# Value of $\int_{x=0}^1\int_{y=\sin^{-1}x}^{\pi/2}f(x,y)\,\,dydx$ when $f(x,y)=\begin{cases}\frac{y}{\sin y} ,&{y\ne0}\\1,&y=0 \end{cases}$

Let $f:\Bbb R^2\rightarrow \Bbb R^2$ be defined by $$f(x,y)=\begin{cases}\frac{y}{\sin y} ,&{y\ne0}\\1,&y=0 \end{cases}$$ Then find the value of the integral $$\int_{x=0}^1\int_{y=\sin^{-1}x}^{\pi/2}f(x,y)\,\,dydx$$

My attempt: I tried changing the order of integration and got $$\int_{y=0}^{\pi/2}\int_{0}^{\sin y}f(x,y)\,\,dxdy$$ but how do I consider the two cases separately when $y=0$ and $y\ne0$?

How do I choose the limits of integration?

Perhaps you can calculate in the way that $\lim_{N\rightarrow\infty}\displaystyle\int_{1/N}^{\pi/2}\int_{0}^{\sin y}f(x,y)dxdy$.

In fact, the question is that why one can do changing the order of integration from $\displaystyle\int_{0}^{1}\int_{\sin^{-1}x}^{\pi/2}f(x,y)dydx$ because here the function $\varphi$ defined by $x\rightarrow\displaystyle\int_{\sin^{-1}x}^{\pi/2}f(x,y)dy$ has some "problem" at $x=0$. However, all these worries can be overcome if we deal with Lebesgue integral, and the Fubini Theorem of Lebesgue integral version is crucial in this issue.

The following are the details:

First show that \begin{align*} \int_{0}^{1}\int_{\sin^{-1}x}^{\pi/2}|f(x,y)|dydx<\infty, \end{align*} this is not hard, because $f$ is bounded on that region. And the region is bounded as well.

Now Fubini Theorem (Lebesgue integral version) says that \begin{align*} \int_{0}^{1}\int_{\sin^{-1}x}^{\pi/2}f(x,y)dydx=\int_{0}^{\pi/2}\int_{0}^{\sin y}f(x,y)dxdy. \end{align*}

Now Lebesgue integral does not matter in a.e. difference, so \begin{align*} \int_{0}^{\pi/2}\int_{0}^{\sin y}f(x,y)dxdy=\int_{0}^{\pi/2}\int_{0}^{\sin y}\dfrac{\sin y}{y}dxdy. \end{align*}

However, there seems to be a singularity at $y=0$ and by Lebesgue Dominated Convergence Theorem one can have \begin{align*} \int_{0}^{\pi/2}\int_{0}^{\sin y}\dfrac{\sin y}{y}dxdy=\lim_{N\rightarrow\infty}\displaystyle\int_{1/N}^{\pi/2}\int_{0}^{\sin y}f(x,y)dxdy. \end{align*}

$f(x,y)$ is independent of $x$ and $y=0$ is measure $0$.

\begin{align}&\int_{y=0}^{\pi/2}\int_{0}^{\sin y}f(x,y)\,\,dxdy \\&= \int_{y=0}^{\frac{\pi}2} f(y)\sin y \, dy \\ &=\int_{y=0}^\frac{\pi}{2} y \, dy\\ &= \frac{1}{2}\left( \frac{\pi}{2}\right)^2\end{align}

Also, note that at $y=0$, $\sin y = y$.