1
$\begingroup$

I had the following double integral in my recent math examination:

$$\int_{0}^{a} \int_{0}^{x} \frac{x}{y} \cosh{y} \; dy \, dx$$ where $$a \in \mathbb{R}$$

I tried changing the order of the integrals, however that didn't help much. Here's what I've got after changing the order:

$$\int_{0}^{a} \int_{y}^{a} \frac{x}{y} \cosh{y} \; dx \, dy$$

In both cases I tried integration by parts, however I couldn't solve the resulting integrals. I'm beginning to think that there might have been an error and the actual task should have been:

$$\int_{0}^{a} \int_{x}^{a} \frac{x}{y} \cosh{y} \; dy \, dx$$

In this case, if I change the order of integration the resulting integral can be easily solved.

I would much appreciate any help. Thanks in advance.

$\endgroup$
2
  • $\begingroup$ Ask your teacher. $\endgroup$ May 24 '13 at 9:52
  • $\begingroup$ Maybe it should be $ \sinh $? $\endgroup$
    – Jon Claus
    May 24 '13 at 19:28
0
$\begingroup$

let $u=\frac{x}{y}$ and $v=y$ $$\left |{\frac{\partial (x,y)}{\partial (u,v)}}\right|=v$$ we have $$\int_{0}^{a} \int_{0}^{x} \frac{x}{y} \cosh{y} \; dy \, dx=\int_{0}^{a} \int_{y}^{a} \frac{x}{y} \cosh{y} \; dx \, dy=\int_{0}^{a} \int_{1}^{\frac{a}{v}}u\,v\,cosh(v)du\,dv$$

$\endgroup$
0
$\begingroup$

I'm not sure what the more rigorous way to say this is, but I'm pretty sure the integral as stated doesn't exist. In the vicinity of $y = 0$, $$ \frac{\cosh y}{y} = \frac{1}{y} \left( 1 + \frac{1}{2} y^2 + \dots \right) = \frac{1}{y} + \frac{1}{2} y + \dots, $$ which implies that $$ \int_\epsilon^x \frac{\cosh y}{y} dy = \ln \frac{x}{\epsilon} + \frac{1}{4} (x^2 - \epsilon^2) + \dots $$ and in the limit as $\epsilon \to 0$, this first term will diverge (while none of the other terms will.) Thus, the integral $$ \int_0^x \frac{\cosh y}{y} dy $$ does not converge, and so the double integral as stated doesn't converge either.

$\endgroup$

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.