$\int_0^\infty \frac{\sin(x^2+ax)}{x}\,dx$ converges

Question

I want to show that the following converges for $a\ge 0$ $$\int_0^\infty \frac{\sin(x^2+ax)}{x}\,dx$$

We can show that $\lim_{x\to 0} \frac{\sin(x^2+ax)}{x} = a$, and with that fact it's straightforward to prove that $$\left|\int_0^\delta\frac{\sin(x^2+ax)}{x}\,dx \right|< \infty$$ where $\delta >0$ is a constant.

However I'm having trouble showing that the function decays quickly enough as $x$ goes to infinity. I know how to prove this when taking the integral of $\sin(ax)/x$, but that $x^2$ is causing me problems.

Any ideas? Thanks.

Answer 1

Write as $$\int_0^\infty \frac{1}{x} \frac{2x+a}{2x+a} \sin(x^2+ax) dx$$ and integrate by parts with $dv = (2x+a) \sin(x^2+ax)$.

Answer 2

One approach is to enforce the substitution $x\to \frac{-a+\sqrt{a^2+4x}}{2}$. Then, we have $x^2+ax\to x$ and for $a\ge 0$ we have

$$\int_0^\infty \frac{\sin(x^2+ax)}{x}\,dx=\frac12 \int_0^\infty \frac{\sin(x)}{x}\,\left(\frac{a+\sqrt{a^2+4x}}{x\sqrt{a^2+4x}}\right)\,dx$$

Appealing to Dirichlet's Test for improper integrals, we see that since for all $L$

$$\left|\int_0^L \sin(x)\,dx\right|\le 2$$

and $\frac{a+\sqrt{a^2+4x}}{x^2\sqrt{a^2+4x}}\ge 0$ monotonically approaches zero as $x\to \infty$, the integral of interest converges.