4
$\begingroup$

This is a question from an old exam qualifier:

Show that the improper integral $\int_{-\infty}^{\infty}\cos(x\log\left|x\right|)dx$ is convergent.

I first notice that \begin{equation*} \int_{-\infty}^{\infty}\cos(x^{2}) \le \int_{-\infty}^{\infty}\cos(x\log\left|x\right|) \le \int_{-\infty}^{\infty}\cos(x) \end{equation*} where $\int_{-\infty}^{\infty}\cos(x)$ diverges and $\int_{-\infty}^{\infty}\cos(x^{2}) < \infty$.

Next, a quick check on Wolfram alpha suggests this integral will diverge.

Regardless, I attempt to explore the integral further to see if I can determine for myself whether or not the integral converges.

My attempt is to use integration by parts as done in this post: Using, \begin{equation*} \int_{-\infty}^{\infty}\cos(x\log\left|x\right|)dx = \int_{-\infty}^{-1}\cos(x\log\left|x\right|)dx + \int_{-1}^{1}\cos(x\log\left|x\right|)dx + \int_{1}^{\infty}\cos(x\log\left|x\right|)dx \end{equation*}

I find that \begin{alignat*}{2} \int_{1}^{\infty}\cos(x\log\left|x\right|)dx &= \int_{1}^{\infty}\frac{\log\left|x\right|+1}{\log\left|x\right|+1}\cos(x\log\left|x\right|)dx \\ &= \frac{\sin(x\log\left|x\right|)}{\log\left|x\right|+1}\Big{\vert}_{1}^{\infty} + \int_{1}^{\infty}\frac{\sin(x\log\left|x\right|)}{x(\log\left|x\right|+1)^{2}}dx \\ &= 0 + \int_{1}^{\infty}\frac{\sin(x\log\left|x\right|)}{x(\log\left|x\right|+1)^{2}}dx \end{alignat*} From here, I am not sure how to determine whether this integral converges or diverges. Thanks in advance.

$\endgroup$
5
$\begingroup$

A direct comparison shows that the final integral is convergent, using the fact that $|\sin(x \log |x|) \le 1$; for we have

$$\int_1^{\infty} \left|\frac{\sin{x \log |x|}}{x (\log |x| + 1)^2}\right| dx \le \int_1^{\infty} \frac{1}{x (\log x + 1)^2} dx$$

Setting $u = \log x + 1$ and integrating, the right-hand integral is equal to $1$.

$\endgroup$
  • $\begingroup$ Thanks. After applying your suggestion, it works out perfectly for me. $\endgroup$ – jpb Jun 19 '14 at 2:37
  • $\begingroup$ Glad to hear that. $\endgroup$ – user61527 Jun 19 '14 at 2:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.