The convergence of an integral, dependant on a parameter I have to say if the integral
$$\int_0^\infty \frac{e^{-xy}}{1+x}\,dx$$
that depends on the parameter $y$ is convergent (I don't know the right terminology, since I'm not studying in english, but I will provide the criterion for such convergence) when 1) $y\in (0,1)$ and 2) $y\in [1,\infty]$. The criterion we use to determine such convergence is as follows:
If
$$\lim_{b\rightarrow +\infty}\sup_{y\in D}\left | \int_b^\infty \frac{e^{-xy}}{1+x} \, dx \right |=0$$
where $D$ is the domain of parameter $y$ then we say that the integral converges properly when $y$ is in that domain. The problem is, I have no idea how to integrate that function. Any ideas? Thank you!
 A: The terminology you are looking for is uniform convergence, and you have provided a necessary and sufficient condition for the uniform convergence of an improper integral depending on a parameter $y \in D$, viz.
$$\lim_{b \to \infty}\sup_{y \in D} \left|\int_b^\infty f(x,y)\, dx\right| = 0$$
In case (2), we have $\left| \frac{e^{-xy}}{1+x}\right| \leqslant e^{-x}$ for all $y \in [1,\infty)$. Since $\int_0^\infty e^{-x} \, dx = 1 < \infty$ we have uniform convergence of the improper intergral for $y \in [1,\infty)$ by the Weierstrass M-test.
In case (1), we have
$$\sup_{y \in (0,1)} \left|\int_b^\infty \frac{e^{-xy}}{1+x}\, dx\right| \geqslant \sup_{y \in (0,1)}\int_b^{2b} \frac{e^{-xy}}{1+x}\, dx \geqslant \sup_{y \in (0,1)} b \cdot \frac{e^{-2by}}{1+2b} \geqslant \underbrace{\frac{b}{1+2b}e^{-2}}_{y = 1/b}, $$
and, hence,
$$\lim_{b\to \infty}\sup_{y \in (0,1)} \left|\int_b^\infty \frac{e^{-xy}}{1+x}\,dx \right|\geqslant \lim_{b \to \infty}\frac{b}{1+2b}e^{-2} = \frac{e^{-2}}{2} \neq 0$$
Therefore, convergence is not uniform for $y \in (0,1)$.
A: I would adress the problem this way: if $y=0$ then the integrand behaves like $$g(x)=\frac{1}{1+x}$$ (you can try to integrate $g$ and use the definition of indefinite integral) and therefore the integral diverges. If $y>0$, since $$\lim_{x \to \infty} \frac{e^{-xy} x^2}{1+x} =0\text{,}$$ then there is $M$ such that for each $x > M$ then
$$\frac{e^{-xy}}{1+x}<\frac{1}{x^2}\text{.}$$
Thus the integral converge (you can try to apply the defintion to $x^{-2}$ and use a suitable squeeze theorem. Do you know about that?). Finally, if $y<0$ then you are integrating
$$\frac{e^{\vert y \vert x}}{1+x}$$ which surely diverges! (Think about that and feel free to ask if you are not sure). Hope this helped.
