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4.2: Complex Line Integrals


Line integrals are also called path or contour integrals. Given the ingredients we define the complex lineintegral (int_{gamma} f(z) dz) by

[int_{gamma} f(z) dz := int_{a}^{b} f(gamma (t)) gamma ' (t) dt. label{4.2.1}]

You should note that this notation looks just like integrals of a real variable. We don’t need the vectors and dot products of line integrals in (R^2). Also, make sure you understand that the product (f(gamma (t)) gamma '(t)) is just a product of complex numbers.

An alternative notation uses (dz = dx + idy) to write

[int_{gamma} f(z) dz = int_{gamma} (u + iv) (dx + idy) label{4.2.2}]

Let’s check that Equations ef{4.2.1} and ef{4.2.2} are the same. Equation ef{4.2.2} is really a multivariable calculus expression, so thinking of (gamma (t)) as ((x(t), y(t))) it becomes

[int_{gamma} f(z) dz = int_a^b [u(x(t), y(t)) + iv (x(t), y(t)] (x'(t) + iy'(t))dt]

but

[u(x(t), y(t)) + iv (x(t), y(t)) = f(gamma (t))]

and

[x'(t) + iy'(t) = gamma '(t)]

so the right hand side of Equation ef{4.2.2} is

[int_{a}^{b} f(gamma (t)) gamma '(t) dt.]

That is, it is exactly the same as the expression in Equation ef{4.2.1}

Example (PageIndex{1})

Compute (int_{gamma} z^2 dz) along the straight line from 0 to (1 + i).

Solution

We parametrize the curve as (gamma (t) = t(1 + i)) with (0 le t le 1). So (gamma '(t) = 1 + i). The line integral is

[int z^2 dz = int_{0}^{1} t^2 (1 + i)^2 (1 + i) dt = dfrac{2i(1 + i)}{3}. onumber]

Example (PageIndex{2})

Compute (int_{gamma} overline{z} dz) along the straight line from 0 to (1 + i).

Solution

We can use the same parametrization as in the previous example. So,

[int_{gamma} overline{z} dz = int_{0}^{1} t(1 - i) (1 + i) dt = 1. onumber]

Example (PageIndex{3})

Compute (int_{gamma} z^2 dz) along the unit circle.

Solution

We parametrize the unit circle by (gamma ( heta) = e^{i heta}), where (0 le heta le 2pi). We have (gamma '( heta) = ie^{i heta}). So, the integral becomes

[int_{gamma} z^2 dz = int_{0}^{2pi} e^{2i heta} i e^{i heta} d heta = int_{0}^{2pi} ie^{3i heta} d heta = dfrac{e^{3i heta}}{3} vert_{0}^{2pi} = 0. onumber]

Example (PageIndex{4})

Compute (int overline{z} dz) along the unit circle.

Solution

Parametrize (C): (gamma (t) = e^{it}), with (0 le t le 2pi). So, (gamma '(t) = ie^{it}). Putting this into the integral gives

[int_{C} overline{z} dz = int_{0}^{2pi} overline{e^{it}} i e^{it} dt = int_{0}^{2pi} i dt = 2pi i. onumber]


Integration using Euler's formula

In integral calculus, Euler's formula for complex numbers may be used to evaluate integrals involving trigonometric functions. Using Euler's formula, any trigonometric function may be written in terms of complex exponential functions, namely e i x > and e − i x > and then integrated. This technique is often simpler and faster than using trigonometric identities or integration by parts, and is sufficiently powerful to integrate any rational expression involving trigonometric functions. [1]


Properties of Fourier Series Spectrum

A signal's Fourier series spectrum ck has interesting properties.

If s(t) is real,

Real-valued periodic signals have conjugate-symmetric spectra.

This result follows from the integral that calculates the c k from the signal. Furthermore, this result means that:

The real part of the Fourier coefficients for real-valued signals is even. Similarly,

The imaginary parts of the Fourier coefficients have odd symmetry. Consequently, if you are given the Fourier coefficients for positive indices and zero and are told the signal is real-valued, you can find the negative-indexed coefficients, hence the entire spectrum. This kind of symmetry,

is known as conjugate symmetry .

If s(-t) = s(t), which says the signal has even symmetry about the origin,

Given the previous property for real-valued signals, the Fourier coefficients of even signals are real-valued. A real-valued Fourier expansion amounts to an expansion in terms of only cosines, which is the simplest example of an even signal.

If s(-t) = - s(t), which says the signal has odd symmetry,

Therefore, the Fourier coefficients are purely imaginary. The square wave is a great example of an odd-symmetric signal.

The spectral coefficients for a periodic signal delayed by &tau, s(t-&tau), are:

where c k denotes the spectrum of s(t).

