We compute integrals of complex functions along contours.
Let be a contour parameterized by and let be a complex function defined along . Then the integral of along is defined by
example 1 Compute where is the line segment from to .
We first parameterize the line segment. Since the equation of the line in the -plane is , one possible parametrization of the line segment is The derivative of is Next, and hence We are now ready to compute the contour integral: \begin{align*} \int _C \overline{z} \, dz &= \int _{-1}^1 \left [t - i\left (\frac{t}{2} + \frac 12\right )\right ]\cdot \left [1 + \frac{i}{2}\right ] \, dt\\[6pt] &= \int _{-1}^1 \left [t + \frac 12\left (\frac{t}{2} + \frac 12\right )\right ] + i\left [-\left (\frac{t}{2} +\frac 12\right ) + \frac{t}{2}\right ] \, dt\\[6pt] &= \int _{-1}^1 \left [\frac 54 t + \frac 14\right ]+i\left [-\frac 12\right ] \, dt\\[6pt] &=\int _{-1}^1 \left [\frac 54 t + \frac 14\right ] \, dt - i\int _{-1}^1 \frac 12 \, dt\\[6pt] &= \frac 12 - i \end{align*}
We first parameterize the line segment. Since the equation of the line in the -plane is , one possible parametrization of the line segment is The derivative of is Next, and hence We are now ready to compute the contour integral: \begin{align*} \int _C \overline{z} \, dz &= \int _{-1}^1 \left [t - i\left (\frac{t}{2} + \frac 12\right )\right ]\cdot \left [1 + \frac{i}{2}\right ] \, dt\\[6pt] &= \int _{-1}^1 \left [t + \frac 12\left (\frac{t}{2} + \frac 12\right )\right ] + i\left [-\left (\frac{t}{2} +\frac 12\right ) + \frac{t}{2}\right ] \, dt\\[6pt] &= \int _{-1}^1 \left [\frac 54 t + \frac 14\right ]+i\left [-\frac 12\right ] \, dt\\[6pt] &=\int _{-1}^1 \left [\frac 54 t + \frac 14\right ] \, dt - i\int _{-1}^1 \frac 12 \, dt\\[6pt] &= \frac 12 - i \end{align*}
A line segment from to can be parameterized by . Since , the contour in the
previous example can thus also be parametrized by
(problem 1a) Use the parametrization given in the remark above to compute the
integral in example 1.
Here is a video solution to problem 1 (parts b and c):
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If the contour has two different parametrizations, with and with , then That is,
the value of a contour integral is independent of the parametrization used to describe
the contour.
What happens if we reverse the direction of a contour , ?
If is a contour parameterized by then the contour which represents the same points as but traced in the opposite direction has parametrization .
Suppose is continuous along a contour and suppose the contour represents the
same points as traced in the opposite direction, then
- Proof
- Using the substitution , we have \begin{align*} \int _{-C} f(z) \, dz &= \int _a^b f(\zeta (s)) \zeta '(s) \, ds \\ &= \int _a^b f\big (x(a+b-s) + iy(a+b-s)\big ) \big [x'(a+b-s)(-1)+ iy'(a+b-s)(-1)\big ]\, ds\\ &= \int _b^a f\big (x(t)+iy(t)\big ) \big [x'(t)+ iy'(t)\big ] \, dt\\ &= -\int _a^b f(\gamma (t)) \gamma '(t) \, dt\\ &= - \int _C f(z) \, dz \end{align*}
example 2 Let be the semi-circle parametrized by . Compute and verify the the
above proposition.
Note that . We begin with the integral along : Next we integrate along . To do this we parametrize as . Note that . Thus From our calculations, we see that thereby verifying the conclusion of the previous proposition.
Note that . We begin with the integral along : Next we integrate along . To do this we parametrize as . Note that . Thus From our calculations, we see that thereby verifying the conclusion of the previous proposition.
(problem 2b) Let be the circle centered at of radius , traversed in the clockwise
direction. Compute
(problem 2c) Let be the circle centered at of radius , traversed in the counter-clockwise
direction. Compute
Compare your answers to problems 2b and 2c. Notice that they verify the proposition.
(problem 2d) Let be the circle centered at of radius , traversed in the counter-clockwise
direction. Compute
Here is a video solution to problem 2b:
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-formula Let be a contour and suppose is continuous along . Then where is the
maximum value of along and is the length of .
- Proof
- Suppose is parametrized by . Since for all , the previous proposition implies But this last integral is exactly , the length of the curve and hence
example 3 Use the -formula to estimate the modulus of the integral: , where is the
circle centered at of radius .
The length of the circle, , is . In polar form, and hence . From the triangle inequality, the modulus of any point on can be estimated by Hence for any on the circle, . Finally, by the -formula,
The length of the circle, , is . In polar form, and hence . From the triangle inequality, the modulus of any point on can be estimated by Hence for any on the circle, . Finally, by the -formula,
(problem 3) Use the -formula to estimate the modulus of the integral: , where is the
circle centered at of radius .
The maximum modulus of on is
The length of is
By the -formula,
The maximum modulus of on is
The length of is
By the -formula,
Here is a video solution to problem 3:
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If the function is analytic at every point on the contour , then the computation of the integral along can be simplified as long as an anti-derivative of can be found.
Fundamental Theorem of Calculus Suppose is a contour with initial point and
terminal point . If is analytic along and if is an anti-derivative of along , i.e., at
every , then
example 4 Let be any contour from to . Compute .
Since is an entire function, it is analytic at every point on . Moreover, since we have an anti-derivative for for any on . Thus
Since is an entire function, it is analytic at every point on . Moreover, since we have an anti-derivative for for any on . Thus
(problem 4) Let be any contour from to that does not cross the negative real axis
or go through the origin. Compute .
Here is a video solution to problem 4:
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