So I wish to find the value of the integral $\int_{C}f$, where $f = \sin(xy)$ and $C = ([-1,1] \times[-1,1]) \backslash \{(x,y):\|(x,y)\|<1 \}$. So $C$ is basically the closed unit cube minus the unit open ball. So the standard trick is to note that $C \subset A = [-1,1]\times [-1,1]$ and $f$ is certainly a continuous function on $A$. So certainly $\int_{A}f$ exists. Further the boundary of $C$ has measure zero. Therefore $\int_{C}f$ exists. Now if $\psi$ is the characteristic function of $C$, then
\begin{equation} \int_{C}f = \int_{A}f\psi \end{equation}
Now invoking Fubini's theorem, we get that $\int_{A} f \psi = \int_{-1}^{1}(\int_{-1}^1f\psi \ dy) \ dx$. Now the first integral boils down to the following
\begin{equation} \int_{-1}^{1}f\psi \ dy = \int_{-1}^{1}\sin(xy)\psi\ dy = \int_{-1}^{-\sqrt{1-x^2}}\sin(xy)\ dy+ \int_{\sqrt{1-x^2}}^{1}\sin(xy)\ dx \end{equation}
Now
\begin{equation} \int_{\sqrt{1-x^2}}^{1}\sin(xy)\ dx = \frac{-\cos(xy)}{x} \Bigg|_{\sqrt{1-x^2}}^{1} = \frac{\cos(x\sqrt{1-x^2})}{x} - \frac{\cos(x)}{x} \end{equation}
Similarly \begin{equation} \int_{-1}^{-\sqrt{1-x^2}}\sin(xy)\ dy = \frac{\cos(x\sqrt{1-x^2})}{x} - \frac{\cos(x)}{x} \end{equation}
So
\begin{equation} \int_{-1}^{1}f\psi \ dy = 2\left( \frac{\cos(x\sqrt{1-x^2})}{x} - \frac{\cos(x)}{x} \right) \end{equation}
But how do I calculate
\begin{equation} \int_{-1}^{1} \frac{\cos(x\sqrt{1-x^2})}{x} - \frac{\cos(x)}{x} \end{equation}
If my calculations are right, this last integral is not defined. Where have I gone wrong?
For $x=0$ your $\int_{-1}^{-\sqrt{1-x^2}}\cdots+\int^{1}_{\sqrt{1-x^2}}\cdots$ part is just $0$.
Unsurprisingly, it is also the case that $\frac{\cos(x\sqrt{1-x^2})}x-\frac{\cos x}x$ extends continuously to a function such that $g(0)=0$.