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

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!

• Doesn't $\int \frac{1}{x^2+\alpha^2}$ converge? I think you should recheck that part. – Paresseux Nguyen Dec 10 '20 at 2:11
• 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 $\alpha>0$? – Jyrki Lahtonen Dec 10 '20 at 9:26
• @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 $\alpha \in D$. – RRL Dec 10 '20 at 20:06
• Thanks for clearing all that @RRL. +1 to your answer. – Jyrki Lahtonen Dec 10 '20 at 20:07

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.

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 $$\int_{0}^{\infty} \frac{\sin \alpha x}{\alpha^2 + x^2} \quad \text{converges}$$

• 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. – RRL Dec 10 '20 at 20:03