Delaying a signal by &tau seconds results in a spectrum having a linear phase shift of [-frac<2pi k au >] in comparison to the spectrum of the undelayed signal. Note that the spectral magnitude is unaffected. Showing this property is easy.

Note that the range of integration extends over a period of the integrand. Consequently, it should not matter how we integrate over a period, which means that:

The complex Fourier series obeys Parseval's Theorem , one of the most important results in signal analysis. This general mathematical result says you can calculate a signal's power in either the time domain or the frequency domain.


Math Insight

In the introduction to scalar line integrals, we derived the formula for $slint$, the line integral of a function $f$ over a curve parametrized by $dllp(t)$ for $a le t le b$: egin slint =int_a^b dlsi(dllp(t))| dllp,'(t) | dt. end

However, the value of this integral should not depend on the specific parametrization $dllp$, as it is designed to capture quantities like the total mass of a wire with density $dlsi$. The integral should only depend on the density function $dlsi(vc)$ and the image curve, denoted by $dlc$, which is the set of points $dllp(t)$ for all values of $t$ in the interval $[a,b]$.

For the reason, we can think of the line integral as being over the curve $dlc$, rather than the particular parametrization given by $dllp(t)$. To reflect this viewpoint, we could write the integral that gives the mass of the slinky as egin dslint = pslint, end where the only difference from above is that we replaced $dllp$ with $dlc$.

The notation $dslint$ makes it explicit that line integrals are independent of the parametrization $dllp(t)$ (after all, the notation does not mention $dllp(t)$). You may remember that the same curve $dlc$ can be parametrized by many functions. For example, we gave two different parametrizations of the unit circle. Above, we parametrized the slinky by $dllp(t) = (cos t, sin t, t)$, for le t le 2pi$. We could have equally well parametrized the same slinky by $adllp(t) = (cos 2t, sin 2t, 2t)$ for le t le pi$. (If $dllp(t)$ and $adllp(t)$ were the positions at time $t$ of two particles traveling along the slinky, the particle given by $adllp(t)$ travels twice as fast, covering the slinky in half the time, compared to the particle given by $dllp(t)$.)

As long as the density $dlsi(vc)$ is unchanged, the mass of the slinky had better be the same, no matter which parametrization we use. Hence, it must be true that the mass is both egin dslint = int_ dlsi , dals = pslint<0><2pi> end and egin dslint = int_ dlsi , dals = pslint<0>. end

Can you see why the integral is the same for both parametrizations $dllp(t)$ and $adllp(t)$, i.e., why egin pslint<0><2pi> = pslint<0>? end

In the first case, you are integrating over an interval that is twice as long (from 0 to $2pi$ versus from 0 to $pi$), but the speed $| dllp'(t)|$ is half the speed $| adllp'(t) |$. The two effects cancel and the integrals are equal.

Since the line integral of a vector field $dlvf$ over the curve is based on the line integral of a scalar function $f = dlvf cdot vc$, where $vc$ is the unit tangent vector of the curve, we expect that line integrals of vector fields should also be independent of the parametrization $dllp(t)$. Indeed, this is the case, with one important exception. Since $vc = dllp'(t)/dllp(t)$, the unit tangent vector, and hence the integral $dlint$, will be independent of the speed $|dllp'(t)|$ of the parametrization. The unit tangent vector, as its name implies will always be of length 1. But, at any point along the curve, there are two unit tangent vectors pointing in opposite directions. If we call one of them $vc$, then the other is $-vc$. The choice of these unit tangent vectors will depend on the direction that $dllp(t)$ transverses the curve $dlc$ as $t$ increases.

We refer to the choice of the unit tangent vector, or equivalently, the choice of the direction to tranverse $dlc$, as the orientation of the curve. Every simple curve has two orientations, one corresponding to one unit tangent vector $vc$ and the other corresponding to its opposite $-vc$. Scalar line integrals are independent of curve orientation, but vector line integrals will switch sign if you switch the orientation of the curve. This make sense intuitively, as the mass of the slinky shouldn't change, but the work done by a force field changes sign if you move in the opposite direction.

The examples of line integrals of scalar functions and vector fields include calculations of the same line integral with different parametrizations.


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Integration in Proton NMR

  • Contributed by Chris Schaller
  • Professor (Chemistry) at College of Saint Benedict/Saint John's University

There is additional information obtained from 1 H NMR spectroscopy that is not typically available from 13 C NMR spectroscopy. Chemical shift can show how many different types of hydrogens are found in a molecule integration reveals the number of hydrogens of each type. An integrator trace (or integration trace) can be used to find the ratio of the numbers of hydrogen atoms in different environments in an organic compound.

