Template:Calculus In mathematics, a line integral (sometimes called a path integral or curve integral; not to be confused with calculating arc length using integration) is an integral where the function to be integrated is evaluated along a curve.

The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighing distinguishes the line integral from simpler integrals defined on intervals. Many simple formulae in physics (for example, W=F·s) have natural continuous analogs in terms of line integrals (W=∫C F· ds). The line integral finds the work done on an object moving through an electric or gravitational field, for example.

## Vector calculus

In qualitative terms, a line integral in vector calculus can be thought of as a measure of the total effect of a given field along a given curve. More specifically, the line integral over a scalar field can be interpreted as the area under the field carved out by a particular curve. This can be visualised as the surface created by z = f(x,y) and a curve C in the x-y plane. The line integral of f would be the area of the "curtain" created when the points of the surface that are directly over C are carved out.

#### Definition

For some scalar field f : URnR, the line integral along a piecewise smooth curve CU is defined as

${\displaystyle \int _{C}f\,ds=\int _{a}^{b}f(\mathbf {r} (t))|\mathbf {r} '(t)|\,dt.}$

where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C.

The function f is called the integrand, the curve C is the domain of integration, and the symbol ds may be intuitively interpreted as an elementary arc length. Line integrals of scalar fields over a curve C do not depend on the chosen parametrization r of C.

#### Derivation

For a line integral over a scalar field, the integral can be constructed from a Riemann sum using the above definitions of f, C and its parametrization r(t). This can be done by partitioning the interval [a,b] into n intervals of length Δt = (ba)/n. Letting ti be the ith point on [a,b], then r(ti) gives us the position of the ith point on the curve. The results so far, is that we have approximated the curve C by a polygonal path, where the straight line pieces are joined at the points r(ti). We then label the distance from the subsequent points on the curve as Δsi. Evaluating f at all the points on the curve and multiplying it by the length of each partition gives us the area of each partition of f on C. Taking the limit of the sum as the size of the partitions go to zero gives us

${\displaystyle I=\lim _{\Delta t\rightarrow 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))\Delta s_{i}}$

We note that the distance between subsequent points on the curve, is

${\displaystyle \Delta s_{i}=|\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})|=|\mathbf {r} '(t_{i})|\Delta t}$

Substituting this in to our above Riemann sum yields

${\displaystyle I=\lim _{\Delta t\rightarrow 0}\sum _{i=1}^{n}f(\mathbf {r} (t_{i}))|\mathbf {r} '(t_{i})|\Delta t}$

which is the Riemann sum for the integral

${\displaystyle I=\int _{a}^{b}f(\mathbf {r} (t))|\mathbf {r} '(t)|\,dt.}$

### Line integral of a vector field

#### Definition

For a vector field F : URnRn, the line integral along a piecewise smooth curve CU, in the direction of r, is defined as

${\displaystyle \int _{C}\mathbf {F} (\mathbf {r} )\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt.}$

where · is the dot product and r: [a, b] → C is a bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C.

A line integral of a scalar field is thus a line integral of a vector field where the vectors are always tangential to the line.

Line integrals of vector fields are independent of the parametrization r in absolute value, but they do depend on its orientation. Specifically, a reversal in the orientation of the parametrization changes the sign of the line integral.

#### Derivation

The trajectory of a particle along a curve inside a vector field. At the bottom are the vectors of the field seen by the particle as it travels along the curve. The sum of the dot products of these vectors with the tangent vector of the curve at each point of the trajectory results in the line integral.

The line integral of a vector field can be derived in a very similar manner as in the case of a scalar field. Again using the above definitions of F, C and its parametrization r(t), we construct the integral from a Riemann sum. Partition the interval [a,b] into n intervals of length Δt = (ba)/n. Letting ti be the ith point on [a,b], then r(ti) gives us the position of the ith point on the curve. However, instead of calculating up the distances between subsequent points, we need to calculate their displacement vectors, Δsi. As before, evaluating F at all the points on the curve and taking the dot product with each displacement vector which gives us the infinitesimal contribution of each partition of F on C. Letting the size of the partitions go to zero gives us a sum

${\displaystyle I=\lim _{\Delta t\rightarrow 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \Delta \mathbf {s} _{i}}$

We see that the displacement vector between subsequent points on the curve is

${\displaystyle \Delta \mathbf {s} _{i}=\mathbf {r} (t_{i}+\Delta t)-\mathbf {r} (t_{i})=\mathbf {r} '(t_{i})\Delta t}$

Substituting this in to our above Riemann sum yields

${\displaystyle I=\lim _{\Delta t\rightarrow 0}\sum _{i=1}^{n}\mathbf {F} (\mathbf {r} (t_{i}))\cdot \mathbf {r} '(t_{i})\Delta t}$

which is the Riemann sum for the integral defined above.

