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!

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

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


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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.