Contour integration

 ( Complex Analysis ) 

A smooth curve is a curve $z:a,b\backslash to\backslash C$ with a non-vanishing, continuous derivative such that each point is traversed only once ( is one-to-one), with the possible exception of a curve such that the endpoints match ($z(a)=z(b)$). In the case where the endpoints match, the curve is called closed, and the function is required to be one-to-one everywhere else and the derivative must be continuous at the identified point ($z\text{'}(a)=z\text{'}(b)$). A smooth curve that is not closed is often referred to as a smooth arc.

The parametrization of a curve provides a natural ordering of points on the curve: $z(x)$ comes before $z(y)$ if directed smooth curve. It is most useful to consider curves independent of the specific parametrization. This can be done by considering equivalence classes of smooth curves with the same direction. A directed smooth curve can then be defined as an ordered set of points in the complex plane that is the image of some smooth curve in their natural order (according to the parametrization). Note that not all orderings of the points are the natural ordering of a smooth curve. In fact, a given smooth curve has only two such orderings. Also, a single closed curve can have any point as its endpoint, while a smooth arc has only two choices for its endpoints.

Now, to define the contour integral, let $f:\backslash C\backslash to\backslash C$ be a continuous function on the directed smooth curve $\backslash gamma$. Let $z:a,b\backslash to\backslash C$ be any parametrization of $\backslash gamma$ that is consistent with its order (direction). Then the integral along $\backslash gamma$ is denoted $$\backslash int\_\backslash gamma\; f(z)\backslash ,\; dz\backslash ,$$ and is given by $$\backslash int\_\backslash gamma\; f(z)\; \backslash ,\; dz\; :=\; \backslash int\_a^b\; f\backslash big(z(t)\backslash big)\; z\text{'}(t)\; \backslash ,\; dt.$$

This definition is well defined. That is, the result is independent of the parametrization chosen. In the case where the real integral on the right side does not exist the integral along $\backslash gamma$ is said not to exist.

• parametrizing the contour
• : The contour is parametrized by a differentiable complex-valued function of real variables, or the contour is broken up into pieces and parametrized separately.
• substitution of the parametrization into the integrand
• : Substituting the parametrization into the integrand transforms the integral into an integral of one real variable.
• direct evaluation
• : The integral is evaluated in a method akin to a real-variable integral.

In evaluating this integral, use the unit circle as a contour, parametrized by , with , then and $$\backslash oint\_C\; \backslash frac\{1\}\{z\}\backslash ,dz\; =\; \backslash int\_0^\{2\backslash pi\}\; \backslash frac\{1\}\{e^\{it\}\}\; ie^\{it\}\backslash ,dt\; =\; i\backslash int\_0^\{2\backslash pi\}\; 1\; \backslash ,\; dt\; =\; i\; \backslash ,\; t\backslash Big|\_0^\{2\backslash pi\}\; =\; \backslash left(2\backslash pi-0\backslash right)i\; =\; 2\backslash pi\; i\; ,$$ which is the value of the integral. This result only applies to the case in which is raised to power of . If the power is not equal to , then the result will always be zero.

Integral theorems such as the Cauchy integral formula or residue theorem are generally used in the following method:

• a specific contour is chosen:
• : The contour is chosen so that the contour follows the part of the complex plane that describes the real-valued integral, and also encloses singularities of the integrand so application of the Cauchy integral formula or residue theorem is possible
• application of Cauchy's integral theorem
• : The integral is reduced to only an integration around a small circle about each pole.
• application of the Cauchy integral formula or residue theorem
• : Application of these integral formulae gives us a value for the integral around the whole of the contour.
• division of the contour into a contour along the real part and imaginary part
• : The whole of the contour can be divided into the contour that follows the part of the complex plane that describes the real-valued integral as chosen before (call it ), and the integral that crosses the complex plane (call it ). The integral over the whole of the contour is the sum of the integral over each of these contours.
• demonstration that the integral that crosses the complex plane plays no part in the sum
• : If the integral can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integral as described above tends to 0, the integral along will tend to the integral around the contour .
• conclusion
• : If we can show the above step, then we can directly calculate , the real-valued integral.

