What does it mean to show that an integral exists? Question
I am slightly unclear on some terminology that I have come across in the following question:

*

*Let $f(x, t) = xe^{−xt}$ Show that the integral $I(x) = \int_0^{\infty} f(x, t)\ dt $ exists for all $x ≥ 0$.
What does it mean to show that an integral exists? Does this simply mean to show that it converges? I looked this up for some clarification but could only find questions rather than a precise description of what this means.
I would be grateful for any explanation.
 A: The (improper Riemann) integral $\int \limits_0^{\infty} g(t) \, \text{d} t$ is said to exist if:

*

*For each $a > 0$ the proper Riemann integral $\int \limits_0^a g(t) \, \text{d} t$ exists, i.e. $g$ is Riemann integrable over $[0, a]$;

*The limit $\lim \limits_{a \to \infty} \int \limits_0^a g(t) \, \text{d} t$ exists.

The integral $\int \limits_0^{\infty} g(t) \, \text{d} t$ is then the value of the limit. Now for your question just apply the definition to $g(t) := f(x, t)$, where $x \geqslant 0$ is fixed.
A: The question is about if the mapping $\displaystyle x\mapsto \int_{0}^{+\infty}xe^{-xt}\, {\rm d}t$ exists over the interval $[0,+\infty]$. So we can answer two question:

*

*Does it exists if $x=0$?


*Does it exists if $x>0$?
If the answer is "yes" in both cases, then the mapping exists over the interval $[0,+\infty]$.
Well,

*

*If $x=0$, so $\displaystyle \int_{0}^{+\infty}0e^{-0\cdot t}\, {\rm d}t=0$ so there exists and equals to the mapping zero.


*If $x>0$, so $\displaystyle \int_{0}^{+\infty}xe^{-xt}\, {\rm d}t=x\lim_{b\to +\infty}\int_{0}^{b}e^{-xt}\, {\rm d}t=x\left(\frac{1}{x}\right)=1$ so there exists and equals to the mapping one.
Therefore since both cases the answer is "yes" we can say: the mapping $\displaystyle x\mapsto \int_{0}^{+\infty}xe^{-xt}\, {\rm d}t$ exists over all $[0,+\infty]$" and converges to the function $I(x)={\bf 1}_{\{x>0\}}$ or strictly saying to $\displaystyle x\mapsto \begin{cases} 1, \quad \text{if}\quad x\in [0,+\infty]\\ 0,\quad \text{if}\quad x=0\end{cases}$
A: The integral exists as an element of $[0, \infty]$ because the integrand is nonnegative and measurable. But sometimes people use "$\int f$ exists" to mean that $\int |f| < \infty$. At other times, people use "$\int f$ exists" to mean that some kind of problem specific limit exists. For example, $\int_{0}^{\infty}\frac{\sin x}{x}\,dx$ does not exist in the first sense, but does exist in the second sense.