### Path independence

If a vector field F is the gradient of a scalar field G, that is,

${\displaystyle \nabla G=\mathbf {F} ,}$

then the derivative of the composition of G and r(t) is

${\displaystyle {\frac {dG(\mathbf {r} (t))}{dt}}=\nabla G(\mathbf {r} (t))\cdot \mathbf {r} '(t)=\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)}$

which happens to be the integrand for the line integral of F on r(t). It follows that, given a path C , then

${\displaystyle \int _{C}\mathbf {F} (\mathbf {r} )\cdot \,d\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\cdot \mathbf {r} '(t)\,dt=\int _{a}^{b}{\frac {dG(\mathbf {r} (t))}{dt}}\,dt=G(\mathbf {r} (b))-G(\mathbf {r} (a)).}$

In other words, the integral of F over C depends solely on the values of G in the points r(b) and r(a) and is thus independent of the path between them.

For this reason, a line integral of a vector field which is the gradient of a scalar field is called path independent.

### Applications

The line integral has many uses in physics. For example, the work done on a particle traveling on a curve C inside a force field represented as a vector field F is the line integral of F on C.

## Complex line integral

The line integral is a fundamental tool in complex analysis. Suppose U is an open subset of C, γ : [a, b] → U is a rectifiable curve and f : UC is a function. Then the line integral

${\displaystyle \int _{\gamma }f(z)\,dz}$

may be defined by subdividing the interval [a, b] into a = t0 < t1 < ... < tn = b and considering the expression

${\displaystyle \sum _{1\leq k\leq n}f(\gamma (t_{k}))(\gamma (t_{k})-\gamma (t_{k-1})).}$

The integral is then the limit of this sum, as the lengths of the subdivision intervals approach zero.

If ${\displaystyle \gamma }$ is a continuously differentiable curve, the line integral can be evaluated as an integral of a function of a real variable:

${\displaystyle \int _{\gamma }f(z)\,dz=\int _{a}^{b}f(\gamma (t))\,\gamma \,'(t)\,dt.}$

When ${\displaystyle \gamma }$ is a closed curve, that is, its initial and final points coincide, the notation

${\displaystyle \oint _{\gamma }f(z)\,dz}$

is often used for the line integral of f along ${\displaystyle \gamma }$.

The line integrals of complex functions can be evaluated using a number of techniques: the integral may be split in to real and imaginary parts reducing the problem to that of evaluating two real-valued line integrals, the Cauchy integral formula may be used in other circumstances. If the line integral is a closed curve in a region where the function is analytic and containing no singularities, then the value of the integral is simply zero, this is a consequence of the Cauchy integral theorem. Because of the residue theorem, one can often use contour integrals in the complex plane to find integrals of real-valued functions of a real variable (see residue theorem for an example).

### Example

Consider the function f(z)=1/z, and let the contour C be the unit circle about 0, which can be parametrized by eit, with t in [0, 2π]. Substituting, we find

${\displaystyle \oint _{C}f(z)\,dz=\int _{0}^{2\pi }{1 \over e^{it}}ie^{it}\,dt=i\int _{0}^{2\pi }e^{-it}e^{it}\,dt}$
${\displaystyle =i\int _{0}^{2\pi }\,dt=i(2\pi -0)=2\pi i}$

where we use the fact that any complex number z can be written as reit where r is the modulus of z. On the unit circle this is fixed to 1, so the only variable left is the angle, which is denoted by t. This answer can be also verified by the Cauchy integral formula.

### Relation between the line integral of a vector field and the complex line integral

Viewing complex numbers as 2-dimensional vectors, the line integral of a 2-dimensional vector field corresponds to the real part of the line integral of the conjugate of the corresponding complex function of a complex variable. More specifically, if ${\displaystyle \mathbf {r} (t)=x(t)\mathbf {i} +y(t)\mathbf {j} }$ and ${\displaystyle f(z)=u(z)+iv(z)}$, then:

${\displaystyle \int _{C}{\overline {f(z)}}\,dz=\int _{C}(u-iv)\,dz=\int _{C}(u\mathbf {i} +v\mathbf {j} )\cdot d\mathbf {r} -i\int _{C}(v\mathbf {i} -u\mathbf {j} )\cdot d\mathbf {r} ,}$

provided that both integrals on the right hand side exist, and that the parametrization ${\displaystyle z(t)}$ of C has the same orientation as ${\displaystyle \mathbf {r} (t)}$.

Due to the Cauchy-Riemann equations the curl of the vector field corresponding to the conjugate of a holomorphic function is zero. This relates through Stokes' theorem both types of line integral being zero.

Also, the line integral can be evaluated using the change of variables.

## Quantum mechanics

The "path integral formulation" of quantum mechanics actually refers not to path integrals in this sense but to functional integrals, that is, integrals over a space of paths, of a function of a possible path. However, path integrals in the sense of this article are important in quantum mechanics; for example, complex contour integration is often used in evaluating probability amplitudes in quantum scattering theory.