I'm trying to solve the following integral:
$$\iint_{R} \left| \frac{y \cos{y}}{\sqrt{x}} \right| dA$$
where R is the rectangle $[0,1] \times [0,3]$.
Using Tonelli's theorem, I can say the following:
$$\iint_{R} \left| \frac{y \cos{y}}{\sqrt{x}} \right| dA = \int_{0}^{1} \int_{0}^{3} \left| \frac{y \cos{y}}{\sqrt{x}} \right| dy dx \int_{0}^{3} \int_{0}^{1} \left| \frac{y \cos{y}}{\sqrt{x}} \right| dx dy $$
I've tried to split the region into positive and negative parts, since when $y$ is less than $\frac{\pi}{2}$, cosine is positive, whereas when it is greater than $\frac{\pi}{2}$ and less than 3, it is negative.
However, I don't know how exactly to deal with the unboundedness of the function on the line $x=0$ (and the points when $y=0$ and $y=\frac{\pi}{2}$ on this line). I know that Fubini's/Tonelli's theorems work for improper integrals too, where the function becomes unbounded) as long as the condition are met (i.e. Lebsgue integrability for Fubini and non-negativity for Tonelli), but that is when one point blows up to infinity, whereas this is a whole line, so I don't really know what difference that would make).
Any help on breaking down this absolute value integral would be appreciated.
Best Answer
Tonelli's Theorem states that if $X$ and $Y$ are $\sigma$-finite measure spaces and $f:X\times Y\to[0,\infty]$ is measurable then $\int_X\int_Yf=\int_Y\int_Xf=\int_{X\times Y}f$ (even if those integrals are infinite). Clearly, $[0,1]$ and $[0,3]$ are $\sigma$-finite, and your $f$ is nonnegative by definition. The fact that $f=\infty$ on a line segment makes no difference.
So $\int_0^1\frac{1}{\sqrt{x}}\,dx=2$. Integration by parts gives us $\int y\cos(y)\,dy=\cos(y) + y\sin(y)$, and then you break up the definite integral as needed to account for the absolute value.