Uniform convergence of $\int_0^\infty e^{-xt}\,\mathrm{d}t$ Does
$$
\int_0^\infty e^{-xt}\,\mathrm{d}t
$$
converge uniformly to $1/x$ on the set $A=(0,\infty)$? Our definition of uniform convergence is the following.
Definition. Let $A\subseteq\mathbf R$. Given $f(x,t)$ defined on $A\times[c,\infty)\subseteq \mathbf R^2$, assume $F(x) = \int_c^\infty f(x,t)\,\mathrm dt$ exists for all $x\in A$. We say the improper integral converges uniformly to $F(x)$ on $A$ if for all $\varepsilon>0$, there exists $M>c$ such that
$$
\bigg|F(x)-\int_c^df(x,t)\,\mathrm dt\bigg|<\varepsilon
$$
for all $d\ge M$ and all $x\in A$.
What I have tried: We have that $F(x) = \int_0^\infty e^{-xt}\,\mathrm dt = 1/x$. If $f(x,t) = e^{-xt}$, then
$$
\bigg|F(x)-\int_0^df(x,t)\,\mathrm dt\bigg| = \frac{1}{xe^{xd}}.
$$
If $x$ is confined to some subinterval such as $[1/2,\infty)$ then the convergence is uniform since
$$
\frac{1}{xe^{xd}} \le \frac{2}{e^{d/2}},
$$
which can be made as small as we'd like, independently of $x\in [1/2,\infty)$ for sufficiently large $d$. Part of my confusion stems from not knowing how to properly negate the definition of uniform convergence. I would phrase the definition in terms of quantifiers as
$$
\forall\varepsilon>0,\:\exists M>0 \:\text{such that}\: d\ge M\:\text{and}\: x\in A\implies \bigg|F(x)-\int_c^df(x,t)\,\mathrm dt\bigg|<\varepsilon,
$$
so that its negation would be
$$
\exists\varepsilon>0,\:\forall M>0, \:d\ge M\:\text{and}\: x\in A\:\text{but}\:\bigg|F(x)-\int_c^df(x,t)\,\mathrm dt\bigg|\ge\varepsilon.
$$
So, it seems that since $x\in(0,\infty)$ can take on arbitrarily small values, $1/xe^{xd}$ can take on arbitrarily large values, and the convergence is not uniform, but it is hard to evaluate since we can also make $d$ arbitrarily large, if I am not mistaken about the negation of our definition. Help would be appreciated.
 A: The negation of uniform convergence is stated as follows.

The convergence of $\displaystyle F(x)=\int_0^\infty f(x,t)\,dt$ fails to be uniform if there exists a number $\displaystyle  \epsilon>0$, such that corresponding to an arbitrary number $\displaystyle  M\in (0,\infty)$, there exists a number $\displaystyle  d\in [M,\infty)$ and a number $\displaystyle  x\in A$ such that
$$\left|\int_d^\infty f(x,t)\,dt\right|\ge \epsilon$$


Here, we have $f(x,t)=e^{-xt}$ so that
$$\left|\int_d^\infty f(x,t)\,dt\right|=\frac{e^{-dx}}{x}$$
We take $\epsilon=e^{-1}$, $d\ge \max(1,M)$ and $x=1/d$.  Then, no matter how large $M$ is, we have
$$\frac{e^{-dx}}{x}=de^{-1}\ge e^{-1}=\epsilon$$
And we are done!
A: Investigate the right hand side of this for different values of c and d:
$\left|F\left(x\right)-\int_{c}^{d}e^{-xt}dt\right|=\left|\int_{0}^{c}e^{-xt}dt+\int_{d}^{\infty}e^{-xt}dt\right|$
The c term clearly converges uniformly (the integrand is uniformly bounded in t on [0,c] for all x>0). The question is, does the same hold for the d term (the tail)?   
