Uniform convergence of the integral $\int_{0}^{\pi} \frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}} dx$ with $0 < \alpha < 2$

Question

I'm studying the uniform convergence of $\int_{0}^{\pi} \frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}} dx$ with $0 < \alpha < 2$

We have problems at $0$ and $\pi$ so let's split the integral at 1. $$\int_{0}^{\pi} \frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}} dx = \int_{0}^{1} \frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}} dx + \int_{1}^{\pi} \frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}} dx$$

Now my idea is to prove each one is bounded by something I can integrate, then the Weierstrass $M$ test will give uniform convergence. Obviously $|\sin(x)| \leq 1$ but that doesn't help.

In $[0,1]$ I realize that $|\frac{\sin(x)}{x^{\alpha} (\pi - x)^{\alpha}}| \leq \frac{100}{x^{\alpha -1} (\pi - x)^{\alpha}}$ but can't see how that helps.

I haven't tried the other integral.

Any hints would be highly appreciated.

Answer 1

The convergence fails to be uniform for $\alpha\in (0,2)$. To show this, let $\displaystyle \epsilon=\frac{1}{\pi^3 e}$.
Then for all $\nu\in (0,1)$, we select $\alpha=2+\frac1{\log(\nu)}$. Using $\sin(x)\ge \frac{2x}{\pi}$ for $x\in [0,\pi/2]$ and $\pi-x\le \pi$ we have the estimates

$$\begin{align} \left|\int_0^{\nu}\frac{\sin(x)}{x^{\alpha}(\pi-x)^\alpha}\,dx\right|&\ge \int_0^\nu \frac{(2x/\pi)}{x^\alpha\pi^\alpha }\,dx\\\\ &=\frac{2\nu^{2-\alpha}}{(2-\alpha)\pi^{1+\alpha}}\\\\ &=\frac{\nu^{-1/\log(\nu)}}{\pi^{3+1/\log(\nu)}}\\\\ &\ge\frac{1}{e\pi^3} \end{align}$$

which negates the uniform convergence for $\alpha\in (0,2)$.

Answer 2

Notice for $x\in(0,\pi/2]$ we have $$ \left\lvert\frac{\sin(x)}{x^\alpha(\pi-x)^{\alpha}}\right\rvert\leq\frac{1}{x^{\alpha}(\pi-\pi/2)^{\alpha}}=\frac{2^\alpha}{\pi^\alpha}\cdot\frac{1}{x^\alpha}$$ Meanwhile, for $x\in[\pi/2,\pi)$ we have $$\left\lvert\frac{\sin(x)}{x^\alpha(\pi-x)^{\alpha}}\right\rvert\leq\frac{1}{(\pi/2)^\alpha(\pi-x)^\alpha}=\frac{2^\alpha}{\pi^\alpha}\cdot\frac{1}{(\pi-x)^\alpha}$$ Both $\int_{0}^{\pi/2}\frac{2^\alpha}{\pi^{\alpha}x^\alpha}dx$ and $\int_{\pi/2}^{\pi}\frac{2^\alpha}{\pi^\alpha(\pi-x)^\alpha}dx$ converge whenever $0<\alpha<1$.

Answer 3

If the problem was limited to the convergence of the integral (not the uniform convergence), it is obvious that the integral will converge for any $\alpha <0$ since the integrand is just the product of a sine function by something looking like a polynomial in $x$.

Now, close to $x=0$, by Taylor we have $$\frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}}=x^{-\alpha } \left(\pi ^{-\alpha } x+\pi ^{-\alpha -1} \alpha x^2+O\left(x^3\right)\right)\sim x^{-\alpha+1 }$$ and the integral would converge as long as $\alpha < 2$.

Close to $x=\pi$, by Taylor we have $$\frac{\sin(x)}{x^{\alpha}(\pi -x)^{\alpha}}=(\pi -x)^{-\alpha } \left(-\pi ^{-\alpha } (x-\pi )+\pi ^{-\alpha -1} \alpha (x-\pi )^2+O\left((x-\pi )^3\right)\right)\sim (\pi-x)^{-\alpha+1 }$$ and, again, the integral would converge as long as $\alpha < 2$.