What is the integral of a vector-valued function?

what does it mathematically mean an integral of a vector ? for example in physics we say that the impulse $\vec{I}$ is the time integral of force $\vec{f}$ :

$\vec{I} = \int_{t_{1}}^{t_{2}} \vec{f} dt$

what does this object $~~ \int_{t_{1}}^{t_{2}} \vec{f} dt$ clearly mean ? is it a vector ? does it have the same direction as $\vec{I}$ ?

Solution 1:

In this case, integrating the vector $\vec{f}(t)$ results in a vector $\vec{I}$. There are a couple ways to think about this. Probably the easiest is to think about $\vec{f}$ as having components $f_x,f_y,f_z$:

$$ \vec{f}(t) = \left[\begin{array}{c}f_x(t)\\f_y(t)\\f_z(t)\end{array}\right] $$Each component is a scalar function, so integrating $\vec{f}(t)$ is the same as integrating each component separately:

$$ \int_{t_1}^{t_2}\vec{f}(t)dt = \left[\begin{array}{c}\int_{t_1}^{t_2}f_x(t)dt\\\int_{t_1}^{t_2}f_y(t)dt\\\int_{t_1}^{t_2}f_z(t)dt\end{array}\right] $$ The other way to think about it is using the Riemann sum picture:

$$ \int_{t_1}^{t_2}\vec{f}(t)dt \approx \sum_{j=1}^n (\vec{f}(t_j)-\vec{f}(t_{j-1}))(t_{j}-t_{j-1}) $$This makes it clear that integrating a vector is just like subtracting, adding and scaling vectors, which is something you should know how to do.

There are many different ways to integrate vectors, in fact - this is just one way. For instance, there are line integrals that integrate a vector field along a curve to produce a scalar. Many other types of vector integrals would be covered in a "multivariate calculus" course.

Solution 2:

One has to specify the dimension of the vector space where $f$ takes its value. It is very different between finite and infinite dimensional vector spaces.

Here is essentially a comment by Terry Tao:

To integrate functions taking values in a finite-dimensional vector space, one can pick a basis for that vector space and integrate each coordinate of the vector-valued function separately; this gives a well-defined notion of integral that is independent of the choice of basis.

For functions taking values in infinite-dimensional vector spaces, things get trickier (much as infinite sums become a more subtle topic in infinite-dimensional vector spaces than in finite-dimensional ones, due to the number of different concepts of convergence for such sums). The two most popular notions of integral here are the Bochner integral and the Pettis integral, with the latter being more general (but weaker) than the former.

Solution 3:

Write

$$\vec{f} =\vec{i}f_0+\vec{j}f_1+\vec{k}f_2.$$

Then by linearity of the summation operator

$$\int_{t_0}^{t_1}\vec{f} dt= \int_{t_0}^{t_1}\vec{i}f_0 dt+\int_{t_0}^{t_1}\vec{j}f_1 dt+\int_{t_0}^{t_1}\vec{k}f_2 dt= \vec{i} \int_{t_0}^{t_1}f_0dt+\vec{j}\int_{t_0}^{t_1}f_1 dt+\vec{k} \int_{t_0}^{t_1}f_2dt.$$

This says all.