A question on the definition of uniform convergence of an improper integral

Question

I am dealing with the integral:

$$\int_{0}^{\infty} e^{-au} \frac{\sin(ut)}{u} du, \quad \alpha \geq 0$$

which according to my book converges uniformly. Could you please explain to me what it is meant by that? A good reference would also help.

The author uses the uniform convergence of the improper integral to justify the interchange of the integral and the limit $\alpha \to 0$. A reference on that result would be helpful as well.

Thank you in advance.

Answer 1

In general, if an improper integral

$$F( \alpha)= \int_0^\infty f(u, \alpha) \, du$$

converges uniformly for $\alpha \geqslant 0$ and $f$ is continuous on $[0,\infty) \times [0,\infty)$, then $F$ is continuous on $[0,\infty)$ and, consequently

$$\lim_{\alpha \to 0}\int_0^\infty f(u, \alpha) \, du = \int_0^\infty \lim_{\alpha \to 0}f(u, \alpha) \, du = \int_0^\infty f(u, 0) \, du . $$

One way to prove this is to notice that convergence of the improper integral implies uniform convergence of the sequence

$$F_n(\alpha) = \int_0^nf(u,\alpha) \, du, \\\lim_{n\to \infty} F_n(\alpha) = F(\alpha).$$

We have the well-know result that a sequence of continuous functions converging uniformly has a continuous limit function.

All of these conditions are satisfied here. Hence, with $t$ fixed,

$$\lim_{\alpha \to 0} F(\alpha) = F(0), \\ \implies \lim_{\alpha \to 0} \int_0^\infty e^{-\alpha u} \frac{\sin (tu)}{u} \, du = \int_0^\infty\frac{\sin (tu)}{u} \, du $$

Uniform convergence for this integral

Uniform convergence follows from Dirichlet's test since $e^{-\alpha u}/u$ is monotone decreasing and uniformly convergent to $0$ as $u \to \infty$ and the integral of $\sin (tu)$ over any interval is uniformly bounded.

For a direct proof we show that the Cauchy criterion is satisfied uniformly, whereby, for any $\epsilon > 0$ there exists $c > 0$ such that for all $c_2 > c_1 \geqslant c$ we have for every $\alpha \geqslant 0$

$$\left|\int_{c_1}^{c_2} e^{-\alpha u} \frac{\sin(tu)}{u} \, du \right| < \epsilon.$$

For this case, in particular, since $e^{-\alpha u}/u$ is decreasing, we can apply the second mean-value theorem for integrals to get for some $\xi$ between $c_1$ and $c_2$,

$$\left|\int_{c_1}^{c_2} e^{-\alpha u} \frac{\sin(tu)}{u} \, du \right| = \left|\frac{e^{-\alpha c_1}}{c_1}\int_{c_1}^{\xi} \sin(tu) \, du \right| \\ \leqslant \frac{1}{c_1}\left|\int_{c_1}^{\xi} \sin(tu) \, du\right| \\ = \frac{1}{c_1}\frac{\left|\cos (t c_1) - \cos (t \xi)\right|}{t} \\ \leqslant \frac{2}{c_1 t}$$

For fixed $t > 0$ the RHS can be made smaller than any $\epsilon > 0$ by choosing $c$ sufficiently large with $c_1 \geqslant c$, independently of $\alpha$. If $t = 0$ then the integral is $0$ and uniform convergence is a trivial consequence. Thus, convergence is uniform for $\alpha \geqslant 0$.

Answer 2

It is meant that for every $\epsilon>0$ there exists a number $M>0$ such that $$\left| \int_M^\infty \mathrm{e}^{-au} \frac{\sin ut}{u} \mathrm{d} u \right|< \epsilon $$ holds for all $\alpha \geq 0$. Here is a reference to the definition.