Convergence of $\int_0^\infty \frac{\sin(x^2+ax)}{x}dx$ I would like to show the convergence for the integral
$\displaystyle \int_0^{\infty} \frac{\sin(x^2+ax)}{x}dx$.
when $a\geqslant 0$.
I can show convergence at $a=0$. Making the substitution $u=x^2$, I can obtain the  integral $\int_0^\infty \frac{\sin(u)}{u} du$ (This one I can show directly using contour integral of $f(z)=e^{iz}/z$ and with Jordan's lemma)
For the $a>0$, a substitution doesn't make it any much easier and trying to evaluate the integral piecewise seems to be difficult as well.
My naive bound was to say that on the interval $\displaystyle \left[\frac{-a+\sqrt{a^2+8k\pi}}{2},\frac{-a+\sqrt{a^2+(8k+4)\pi}}{2}\right]$, we get the trivial bound $\displaystyle\left|\frac{\sin(x^2+ax)}{x}\right|\leqslant \frac{12}{-a+\sqrt{a^2+8k\pi}}$ and sum these positive parts and neglect the negative ones. 
This naive bound fails unfortunately, as I can only bound it above by the harmonic series (which doesn't help much). 
The other thing I was thinking was to show convergence of the integral 
$\displaystyle \int_0^\infty\frac{e^{ix^2}}{x}e^{iax}dx$ but the numerator makes it tricky to use Jordan's lemma. 
How can I use that $a\geqslant 0$ to show the convergence? (Is it divergent $a<0$).
 A: The integrand has a removable singularity at $x=0$, so that the integral exists on all finite intervals $[0,\alpha]$. Consider thus the remaining part over $[α,\infty)$, for simplicity use $α=1$.
The usual trick now is to use partial integration to get an absolutely converging integral.
$$
\int_1^\infty \frac{\sin(x^2+ax)(2x+a)}{x(2x+a)}dx 
= \left[-\frac{\cos(x^2+ax)}{(2x^2+ax)}\right]_1^\infty 
 - \int_1^\infty\frac{\cos(x^2+ax)(4x+a)}{(2x^2+ax)^2}dx 
$$
Now the integrand is asymptotically $\sim x^{-3}$ using the boundedness of the cosine, and thus absolutely integrable.
For the case of negative $a$, one would have to extend the first part from $[0,1]$ so that $2x+a>0$ for the partial integration. I do not see where this could be problematic.
A: In the case $a=0$ we have
$$ \int_{0}^{+\infty}\frac{\sin(x^2)}{x}\,dx \stackrel{x\mapsto\sqrt{z}}{=} \frac{1}{2}\int_{0}^{+\infty}\frac{\sin z}{z}\,dz = \frac{\pi}{4}.$$
If $a>0$, $x^2+ax$ is an unbounded, increasing function mapping $0$ into itself. By denoting as $g(x)$ the inverse function of $f(x)=x^2+ax$ we have
$$ \int_{0}^{+\infty}\frac{\sin(f(x))}{x}\,dx \stackrel{x\mapsto g(z)}{=}\int_{0}^{+\infty}\frac{\sin(z)g'(z)}{g(z)}\,dz =\int_{0}^{+\infty}\sin(z)\left[2-\frac{2a}{\sqrt{a^2+4z}}\right]\,dz.$$
$\sin(z)$ has a bounded primitive and $2-\frac{2a}{\sqrt{a^2+4z}}$ is decreasing to zero as $z\to +\infty$, so the RHS is convergent by Dirichlet's test. In explicit terms the general integral depends on the Fresnel integrals $C$ and $S$ (and their squares) evaluated at $\frac{a}{\sqrt{2\pi}}$. The general integral is an increasing function of the $a$-variable and
$$ \lim_{a\to +\infty}\int_{0}^{+\infty}\frac{\sin(x^2+ax)}{x}\,dx = \frac{\pi}{2}.$$
