Show that double integral does not exist

[bombz]

Homework Statement

Here are my problems.[PLAIN]http://img256.imageshack.us/img256/2254/whatua.jpg

Homework Equations

possible conversions to polar coordinates but I doubt that's needed. Fubini's theorem?

The Attempt at a Solution

So for the h(x,y) integral does not exist, I got this:

i took the first integral with respect to y and I came up with (Integral)(bound 0 to infinity) of [LN (1+x^2 +y^2)/(2*y)] evaluated from 0 to infinity. As that is not possible, the limit does not exist.. correct? I have no clue how to show it properly.

The second part, I have no clue at all. The stupid piecewise functions i do not understand

 

[Mark44]

Homework Statement

Here are my problems.


[PLAIN]http://img256.imageshack.us/img256/2254/whatua.jpg

Homework Equations

possible conversions to polar coordinates but I doubt that's needed. Fubini's theorem?

The Attempt at a Solution

So for the h(x,y) integral does not exist, I got this:

i took the first integral with respect to y and I came up with (Integral)(bound 0 to infinity) of [LN (1+x^2 +y^2)/(2*y)] evaluated from 0 to infinity. As that is not possible, the limit does not exist.. correct? I have no clue how to show it properly.

This looks wrong to me. I think you are trying to say something like this:$$\int \frac{dx}{1 + x^2} = ln(1 + x^2)$$

It is not at all true that the integral of 1/<whatever> = ln(whatever).

The second part, I have no clue at all. The stupid piecewise functions i do not understand

 

[jackmell]

I immediately think polar coordinates when I see that:

$$\int_0^{\infty}\int_0^{\pi/2} \frac{r}{1+r^2}drdt$$

$$\pi/2\int_0^{\infty} \frac{r}{1+r^2}dr$$

$$\pi/4\int_1^{\infty}\frac{1}{u}du$$

$$\pi/4(\infty-0)$$

bingo-bango.

 

[Johnson04]

You may also integrate in the following way:

$$I = \int\limits_{0}^{\infty}\frac{1}{\sqrt{1+y^2}}\int\limits_{0}^{\infty}\frac{1}{1+(\frac{x}{\sqrt{1+y^2}})^2}\frac{1}{\sqrt{1+y^2}}dxdy$$

$$= \int\limits_{0}^{\infty}\frac{1}{\sqrt{1+y^2}}[\arctan(\frac{x}{\sqrt{1+y^2}})]|_{0}^{\infty}dy$$

$$= \frac{\pi}{2}\int\limits_{0}^{\infty}\frac{1}{ \sqrt{1+y^2}}dy$$

Let $y = \tan\theta$, then $\frac{1}{\sqrt{1+y^2}} = \cos\theta$, thus

$$I = \frac{\pi}{2}\int\limits_{0}^{\infty}\cos\theta \sec\theta \tan\theta d\theta$$

$$= \frac{\pi}{2}\int\limits_{0}^{\infty}\tan\theta d\theta = -\frac{\pi}{2}[\ln\cos\theta]|_{0}^{\frac{\pi}{2}} = -\frac{\pi}{2} [-\infty - 0]$$ Done.