# Thread: Uniform Convergence of Inegral

1. ## Uniform Convergence of Inegral

Prove that the integral $\int_{0}^{1}\frac{\sin\alpha x}{\sqrt{|x-\alpha|}}\,dx$ converges uniformly for $\alpha\in[0,1]$.

2. ## Re: Uniform Convergence of Inegral

I'm not sure what "converges uniformly" means in this situation. Absolute convergence, as in the integral of the absolute value of the function converges? If so, then it seems the answer is yes.

With the exception of $\alpha \in \{0, 1\}$, you split the integral up into two integrals having its "bad" point as an endpoint, and then each of those integrals converges in absolute value. Write:

$\int_{0}^{1} \frac{\sin(\alpha x)}{\sqrt{|\alpha - x|}} dx = \int_{0}^{\alpha} \frac{\sin(\alpha x)}{\sqrt{\alpha - x}} dx + \int_{\alpha}^{1} \frac{\sin(\alpha x)}{\sqrt{x - \alpha }} dx$,

and then substitute $u = 1 - \frac{x}{\alpha}$ for the first, and $u = \frac{x}{\alpha} - 1$ for the second. Then simplify and use the trig angle sum formulas.

What you get is that, for each $\alpha \neq 0$, after these standard integral manipulations, you're left asking if

$\int_{0}^{whatever} \frac{|\sin(u)|}{\sqrt{u}} du$ and $\int_{0}^{whatever} \frac{|\cos(u)|}{\sqrt{u}} du$ converge. They do.

3. ## Re: Uniform Convergence of Inegral

If that felt "too Riemannian", consider the function $g(x) = \frac{1}{\sqrt{-x}}$ for $-1 \le x < 0, g(0) = 1$, and $\frac{1}{\sqrt{x}}$ for $0 < x \le 1$.

Then $g \in L^1([-1, 1])$, so $h(x) = g(x-\alpha) \in L^1([-1+\alpha, 1+\alpha])$, so $\tilde{h} = h|[0, 1] \in L^1([0, 1])$.

Then on $x \in [0,1], \ \frac{\sin(\alpha x)}{\sqrt{|\alpha-x|}} = \sin(\alpha x)g(x - \alpha) = \sin(\alpha x)\tilde{h}(x)$, so

$\int_{0}^{1}| \ \frac{\sin(\alpha x)}{\sqrt{|\alpha-x|}}\ | \ dx = \int_{0}^{1}| \ \sin(\alpha x)\tilde{h}(x) \ | \ dx \le \int_{0}^{1}|\tilde{h}(x)| dx < \infty$.