contradiction to Fubini's theorem

Question

The function $f(x) = cos(xy)$ is continuous on the rectangle $ R = \displaystyle {[0, \frac{\pi}{2}] \times [0, \frac{\pi}{2}]}$ so $ \displaystyle \int\int _{R}\cos(xy)dA$ is defined and equal to the iterated integral $\displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \cos(xy)dxdy$ by Fubini's Theorem. However the innermost integral $\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(xy)dx$ evaluates to $\frac{\sin(\frac{\pi}{2}y)}{y}$ which is undefined when $y=0$. Can someone help explain what is going here and why this is not a contradiction to Fubini's Theorem. What I'm thinking is that this has to do with improper integrals possibly and $\displaystyle \int \frac{\sin(x)}{x}dx$ but i'm not sure. As a small psa I don't know anything about measure theory. Any help would be greatly appreciated.

Answer 1

When $y=0,$ then $xy=0$ for all $x$, and therefore $\cos(xy)=1.$

The formula you gave for the integral only applies when $y\neq0.$ For $y=0$ the inner integral is $\frac\pi2.$

Answer 2

The set of $y$ with $y=0$ has measure $0$ and so there is no problem whatsoever. For Lebesgue integral it makes no difference as to whether you are integrating over an open interval or a closed interval.