Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

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

3 Answers 3

up vote 1 down vote accepted

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 $\mathbf{dx}=\mathrm{d}x\ \mathbf{i}+\mathrm{d}y\ \mathbf{j}+\mathrm{d}z\ \mathbf{k}$.

share|improve this answer

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 $$\int_{C}\mathbf{f}\cdot d\mathbf{x}$$ ?

share|improve this answer

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

share|improve this answer
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 $\int \vec f(\vec x)\cdot\mathrm d\vec x$... –  J. M. 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? –  J. M. Sep 22 '11 at 11:02
@J.M. sure, just pointed out that it is not necessary. –  Ilya Sep 22 '11 at 11:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.