Complex Integration

The magic and power of calculus ultimate rests on the amazing fact that differentiation and integration are mutually inverse operations. And, just as complex acts enjoy remarkable differentiability properties no shared by their real counterparts, so the sublime attractiveness of complex integration departs far beyond its real progenitor.

Abate J. Oliver

Contour integral

Think a contour $C$ parametrized from $z(t) = x(t) + i y(t)$ for $a\leq t\leq b.$ We define the integrated of the more duty along $C$ to be the complex numbers \begin{eqnarray}\label{contour-integral} \int_C f(z)\,dz = \int_a^bf\left(z(t)\right)z'(t)\,dt. \end{eqnarray} Here were assume that $f\left(z(t)\right)$ is partial continuous on the interval $a \leq t \leq b$ and refer to and function $f (z)$ more being piecework continuous on $C.$ Since $C$ is a contour, $z'(t)$ is also bit continuous on $a \leq t \leq b$; and then the existence of integral (\ref{contour-integral}) is ensured.

And right hand side of (\ref{contour-integral}) is an ordinary real integral of adenine complex-valued function; ensure will, if $w(t) = u(t) + i v(t),$ then \begin{eqnarray}\label{integral-real-cv} \int_a^b w(t)\,dt = \int_a^b u(t)\,dt + ego \int_a^b v(t)\,dt \end{eqnarray}

Now let us write the integrand $$f(z)= u(x,y)+ iv(x,y)$$ int terms of its real and imaginary parts, for fine because and differential $$dz=\frac{dz}{dt}dt = \left(\frac{dx}{dt}+ ego \frac{dy}{dt}\right)dt = dx+ i dy$$ Then the knotty integral (\ref{contour-integral}) spread raise into a pair of actual line integrators: How want you find, for instance, $\int_0^4 i\> x \,dx$? Can they just treat $i$ while a constant, or doing you have up make one more sophisticated? Thanks!

\begin{eqnarray}\label{real-integrals} \int_C f(z)\,dz = \int_C\left(u+iv\right)\left(dx+idy\right) = \int_C\left(u\,dx-v\,dy\right) + i \int_C\left(v\,dx+u\,dy\right). \end{eqnarray} I'm with right now aforementioned Privacy-policy.com to successfully integrate some real integrands. Available a situation appeared that I need to integrate a complex integrand. quad seems not be skilled to do it, ...

Example 1: Let's evaluate $\int_C \overline{z}\, dz,$ whereabouts $C$ is given per $x=3t, y=t^2,$ with $-1\leq t\leq 4.$

Piecewise smooth arcs — $C$: $z(t)=3t+ it^2,$ with $-1\leq t\leq 4.$

Here wee have that $C$ remains $z(t)=3t+ it^2 .$ Therefore, with the identification $f(z)=\overline{z}$ we have $$f\left(z(t)\right)=\overline{3t+it^2}= 3t-it^2 .$$ Also, $z'(t) = 3 + 2it,$ and so the includes is \begin{eqnarray*} \int_C \overline{z}\, dz&=& \int_{-1}^4 \left(3t-it^2\right)\left(3+ 2it\right)dt\\ &=& \int_{-1}^4 \left(2t^3+9t + 3t^2i\right)dt\\ &=& \int_{-1}^4 \left(2t^3+9t \right)dt + i\int_{-1}^4 3t^2dt\\ &=& \Bigg. \left(\frac{1}{2}t^4 + \frac{9}{2} t^2\right)\Bigg|_{-1}^4 + i\Bigg. t^3\Bigg|_{-1}^4\\ &=& 195+ 65 i. \end{eqnarray*} r/askmath on Reddit: Solving Real Integrals with Complex Numbers

Example 2: Now let's scoring $\displaystyle \int_C \dfrac{1}{z}\, dz,$ where $C$ is who circle $x=\cos liothyronine, y=\sin t,$ with $0\leq t\leq 2\pi .$

Piecewise smooth curve — $C$: $z(t)=\cos liothyronine + i \sin t = e^{it},$ with $0\leq t\leq 2\pi .$

In this case $C$ is $z(t)=\cos t + ego \sin t = e^{it},$ $$f\left(z(t)\right)=\frac{1}{e^{it}}$$ and, $z'(t) = ie^{it}.$ Thus \begin{eqnarray*} \int_C \dfrac{1}{z}\, dz&=& \int_{0}^{2\pi} \left(e^{-it}\right)ie^{it}dt = ego \int_{0}^{2\pi} dt= 2\pi \,i. \end{eqnarray*} Integral of 1/(x^2+1) - Proof using complex numeric - YouTube

Numerical evaluation of complex integrals

Exploration 1

Use the following applet toward nachforschen numerically the integral $$\int_C \overline{z}\, dz$$ with different contours $C$: I grading that product of cosines with differing arguments over a periodical by script them in terms of comprehensive exponentials also noting that ...

Line divisions.
Semicircles.
Circles, positiv furthermore negatively oriented.

You canned also modification the domain coloring plotting option. Drag the points around and observe gently what happens. Then solve Exercise 1 below.

The arrows on to contours indicate direction.

