Making a function continuous by choosing a value at a point

Question

Consider the function $h:\mathbb R ^2\to\mathbb R$ where $$ h(x,y)=\begin{cases} \displaystyle \int_0^x\frac\pi x \sin \left (\frac\pi x y\right)\,dy & \text{if } x\neq 0\\[5pt] h(0,y)&\text{if }x=0\end{cases} $$ The task is to choose $h(0,y)$ s.t. $h$ is continuous. My first thought would be choosing $h(0,y)=0$ relating to $$ \int_c^cf=0 $$ What troubles me is that I don't know how to make sense of the expression for $x\neq 0$. $y$ is a parameter of the function $h$ but isn't $y$ also bound by the integral expression. Wouldn't this mean that for $x\neq 0$, $h(x,y)=h(x,y')$ for any $y,y'$ as they don't contribute to the value?

Answer 1

Assuming that the given information is correct (which would be strange because $h(x,y)$ has no function of $y$) we have \begin{align} \int_0^x \frac\pi x\sin\left(\frac\pi xy\right)\,\mathrm{d}y&=-\cos\left(\frac\pi xy\right)\bigg\rvert_0^x\\ &=-\cos\pi+1=2 \end{align} So at every $x$ value (excluding 0) the function is simply constant 2. The choice of $h(0,y)$ must then also be 2. However as I mentioned before, something does appear off about this question since $h$ is not a function of $y$