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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.

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Ask your teacher. – Gerry Myerson May 24 '13 at 9:52
    
Maybe it should be $ \sinh $? – Jon Claus May 24 '13 at 19:28

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

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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.

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