An integrator trace is a computer generated line which is superimposed on a proton NMR spectra. In the diagram, the integrator trace is shown in red.

An integrator trace measures the relative areas under the various peaks in the spectrum. When the integrator trace crosses a peak or group of peaks, it gains height. The height gained is proportional to the area under the peak or group of peaks. You measure the height gained at each peak or group of peaks by measuring the distances shown in green in the diagram above - and then find their ratio.

For example, if the heights were 0.7 cm, 1.4 cm and 2.1 cm, the ratio of the peak areas would be 1:2:3. That in turn shows that the ratio of the hydrogen atoms in the three different environments is 1:2:3.

Figure NMR16. 1 H NMR spectrum of ethanol with solid integral line. Source: Spectrum taken in CDCl3 on a Varian Gemini 2000 Spectrometer with 300 MHz Oxford magnet.

Looking at the spectrum of ethanol, you can see that there are three different kinds of hydrogens in the molecule. You can also see by integration that there are three hydrogens of one type, two of the second type, and one of the third type -- corresponding to the CH3 or methyl group, the CH2 or methylene group and the OH or hydroxyl group. That information helps narrow down the number of possible structures of the sample, and so it makes structure elucidation of an unknown sample much easier.

Integral data can be given in different forms. You should be aware of all of them. In raw form, an integral is a horizontal line running across the spectrum from left to right. Where the line crosses the frequency of a peak, the area of the peak is measured. This measurement is shown as a jump or step upward in the integral line the vertical distance that the line rises is proportional to the area of the peak. The area is related to the amount of radio waves absorbed at that frequency, and the amount of radio waves absorbed is proportional to the number of hydrogen atoms absorbing the radio waves.

Sometimes, the integral line is cut into separate integrals for each peak so that they can be compared to each other more easily.

Figure NMR17. 1 H NMR spectrum of ethanol with broken integral line. Source: Spectrum taken in CDCl3 on a Varian Gemini 2000 Spectrometer with 300 MHz Oxford magnet.

Often, instead of displaying raw data, the integrals are measured and their heights are displayed on the spectrum.

Figure NMR18. 1 H NMR spectrum of ethanol with numerical integrals.

Source: Spectrum taken in CDCl3 on a Varian Gemini 2000 Spectrometer with 300 MHz Oxford magnet.

Sometimes the heights are "normalized". They are reduced to a lowest common factor so that their ratios are easier to compare. These numbers could correspond to numbers of hydrogens, or simply to their lowest common factors. Two peaks in a ratio of 1H:2H could correspond to one and two hydrogens, or they could correspond to two and four hydrogens, etc.

Figure NMR19. 1 H NMR spectrum of ethanol with normalized integral numbers.

Source: Spectrum taken in CDCl3 on a Varian Gemini 2000 Spectrometer with 300 MHz Oxford magnet.


4.2: Complex Line Integrals

In the descriptions of the following functions, z is the complex number x + i y , where i is defined as sqrt (-1) .

Compute the magnitude of z .

The magnitude is defined as | z | = sqrt (x^2 + y^2) .

: arg ( z ) : angle ( z )

Compute the argument, i.e., angle of z .

This is defined as, theta = atan2 ( y , x ) , in radians.

Return the complex conjugate of z .

The complex conjugate is defined as conj ( z ) = x - i y .

: cplxpair ( z ) : cplxpair ( z , tol ) : cplxpair ( z , tol , dim )

Sort the numbers z into complex conjugate pairs ordered by increasing real part.

The negative imaginary complex numbers are placed first within each pair. All real numbers (those with abs (imag ( z ) / z ) < tol ) are placed after the complex pairs.

tol is a weighting factor which determines the tolerance of matching. The default value is 100 and the resulting tolerance for a given complex pair is 100 * eps (abs ( z (i))) .

By default the complex pairs are sorted along the first non-singleton dimension of z . If dim is specified, then the complex pairs are sorted along this dimension.

Signal an error if some complex numbers could not be paired. Signal an error if all complex numbers are not exact conjugates (to within tol ). Note that there is no defined order for pairs with identical real parts but differing imaginary parts.


Integration segment limits

Depending on the endpoints used by an integration method, open or closed rules are distinguished.

Open rules do not use endpoints. The open integration methods can be used in cases where the integrand function is undefined in some points.
E.g. using rectangle method we can approximate ln(x) definite integral value on (0,1) line segment, in spite of ln(0) is undefined.

In opposite, Closed rules use endpoints as well as midpoints to evaluate integrand function values.

Half-opened rules (e.g., left rectangle rule or right rectangle rule) can also be used to approximate integral on the line segment opened from only one side.


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Watch the video: Complex line integrals. Math. Chegg Tutors (December 2021).