2
$\begingroup$

From Munkres Section 15 Question 2, I am having a problem in proving the integral exists, I know how to compute the integral by repeated integration (provided it does actually exist). The question is as follows:

Compute $$\int_A{\frac{1}{x\sqrt{y}}}dx\, dy,$$ where $$A=\left\{(x,y)\in \mathbb{R}^2|x>1,\ 0<y<1/x\right\}.$$

Now from repeated integration I calculated the definite integral as being 4, and Wolfram Alpha confirmed my answer for me. The fact that it did confirm my answer leads me to believe the fucntion is integrable over A, however I do not know how to prove that, any help would be greatly appreciated.

$\endgroup$
3
$\begingroup$

$f(x,y)=\frac{1}{x\sqrt{y}}$ is measurable and $ \ge 0$ and, by Fubini,

$\int_A{\frac{1}{x\sqrt{y}}}d(x,y)= \int_1^{\infty}(\int_0^{1/x}\frac{1}{x\sqrt{y}}dy)dx$.

The integral on the RHS $=4 < \infty$, hence $f$ is integrable.

$\endgroup$

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.