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Study the uniform convergence of the integral $$\int_0^\infty\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x,\quad \alpha\in(0,+\infty)$$

I have tried the Weierstrass criterion, but $\int_0^\infty\frac{1}{\alpha^2+x^2}\mathrm{d}x$ does not converge. Then I tried the Dirichlet criterion, but $\int_0^A \sin \alpha x$ are not bounded uniformly. So I turned to prove it does not converge uniformly. My attempt are as follows

$\forall A>1$, choose $\alpha=\frac{\pi}{4A}$ then $\forall x\in[A, 2A]$ $$ \left|\int_A^{2A}\frac{\sin \alpha x}{\alpha^2+x^2}\mathrm{d}x\right|>\frac{\sqrt{2}}{2}\frac{1}{\alpha}(\arctan\frac{2A}{\alpha}-\arctan\frac{A}{\alpha}) $$ but I cannot find a lower bound of RHS, so the method seems do not work.

Is my way wrong? You can also suggest your way. Appreciate any help!

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    $\begingroup$ Doesn't $\int \frac{1}{x^2+\alpha^2}$ converge? I think you should recheck that part. $\endgroup$ – Paresseux Nguyen Dec 10 '20 at 2:11
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    $\begingroup$ If we fix any number $a>0$, then the integral $$\int_0^\infty\frac1{a^2+x^2}\,dx$$ converges, giving you uniform convergence on the subinterval $\alpha\in[a,\infty)$. I don't know whether the convergence is uniform on all of $(0,\infty)$. The trouble being what happens near $x=0$. Wait?! Exactly what kind of uniform convergence are we looking for given that there is nothing improper about the integral at $x=0$ when $\alpha>0$? $\endgroup$ – Jyrki Lahtonen Dec 10 '20 at 9:26
  • $\begingroup$ @JyrkiLahtonen: That definately proves uniform convergence for $a > 0$ but not for $\alpha \in (0,\infty)$. The type of uniform convergence we are talking about is uniform convergence of $\int_0^c f(x, \alpha) \, dx \to \int_0^\infty f(x,\alpha) \, dx$ as $c \to \infty$ with respect to $\alpha \in D$. $\endgroup$ – RRL Dec 10 '20 at 20:06
  • $\begingroup$ Thanks for clearing all that @RRL. +1 to your answer. $\endgroup$ – Jyrki Lahtonen Dec 10 '20 at 20:07
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The improper integral

$$I(\alpha) = \int_0^\infty \frac{\sin \alpha x}{\alpha^2 + x^2} \, dx$$

converges uniformly by the Weierstrass-M test on any interval $[A,\infty)$ where $A > 0$ since for all $\alpha > A$,

$$\left|\frac{\sin \alpha x}{\alpha^2 + x^2} \right| \leqslant\frac{1}{A^2 + x^2} \,\,\text{and }\,\, \int_0^\infty\frac{dx}{A^2 + x^2 }< \infty$$

We can also show that convergence is uniform on $(0,A]$ and, hence, on $(0,\infty]$ using the Dirichlet test.

We can rewrite the integral as

$$I(\alpha) = \int_0^\infty \frac{x}{\alpha^2+x^2} \frac{\sin \alpha x}{x} \, dx$$

With the change of variables $t = \alpha x$ we have $$\int_0^c \frac{\sin \alpha x}{x} \, dx = \int_0^{\alpha c} \frac{\sin t}{t} \, dt = \mathrm{Si}(\alpha c) $$

and the integral (see sine integral ) is uniformly bounded for all $\alpha,c \in [0,\infty)$.

We also have the uniform convergence $f(x,\alpha) = \frac{x}{\alpha^2+x^2} \leqslant \frac{1}{x} \to 0$ as $x \to \infty$. Furthermore, $f(x, \alpha)$ is monotonically decreasing with respect to $x$ for all $x \geqslant A$ independent of $\alpha \in (0,A]$. Note that $f(0,\alpha) = 0$, there is a maximum for $x = \alpha$, and then $f(x,\alpha)$ is decreasing to $0$ for all $x > \alpha$. Therefore, the improper integral is uniformly convergent for $\alpha \in (0,A]$.


Note that we only need eventual monotonicity of $f$ (independent of the parameter) to retain applicability of the Dirichlet test. To see this, suppose $f(x,\alpha) \searrow 0$ uniformly and $f(x,\alpha)$ is decreasing with respect $x$ when $x \geqslant A$.

For all $c_2 > c_1 >A$, by the second mean value theorem for integrals (which requires monotonicity of $f$), there exist $\xi \in (c_1,c_2)$ such that

$$\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx = f(c_1,\alpha) \int_{c_1}^\xi g(x, \alpha) \, dx + f(c_2,\alpha) \int_{\xi}^{c_2} g(x, \alpha) \, dx, $$

and

$$\left|\int_{c_1}^{c_2} f(x,\alpha) g(x,\alpha) \, dx\right| \leqslant |f(c_1,\alpha)| \left|\int_{c_1}^\xi g(x, \alpha) \, dx \right|+ |f(c_2,\alpha)| \left|\int_{\xi}^{c_2} g(x, \alpha) \, dx\right|$$

Using the uniform boundedness of the integral of $g$ and the uniform convergence of $f(x,\alpha)$ to zero, we can show that for all $c_2 > c_1$ sufficiently large (and independent of $\alpha \in (0,A]$), the Cauchy criterion for uniform convergence is satisfied.

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You can use Dirichlet's Criterion: Let $f\colon (a, b) \to \mathbb{R}$ a monotone function on $(a, b)$ and $g\colon (a, b)\to \mathbb{C}$ an integrable function in $(a, t]$, for each $t\in (a, b)$. If $\displaystyle \lim_{x\to b} f(x) = 0$, and $\int_{a}^{t} g(x) dx$ is bounded for each $t \in (a, b)$, then $\int_{a}^{\to b} fg $ converges.

Taking $a = 0$, $b = \infty$, $f = \frac{1}{\alpha^2 + x^2}$ and $g = \sin \alpha x$ we have $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha^2 + x^2} \quad \text{converges}$$

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  • $\begingroup$ You need $f(x,\alpha) \to 0$ monotonically and uniformly and for $\int_0^t g(x,\alpha) \, dx$ to be uniformly bounded forall $t, \alpha$ for this the Dirichlet test to imply uniform convergence. That is not the case here as the OP already pointed out. $\endgroup$ – RRL Dec 10 '20 at 20:03

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