(\pi/2)^\alpha}\} = M_2(\alpha) < \infty$$
$$ \frac{2M_1(\alpha)}{\pi}\frac{1}{x^\frac12} \le \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} \le \frac{M_2(\alpha)}{x^\frac12 }$$
so
$$ \int_0^{\pi/2} \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx \text{ is convergent} \Leftrightarrow \int_0^{\pi/2} \frac{1}{x^\frac12} dx \text{ is convergent} \Leftrightarrow \alpha \in \mathbb R$$
since
$$\int_0^{\pi/2} \frac{1}{x^\frac12} dx = \Big[ 2x^\frac12\Big] \Big|_{x=0}^{x=\pi/2} = \sqrt{2\pi}$$
is convergent, then this part of integral is convergent for all $\alpha\in\mathbb R$.
For $x\in[\pi/2,\pi]$ we have $$ \frac{2}{\pi} \le \frac{\sin x}{\pi -x} = \frac{\sin(\pi-x)}{\pi-x} \le 1$$
$$ \frac{1}{\pi^\frac32} \le \frac{1}{x^\frac32} \le \frac{1}{(\pi/2)^\frac32}$$
$$ \frac{2}{\pi^\frac52}\frac{1}{ (\pi-x)^{\alpha-1}} \le \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} \le \frac{1}{(\pi/2)^\frac32}\frac{1}{(\pi-x)^{\alpha-1}}$$
so
$$ \int_{\pi/2}^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx \text{ is convergent} \Leftrightarrow \int_{\pi/2}^\pi \frac{1}{(\pi-x)^{\alpha-1}} dx \text{ is convergent} $$
and we have $$ \int_{\pi/2}^\pi \frac{1}{(\pi-x)^{\alpha-1}} dx = \Big[\frac{-1}{2-\alpha}(\pi-x)^{2-\alpha}\Big]\Big|_{x=\pi/2}^{x=\pi}$$
which is finite iff $\alpha<2$.
In other words we use the comparison $$ \sin x < \frac{\pi}{2} - |x-\frac{\pi}{2}| = \left\{\begin{array}{ll} x & \text{for } 0 \le x \le \frac{\pi}{2}\\ \pi - x & \text{for } \frac{\pi}{2} \le x \le\pi\end{array}\right. $$
