Uniformly continuous integral example

Question

Can someone, please, help me with an example of improper integral of some function $$\int_{0}^{+\infty} f(x, y)dx,$$ $c \leq y \leq d$, which is uniformly continuous on $c \leq y \leq d$, but for which we cannot apply Weierstrass M-test?

Answer 1

For $0\le y\le {1\over 2},$ let $f(x,0)=0$ and $$f(x,y)= \begin{cases} -{y\over \log y}, & 0\le x\le {1\over y}\\ 0, & x>{1\over y} \end{cases}$$ for $0<y\le {1\over 2}.$ Then $$\int\limits_0^\infty f(x,0)\,dx=0,\quad \int\limits_0^\infty f(x,y)\,dx=-\int\limits_0^{1/y}{y\over \log y}\,dx =-{1\over \log y},\quad 0<y\le {1\over 2}$$ The function $$g(y):=\int\limits_0^\infty f(x,y)\,dy=\begin{cases}0 & y=0\\ -{1\over \log y} & 0<y\le {1\over 2}\end{cases}$$ is continuous, therefore uniformly continuous on $[0,{1\over 2}].$ Moreover for fixed $x\ge 2$ we have $$\sup_{0\le y\le {1\over 2}}f(x,y)\ge f\left (x,{1\over x}\right )={1\over x\log x} $$ and the function $(x\log x)^{-1}$ is not integrable on $[2,\infty).$ Hence the $M$-test fails.

Answer 2

Take $f:\mathbb{R}^2 \to \mathbb{R}, f(x,y)=1$. Then f is continuous and in particular uniformly continuous but you cannot dominate it with an integrale function.