I have the following sum:

$$S = \int f_1(x_1) dx_1 + \int f_2(x_2) dx_2 + \int f_3(x_3) dx_3$$

Letting $x = (x_1,x_2, x_3)$ and $f = (f_1(x_1), f_2(x_2), f_3(x_3))$ can I rewrite

$$S = \int f(x)\cdot dx$$

where $\cdot$ is the dot product ?

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You have the sum of 3 single integrals. –  Américo Tavares Sep 22 '11 at 13:40

I have never seen such a notation used for a sum of indefinite integrals of a single variable, only for line integrals.

A line integral of a scalar function $f(x_{1},x_{2},x_{3})$ with respect to $x_{i}$ along an oriented curve $C$ is denoted as

$$\int_{C}f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{i},\qquad i=1,2,3.$$

A sum of line integrals

$$\int_{C}f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{1}+\int_{C}f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{2}+ \int_{C}f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{3}$$

may be written as

$$\int_{C}f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{1}+f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{2}+f(x_{1},x_{2},x_{3})\ \mathrm{d}x_{3}.$$

In your case if you had a sum of 3 line integrals each of a single variable scalar function $f_{i}(x_{i})$, with $i=1,2,3$ you could write $$\begin{eqnarray*} S_{C} &=&\int_{C}f_{1}(x_{1})\ \mathrm{d}x_{1}+\int_{C}f_{2}(x_{2})\ \mathrm{d}x_{2}+ \int_{C}f_{3}(x_{3})\ \mathrm{d}x_{3} \\ &=&\int_{C}f_{1}(x_{1})\ \mathrm{d}x_{1}+f_{2}(x_{2})\ \mathrm{d}x_{2}+f_{3}(x_{3})\ \mathrm{d}x_{3}. \end{eqnarray*}$$

The dot product is a compact way of writing the integrand of a line integral of a vector function. The line integral along an oriented curve with initial point $\left( x_{0},y_{0},z_{0}\right)$ and terminal point $\left( x_{1},y_{1},z_{1}\right)$

$$\int_{C}P\left( x,y,z\right) \ \mathrm{d}x+Q\left( x,y,z\right) \ \mathrm{d}y+R\left( x,y,z\right) \ \mathrm{d}z$$

can be expressed as

$$\int_{C}P\ \mathrm{d}x+Q\ \mathrm{d}y+R\ \mathrm{d}z=\int_{C}\ \mathbf{F}\cdot \mathbf{T}\ \mathrm{d}s$$

or, sometimes,

$$\int_{C}P\ \mathrm{d}x+Q\ \mathrm{d}y+R\ \mathrm{d}z=\int_{C}\mathbf{F}\cdot \mathbf{ds} ,$$

where $\mathbf{F}\left( x,y,z\right) =P\ \mathbf{i}+Q\ \mathbf{j}+R\ \mathbf{k}$ is a vector function with components $P,Q,R$ in the $xyz$ coordinate system, $\mathbf{T}=\frac{\mathrm{d}x}{\mathrm{d}s}\mathbf{i}+\frac{\mathrm{d}y}{\mathrm{d}s}\mathbf{j}+\frac{\mathrm{d}z}{\mathrm{d}s} \mathbf{k}$ is the unit vector tangent to $C$ in the positive sense, $s$ is the arc length along $C$, with $s=0$ at $\left( x_{0},y_{0},z_{0}\right)$ and $s=\ell$ at $\left( x_{1},y_{1},z_{1}\right)$, and $\mathbf{ds}=\mathrm{d}x\ \mathbf{i}+\mathrm{d}y\ \mathbf{j}+\mathrm{d}z\ \mathbf{k}$.

In your case if you had a line integral you could write

$$\int_{C}f_{1}(x_{1})\ \mathrm{d}x_{1}+f_{2}(x_{2})\ \mathrm{d}x_{2}+f_{3}(x_{3})\ \mathrm{d}x_{3}=\int_{C} \mathbf{f}\cdot \mathbf{dx},$$

where $\mathbf{f}\left( x_1,x_2,x_3\right) =f_1(x_1)\ \mathbf{i}+f_2(x_2)\ \mathbf{j}+f_2(x_3)\ \mathbf{k}$, $\mathbf{x}=x_1\ \mathbf{i}+x_2\ \mathbf{j}+x_3\ \mathbf{k}$ and $\mathbf{dx}=\mathrm{d}x\ \mathbf{i}+\mathrm{d}y\ \mathbf{j}+\mathrm{d}z\ \mathbf{k}$.

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Some notation insights can be found here. I think it is a usual notation and will not lead to misunderstanding.

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However, you will want to make sure scalars and vectors are distinguished. In the OP's case, either $\int \mathbf f(\mathbf x)\cdot\mathrm d\mathbf x$ or $\int \vec f(\vec x)\cdot\mathrm d\vec x$... –  Guess who it is. Sep 22 '11 at 10:54
+1, nice observation - though it depends a bit on the context. sometimes there is no reason to make a distinction between vectors and scalars. –  Ilya Sep 22 '11 at 10:59
But here, better safe than sorry, right? –  Guess who it is. Sep 22 '11 at 11:02
@J.M. sure, just pointed out that it is not necessary. –  Ilya Sep 22 '11 at 11:10

Thanks for the detailed examples. I probably need to think more about what I want to write exactly.

Would you then write $$\int_{C}\mathbf{f}\cdot \mathbf{dx}$$ or $$\int_{C}\mathbf{f}\cdot d\mathbf{x}$$ ?

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