2
$\begingroup$

If $f(x)\nearrow\infty$ in $[a, \infty)$ and $f'(x)$ is continuous in $[a, \infty)$ then $\int_{a}^{\infty}\frac{f'(x)}{f(x)}\sin(f(x))$ converges

I'm not exactly sure what to try here. I thought of Dirichlets test but I can't seem to identify the right functions to make it work.

$\endgroup$
  • 1
    $\begingroup$ I've never seen that notation before; what does $f(x)\nearrow\infty$ mean? Or did you just mean $f(x)\to\infty$? $\endgroup$ – Adrian Petrescu Feb 9 '11 at 1:52
  • 1
    $\begingroup$ @Adrian: I think it means $f(x)$ is increasing in the interval and $\lim_{x\to\infty}f(x)=\infty$. $\endgroup$ – daniel.jackson Feb 9 '11 at 7:24
3
$\begingroup$

Try a change of variables to reduce it to an integral of a sinc function.

$\endgroup$
  • 1
    $\begingroup$ Thanks. I got $$\int_{f(a)}^{\infty}\frac{\sin t}{t}dt$$ which does converge according to Dirichlets test. Why do I need the information regarding continuity of $f'(x)$ and that $f(x)$ is increasing? $\endgroup$ – user6801 Feb 8 '11 at 19:19
  • $\begingroup$ I don't think that $f$ being increasing is necessary. Continuity of $f'$ is a standard assumption in the proof that integration by substitution works, because the elementary calculus version of FTC is usually done with continuously differentiable functions. The hypotheses required to generalize are more complicated. E.g., I believe it would suffice for $f$ to be absolutely continuous on bounded intervals, as long as you're willing to trade in Riemann integrals for Lebesgue integrals. $\endgroup$ – Jonas Meyer Feb 8 '11 at 19:32
0
$\begingroup$

Integration by parts gives

$$\int_a^b \frac{f'(x)}{f(x)}\sin(f(x)) \, dx = \frac{-\cos(f(x))}{f(x)}\Bigg \vert_a^b - \int_a^b \frac{f'(x)}{f(x)^2}\cos(f(x)) \, dx $$

But now

$$\left| \int_a^b \frac{f'(x)}{f(x)^2}\cos(f(x)) \, dx \right| \le \int_a^b \frac{f'(x)}{f(x)^2} \, dx = \frac{1}{f(x)}\Bigg \vert_b^a$$

Edit: Looking at it again, there might be some trouble with f being zero somewhere...

$\endgroup$
  • $\begingroup$ I thought of integration by parts, but instead tried taking $u=\ln(f(x)),\ v=\sin(f(x))$ but that didn't work. Hadn't thought of this though. $\endgroup$ – user6801 Feb 8 '11 at 19:26
  • $\begingroup$ Yeah, that's certainly the first impulse seein $f'/f$. The possible zero of $f$ shouldnt be a problem btw. Since $\sin(f(x))/f(x)$ can be extended continuously. So you can split up the integral over $[a, \infty)$ in one over $[a,1]$ and $[1, \infty)$ if necessary. $\endgroup$ – Sam Feb 8 '11 at 19:36
  • $\begingroup$ For this to work, note that $f' \ge 0$ is required. (As opposed to the other answer.) $\endgroup$ – Glen Wheeler Feb 9 '11 at 19:37

Your Answer

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