Measurability/Lebesgue Integration

[Ted123]

Homework Statement

Homework Equations

For part (a):

The Attempt at a Solution

I think I've done all of this question except for the very last part of (c).

For (a) I've found the integral $$\int \frac{x^2-y^2}{(x^2+y^2)^2}\;dx$$ using Wolfram Alpha but how would I integrate it bare handedly?

Since the double integral wrt x then y does not equal the double integral wrt y then x, f is not Lebesgue integrable over [0,1]x[0,1]. However if 0<a<b then f is Lebesgue integrable over [a,b]x[a,b] and the integral is 0 by Fubini.

The last part of (c) is what I'm not sure about: deducing that $$\int f = |S^+| - |S^-|.$$ Am I going about it the right way? We can write $f=f^+ - f^-$ where $f^+ = \max (f,0)$ and $f^- = -\min (f,0)$ (i.e. the +ve and -ve parts of f respectively).

Then we know if $f\in L^1 (\mathbb{R})$ then $f^+ , f^- \in L^1 (\mathbb{R})$ and $f^+ , f^- \geq 0$.

Now is the following correct? $$\int f = \int (f^+ - f^-) = \int f^+ - \int f^- = |S^+| - |S^-|$$

 

[morphism]

For (a) I've found the integral $$\int \frac{x^2-y^2}{(x^2+y^2)^2}\;dx$$ using Wolfram Alpha but how would I integrate it bare handedly?

If guess you could notice that it might be d/dx of something over (x^2+y^2) (because of the quotient rule), and then guess what the numerator should be.

Or, you could proceed as follows: $$\begin{align*} \int \frac{x^2-y^2}{(x^2+y^2)^2}\;dx &= \int \frac{x^2}{(x^2+y^2)^2}\;dx - y^2 \int \frac{1}{(x^2+y^2)^2}\;dx \\
&= \int \frac{1}{x^2+y^2} - \frac{y^2}{(x^2+y^2)^2}\;dx - y^2 \int \frac{1}{(x^2+y^2)^2}\;dx \\
&= \int \frac{dx}{x^2+y^2} - 2y^2 \int \frac{1}{(x^2+y^2)^2}\;dx. \end{align*}$$

To deal with the second integral on the RHS, all you need to know is how to integrate
$$ \int \frac{dv}{(v^2+1)^2}. $$ But this is easy (though slightly tedious). E.g. start off with the sub v=tan(t).

The last part of (c) is what I'm not sure about: deducing that $$\int f = |S^+| - |S^-|.$$ Am I going about it the right way? We can write $f=f^+ - f^-$ where $f^+ = \max (f,0)$ and $f^- = -\min (f,0)$ (i.e. the +ve and -ve parts of f respectively).

Then we know if $f\in L^1 (\mathbb{R})$ then $f^+ , f^- \in L^1 (\mathbb{R})$ and $f^+ , f^- \geq 0$.

Now is the following correct? $$\int f = \int (f^+ - f^-) = \int f^+ - \int f^- = |S^+| - |S^-|$$

Yes, of course. This follows from the previous part applied to $f^+$ and $f^-$.