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Prove $$I(\alpha)=\int_\limits{0}^{\infty}\frac{\sin(\alpha x)}{x}dx$$ converges uniformly for $0<a\leqslant \alpha\leqslant b$ and it does not converge uniformly for $0\leqslant \alpha\leqslant b$

I know that $\int_\limits{0}^{\infty}\frac{\sin( x)}{x}dx=\frac{\pi}{2}$. If I take $I(\alpha)=\int_\limits{0}^{\infty}\frac{\sin(\alpha x)}{x}dx$ and do the following substitution:$ u=\alpha x$, I get $\int_\limits{0}^{\infty}\frac{\sin(u)}{u}du=\frac{\pi}{2}$. However I am not proving what I intended. I have thought of comparison test but I have not come up with a solution.

Question:

What do you think of the problem? How should I solve it?

Thanks in advance!

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    $\begingroup$ Where are $a,\,b$ in your problem? It's one thing to prove $I(\alpha)$ has an $\alpha$-dependent behaviour, but you can't expect the details of that dependence to be $a,\,b$-dependent when $a,\,b$ play no role in the definition of $I(\alpha)$. Is it possible $a,\,b$ are supposed to be the integration limits, or even that the integrand's behaviour for $x\in (a,\,b)$ is of interest? $\endgroup$ – J.G. Apr 25 '18 at 15:02
  • $\begingroup$ @J.G. Thanks for the reply! The exercise is defined as I stated. There is no more information provided. $\endgroup$ – Pedro Gomes Apr 25 '18 at 15:10
  • $\begingroup$ @PedroGomes the exercise is ill-defined. $\endgroup$ – Mike Hay Apr 25 '18 at 16:03
  • $\begingroup$ If $\alpha>0$ then $f(\alpha)=$ $\int_0^{\infty}\frac {\sin \alpha x}{\alpha x}d(\alpha x)=$ $\int_0^{\infty}\frac {\sin y}{y}dy$ is constant. Perhaps there is a typo in the book. $\endgroup$ – DanielWainfleet Apr 25 '18 at 16:16
  • $\begingroup$ @DanielWainfleet Professor's exercise. $\endgroup$ – Pedro Gomes Apr 25 '18 at 16:17
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Firstly, we have to be careful. The observation $$\int_0^{+\infty} \frac{\sin \alpha x}{x} = \int_0^{+\infty} \frac{\sin x}{x} = \frac{\pi}{2}$$ only holds if $\alpha \ne 0$ since we cannot divide by $0$. Otherwise, $$\int_0^{+\infty} \frac{\sin \alpha x}{x} = 0$$ as we are integrating the zero function. In fact, this very degeneracy will help us to establish the result.


Now the improper integral $$\int_0^{+\infty} \frac{\sin \alpha x}{x} = \frac{\pi}{2} \text{ or } 0$$ In particular it exists, so we may take any $(c_n) \to 0$ and $(d_n) \to +\infty$ with $0 < c_n < d_n$ and conclude $$\lim_n \int_{c_n}^{d_n} \frac{\sin \alpha x}{x} = \int_0^{+\infty} \frac{\sin \alpha x}{x}$$ Therefore, we can reformulate the question more precisely as follows:

Let $0 < a < b$ and $(c_n) \to 0$, $(d_n) \to +\infty$ with $0 < c_n < d_n$ such that $$f_n(\alpha) = \int_{c_n}^{d_n} \frac{\sin \alpha x}{x}$$ and $$f(\alpha) = \lim_n \int_{c_n}^{d_n} \frac{\sin \alpha x}{x}$$ Then $(f_n) \rightrightarrows f$ on $[a, b]$ but not on $[0, b]$


To show $(f_n) \rightrightarrows f$ on $[a, b]$, it suffices to show $(||f_n - f||_\sup) \to 0$

Now $$\begin{align} ||f_n - f||_\sup &= \sup_{\alpha \in [a, b]} |f_n(\alpha) - f(\alpha)| \\ &= \sup_{\alpha \in [a, b]} \left| \int_{c_n}^{d_n} \frac{\sin \alpha x}{x} - \int_0^{+\infty} \frac{\sin \alpha x}{x} \right| \\ &= \sup_{\alpha \in [a, b]} \left| \int_{c_n}^{d_n} \frac{\sin x}{x} - \frac{\pi}{2} \right| \\ &= \left| \int_{c_n}^{d_n} \frac{\sin x}{x} - \frac{\pi}{2} \right| \to 0 \text{ as } n \to +\infty \end{align}$$ The above argument uses change of variables, $\alpha > a > 0$ and the fact that $$\left( \int_{c_n}^{d_n} \frac{\sin x}{x} \right) \to \frac{\pi}{2}$$


It remains to show $(f_n) \not \rightrightarrows f$ on $[0, b]$

Observe that $$f(\alpha) = \begin{cases} \frac{\pi}{2} & \text{ on } (0, b] \\ 0 & \text{ at } 0 \end{cases}$$ is discontinuous. Since uniform limit of continuous functions is continuous, it suffices to show $(f_n)$ are continuous, i.e. for all $\beta \in [0, b], \varepsilon > 0$ we can find $\delta > 0$ such that for all $\alpha \in [0, b] \cap (x - \delta, x + \delta)$ we have $$\left| \int_{c_n}^{d_n} \frac{\sin \alpha x}{x} - \int_{c_n}^{d_n} \frac{\sin \beta x}{x} \right| < \varepsilon$$ Maybe there is a cleverer method but I cannot think of one now. I will leave you to check this. You may need to use $\sin \alpha - \sin \beta = 2 \sin \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2}$ and $|\sin x| \le |x|$ to bound stuff, feel free to ask for help. So we are done!

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