How do I show that $\int _0^\infty \! \frac{x\cos^3(x)}{t^2+x^2} \, \mathrm{d}x$ converges uniformly for all t > 0?

Question

Until now, I've tried to solve by Dirichlet Criteria and Weierstrass M-test, but couldn't find an answer. Also tried to do inequalities, but didn't get anywhere. Can somebody help me on this please?

Answer 1

According to the Dirichlet test for uniform convergence, the improper integral $\int_0^\infty f(x,t) g(x,t) \, dx$ is uniformly convergent for $t \in D$ if $\int_0^x g(y,t) \, dy$ is uniformly bounded for all $x >0$ and $t \in D$ and $f(x,t) \searrow 0$ as $x \to \infty$ monotonically with respect to $x$ and uniformly for $t \in D$.

In this case, the first hypothesis is true for all $x > 0$ and $t \in (0,\infty)$, since

$$\left|\int_0^x \cos^3y \, dy\right| = \left|\frac{9}{12}\sin x + \frac{1}{12}\sin 3x \right| \leqslant \frac{10}{12}$$

Considering the behavior of $f(x,t) = \frac{x}{x^2 +t^2}$ we first note that $|f(x,t)| \leqslant \frac{1}{x}$ which gives us uniform convergence to $0$ as $x \to \infty$ for all $t \in (0,\infty)$. However, $f(x,t)$ is monotone nonincreasing only when $x \geqslant t$ and we don't fulfill the uniformity of the monotonicity condition for all $t \in (0,\infty)$.

If we restrict the domain to $t \in (0,\alpha]$ for any finite $\alpha > 0$, then we have uniform convergence because

$$\int_0^\infty \frac{x}{x^2 + t^2} \cos ^3 x \, dx = \int_0^\alpha \frac{x}{x^2 + t^2} \cos ^3 x \, dx+ \int_\alpha^\infty \frac{x}{x^2 + t^2} \cos ^3 x \, dx,$$

and the first integral on the RHS is proper and for the second we have all the conditions of the Dirichlet test fulfilled as $x \geqslant \alpha \geqslant t$ for all $t \in (0,\alpha]$.

There remains the question of uniform convergence for all $t > 0$.

Answer 2

Around $x=0$ $$\frac{x\cos^3(x)}{t^2+x^2}=\frac{x}{t^2}-\left(\frac{1}{t^4}+\frac{3}{2 t^2}\right) x^3+O\left(x^4\right)$$ so no problem at the lower bound.

You can also compute the antiderivative writing $$\frac{x}{t^2+x^2}=\frac{x}{(x+it)(x-it)}=\frac{1}{2 (x+i t)}+\frac{1}{2 (x-i t)}$$ and use $$4\cos^3(x)=\cos(3x)+3\cos(x)$$ to face rather simple integrals.