I know from complex analysis that $\int_0^\infty \frac{\cos(x)}{\sqrt{x}}dx = \sqrt{\frac{\pi}{2}}$, so the function is improper Riemann integrable, but I am having trouble showing it without using complex analysis. I think the function is not Lebesgue integrable, but I'm not sure about my proof. $\int_{(n-1)\pi}^{n\pi} \frac{|\cos(x)|}{\sqrt{x}}>\sqrt{\frac{1}{n\pi}}\int_{(n-1)\pi}^{n\pi}|\cos(x)|>2\sqrt{\frac{1}{n\pi}}$, so $\sum_{n=1}^\infty\int_{(n-1)\pi}^{n\pi} \frac{|\cos(x)|}{\sqrt{x}}$ doesn't converge. Does that seem right? How would I show its improper Riemann integrable? Thanks a lot.

  • 2
    $\begingroup$ That's about right. The Lebesgue integral does not exist. $\endgroup$ – Cameron Williams Apr 7 '17 at 3:10
  • $\begingroup$ How would I should its improper Riemann integrable? Any hints would be greatly appreciated. $\endgroup$ – user0617 Apr 7 '17 at 4:09

To show that the integral $I=\int_0^\infty \frac{\cos(x)}{\sqrt{x}}\,dx$ exists as an improper Riemann integral, we proceed as follows. We express the integral of interest as the sum

$$\int_0^L \frac{\cos(x)}{\sqrt x}\,dx=\int_0^1 \frac{\cos(x)}{\sqrt x}\,dx+\int_1^L \frac{\cos(x)}{\sqrt x}\,dx\tag 1$$

The first integral on the right-hand side of $(1)$ is absolutely integrable such that $2\cos(1)\le \int_0^1 \frac{\cos(x)}{\sqrt x}\,dx\le 2$.

For the second integral on the right-hand side of $(1)$, we integrate by parts with $u=x^{-1/2}$ and $v=\sin(x)$ to reveal

$$\int_1^L \frac{\cos(x)}{\sqrt x}\,dx=\left(\frac{\sin(L)}{\sqrt L}-\sin(1)\right)+\frac12\int_1^L \frac{\sin(x)}{x^{3/2}}\,dx\tag 2$$

Inasmuch as the second integral on the right-hand side of $(2)$ absolutely converges as $L\to \infty$, the integral on the left-hand side converges (it does not absolutely converge).

Putting it together, we have shown that the integral in $(1)$ converges and exists as an improper Riemann integral.


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.