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.

  • 1
    $\begingroup$ You have a limit for $n\to 0$ of an expression that doesn't contain $n$? $\endgroup$ – Henrik Jul 15 '16 at 20:40
  • $\begingroup$ Your second displayed expression: You should have the absolute values on the inside. $\endgroup$ – zhw. Jul 15 '16 at 22:23

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

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    $\begingroup$ Shouldn't it be $dv=(2x+a)\sin{(x^2+ax)}$? $\endgroup$ – Will Sherwood Jul 15 '16 at 20:49
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    $\begingroup$ When I integrated by parts, I got $$\frac{-\cos(x^2+ax)}{x(2x+a)}\Bigr|_0^\infty - \int_0^\infty \frac{4x+a}{x^2(2x+a)^2}\cos(x^2+ax)$$ The integral on the right converges, but the left term does not. But I suppose if you combined the argument I posted in my question and just use integration by parts on the integral in your post, but with the limits $\delta$ to $\infty$. Thoughts? $\endgroup$ – Kurt Jul 15 '16 at 21:00
  • $\begingroup$ Also, how did you think of introducing those $2x+a$ terms to integrate by parts? Any thought process there? Or just experience? $\endgroup$ – Kurt Jul 15 '16 at 21:00
  • $\begingroup$ @Kurt Indeed, subdividing the integral on as $[0,1]\cup (1\infty)$ solves that problem, as then the two evaluations will be bounded. $\endgroup$ – b00n heT Jul 15 '16 at 21:06
  • $\begingroup$ @Kurt Because $2x+a$ is the derivative of $x^2+ax$ inside the sine, so that way the chain rule is "accounted for" $\endgroup$ – imranfat Jul 15 '16 at 21:42

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.


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