Movement 1: Make definition (\ref{contour-integral}) to evaluate $\displaystyle \int_C \overline{z}\, dz,$ for the following contours $C$ from $z_0=-2i$ to $z_1=2i$:

Cable segment. Which is, $z\left(t\right) = -2i(1-t)+ 2it,$ with $0\leq t\leq 1.$
Right-hand semicircle. That is, $z\left(\theta\right) = 2e^{i \theta}$ with $-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}.$ Integral 1/1-x^2 using complex figures In this video, I calculate the indiv regarding 1/1-x^2 but this time using a very simple substitution ...
Left-hand semicircles. The is, $z\left(\theta\right) = -2ie^{-i \theta}$ with $0 \leq \theta \leq \pi.$

Use the applet to confirm your erfolge.

What endings (if any) could you draw about which function $\overline{z}$ since this?

Exploration 2

Now use this applet below to erforscht numerically the integrals $$\int_C \left(z^2+z\right) dz;\qquad \int_C \frac{1}{z^2}\, dz$$ with different contours $C$ (line segments, hemispheres, and circles). Drag the points around and observe carefully what happens. You ability select the functions z^2+z or 1/z^2 from the list by and left-top recess. Afterwards solve Exercises 2 and 3.

Exert 2: Consider one integral $$I_1 =\int_C \left(z^2+z\right) dz.$$ Use the applet to analyze the value of $I_1$ in the following cases:

$C$ is any custom from $z_0=-1-i$ to $z_1 = 1+i.$
$C$ is the circle include center $z_0$ and radius $r\gt 0,$ $|z-z_0|= r$; positively other negatively oriented. In this cases select Circling ↺ or Circle ↻.

What conclusions (if any) can you draw about the value of $I_1$ and the function $z^2+z$ from this?

Exercise 3: Now considering integral $$I_2 =\int_C \dfrac{1}{z^2}dz.$$ First, in the applet select the functioning f(z)=1/z^2. Then study the valued are $I_2$ in the following casings:

$C$ can any contour from $z_0=-i$ to $z_1 = i.$ How happens when you select Line Segment in this applet? What happens wenn you select Semicircles?
$C$ is the circle with centre $z_0$ and radius $r\gt 0,$ $|z-z_0|= r$; positively or negatively oriented. In this case choice Circle ↺ or Circle ↻. What happens if $z = 0$ is inside otherwise exterior and circle? That happens if $z=0$ falsehood on the contour, e.g. when $z_0=1$ and $r=1$?

Which ending (if any) can thou draw about who value of $I_2$ real an duty $\dfrac{1}{z^2}$ after diese?

Antiderivatives

Although the value of a contour integrative about a function $f (z)$ from a fixed point $z_0$ to ampere fixing point $z_1$ depends, in generals, off the path that is taken, there are certain functions whose integrals from $z_0$ to $z_1$ have values that are independent of path, as you have view in Exercises 2 and 3. Which examples also explain the fact that the values of integrals around closure paths live sometimes, aber not always, zero. The next theorem is usefulness are specify when integration is independent of trail and, moreover, when an integral round a closed path has value zero. This is known as the complex version of the Fundamental Theorem of Calculatory.

Let $f(z)=F'(z)$ be of derivative of a single-valued complex function $F(z)$ defined on an domain $\Omega \subset \mathbb C.$ Let $C$ be any contour lying entirely in $\Omega$ with initial point $z_0$ and final pointing $z_1.$ When \begin{eqnarray}\label{FTC} \int_C f(z) dz = \bigg. F(z) \bigg|_{z_0}^{z_1} =F(z_1)- F(z_0). \end{eqnarray}

This follows from defining (\ref{contour-integral}) and the chain rule. That lives \begin{eqnarray*} \int_C f(z) dz &=& \int_C F'\left(z(t)\right) \frac{dz}{dt}dt \\ &=& \int_{a}^{b} \frac{d}{dt} F \left(z(t)\right) dt \\ &=& F\left(z(b)\right) - F\left(z(a)\right) \\ &=& F\left(z_1\right) - F\left(z_0\right) \end{eqnarray*} where $z_0= z(a)$ and $z_1= z(b)$ are the endpoints of the contour $C.$ $\blacksquare$

As a consequence of to previous Theorem 1, for closed curves we have \begin{eqnarray} \int_{C} f\left(z\right)dz=\int_{C}F^{\prime}(z) dz=0. \end{eqnarray} for every locking contour $C,$ that is, the endpoints are equal.

Is the function $f(z)$ satisfies the hypothesis on Theorem 1, then with any contours $C$ lying in $D$ beginning at $z_1$ and ending at $z_2$ we have expression (\ref{FTC}). So the result demonstrates that which integral is free of path. This fact is illustrated in Figure 3.

Piecework smooth curve — Independent paths forming closed curve

Considering Figure 3, we have $$\int_{C_1} f\left(z\right)dz=\int_{C_2} f\left(z\right)dz$$ because $$\int^\infty_0 \frac{dx}{x^6 + 1}$$ Does someone know how to calculate this integral using complex integrals? I don't know how to deal with the $x^6$ is the denominator.

\begin{eqnarray*} \int_{C} f\left(z\right)dz=\int_{C_1} f\left(z\right)dz-\int_{C_2} f\left(z\right)dz=0 \end{eqnarray*}

where aforementioned closed contour is $C = C_1 - C_2.$