Understanding the definition of a Integral 
Definition: Let $X$ be a Banach space and $I$ the identity operator on $X$. A family $\{T(t)\}_{t\geq 0}$ of bounded linear operators from $X$ into $X$ is a semigroup of bounded linear operator on $X$ if
(i) $T(0)=I$;
  (ii) $T(t + s)= T(t)T(s)$ for every $t,s\geq 0$.

Now my question: What's the definition of the integral
$$ \int_0^{h} T(s) \, \text{d}s, \ \text{for} \ h > 0?$$
I don't know how to imagine this integral because for each $s \geq 0$, $T(s)$ is a linear bounded operator.
 A: You can use the usual Riemann sum definition; for example, see Lang's Analysis, or Lorch's Spectral Theory. The semigroup aspect doesn't really matter; all you need
is a continuous map from an interval $[a,b]$ to a Banach space $E$. (The space of bounded linear operators is a Banach space.)
A few details, taken from Lang (his discussion is more detailed than Lorch's). First we define step maps; these are maps where you partition $[a,b]$, $a=a_0\leq a_1\leq\ldots\leq a_n=b$, and the step map (call it $f$) is constant on each subinterval $[a_i,a_{i+1}]$. Then define the integral of $f$ with respect to a partition $P$, $$\int_P f = \sum_{i=1}^n (a_i-a_{i-1})w_i$$ where $w_i$ is the value of $f$ on the subinterval $[a_i,a_{i-1}]$. Lang then defines refinement of partitions, shows that refining a partition does not change the above sum, and concludes (by looking at common refinements) that the integral of a step function is independent of the partition.
Let $\mathcal{S}$ be the vector space of step functions from $[a,b]$ to $E$. The integral, as defined above, is a linear map from $\mathcal{S}$ to $E$. It can be extended to the closure of $\mathcal{S}$ in the Banach space of continuous functions from $[a,b]$ to $E$ (under the uniform norm) in a pretty straightforward manner. Lang gives the details in general, and dubs this the Linear Extension Theorem.
A: Thank you for your answer. I will try to write it down with my own words, I hope it is correct:
If $X$ is a banach-space, $[a,b] \subset \mathbb{R}$ and $f \colon [a,b] \longrightarrow X$  is a banach-space-valued function I understand how to define the integral $$\int_{a}^{b} f(x) \, \text{d}x. $$
It is - as you say - the usual Riemann sum definition. Now I am not sure if I am correct - but I would say: 
If we define $\mathcal{X}$ as the banach space of bounded linear operators on an other banach space $X$ we can define the function
$$ F_T \colon [a,b] \longrightarrow \mathcal{X} \colon s \longmapsto F_T (s):=T(s) $$
where for each $s \in [0,\infty)$, $T(s)$ is a linear operator from $X \longrightarrow X$, and where $X$ is another banach-space. 
So we are allowed to define the integral 
$$\int_{a}^{b} F_T(s) \, \text{d}s=\int_{a}^{b} T(s) \, \text{d}s =:I.$$
Now if I see $I$ as an operator from $X \longrightarrow X$, I am not sure how to interpret it. I thought it might be:
$$ I \colon X \longrightarrow X \colon x \longmapsto Ix:=I(x):=\int_{a}^{b} T(s) \, \text{d}s(x)=\int_{a}^{b} T(s)(x) \, \text{d}s.$$
Then the question remains: What do we mean with $\int_{a}^{b} T(s)(x) \, \text{d}s$.
$\textbf{EDIT:}$
Which norm $\| \cdot \|$ should be used on the banach space of bounded linear operators $\mathcal{X}$ ? I think it should be the operator-norm.