To evaluate this integral, we look at the complex-valued function $$f(z)=\backslash frac\{1\}\{\backslash left(z^2+1\backslash right)^2\}$$

which has singularities at and . We choose a contour that will enclose the real-valued integral, here a semicircle with boundary diameter on the real line (going from, say, to ) will be convenient. Call this contour .

There are two ways of proceeding, using the Cauchy integral formula or by the method of residues:

Furthermore, observe that $$f(z)=\backslash frac\{1\}\{\backslash left(z^2+1\backslash right)^2\}=\backslash frac\{1\}\{(z+i)^2(z-i)^2\}.$$

Since the only singularity in the contour is the one at , then we can write $$f(z)=\backslash frac\{\backslash frac\{1\}\{(z+i)^2\}\}\{(z-i)^2\},$$

which puts the function in the form for direct application of the formula. Then, by using Cauchy's integral formula, $$\backslash oint\_C\; f(z)\backslash ,dz\; =\; \backslash oint\_C\; \backslash frac\{\backslash frac\{1\}\{(z+i)^2\}\}\{(z-i)^2\}\backslash ,dz\; =\; 2\backslash pi\; i\; \backslash ,\; \backslash left.\backslash frac\{d\}\{dz\}\; \backslash frac\{1\}\{(z+i)^2\}\backslash right|\_\{z=i\}\; =2\; \backslash pi\; i\; \backslash left\backslash frac\{-2\}\{(z+i)^3\}\backslash right\_\{z\; =\; i\}\; =\backslash frac\{\backslash pi\}\{2\}$$

We take the first derivative, in the above steps, because the pole is a second-order pole. That is, is taken to the second power, so we employ the first derivative of . If it were taken to the third power, we would use the second derivative and divide by , etc. The case of to the first power corresponds to a zero order derivative—just itself.

We need to show that the integral over the arc of the semicircle tends to zero as , using the estimation lemma $$\backslash left|\backslash int\_\backslash text\{Arc\}\; f(z)\backslash ,dz\backslash right|\; \backslash le\; ML$$

where is an upper bound on along the arc and the length of the arc. Now, $$\backslash left|\backslash int\_\backslash text\{Arc\}\; f(z)\backslash ,dz\backslash right|\backslash le\; \backslash frac\{a\backslash pi\}\{\backslash left(a^2-1\backslash right)^2\}\; \backslash to\; 0\; \backslash text\{\; as\; \}\; a\; \backslash to\; \backslash infty.$$ So $$\backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash frac\{1\}\{\backslash left(x^2+1\backslash right)^2\}\backslash ,dx\; =\; \backslash int\_\{-\backslash infty\}^\backslash infty\; f(z)\backslash ,dz\; =\; \backslash lim\_\{a\; \backslash to\; +\backslash infty\}\; \backslash int\_\{-a\}^a\; f(z)\backslash ,dz\; =\; \backslash frac\{\backslash pi\}2.\backslash quad\backslash square$$

(See the sample Laurent calculation from Laurent series for the derivation of this series.)

It is clear by inspection that the residue is , so, by the residue theorem, we have $$\backslash oint\_C\; f(z)\backslash ,dz\; =\; \backslash oint\_C\; \backslash frac\{1\}\{\backslash left(z^2+1\backslash right)^2\}\backslash ,dz\; =\; 2\; \backslash pi\; i\; \backslash ,\backslash operatorname\{Res\}\_\{z=i\}\; f(z)\; =\; 2\; \backslash pi\; i\; \backslash left(-\backslash frac\{i\}\{4\}\backslash right)=\backslash frac\{\backslash pi\}2\; \backslash quad\backslash square$$

Thus we get the same result as before.

This does not affect the use of the method of residues by series.

(which arises in probability theory as a scalar multiple of the characteristic function of the Cauchy distribution) resists the techniques of elementary calculus. We will evaluate it by expressing it as a limit of contour integrals along the contour that goes along the real number line from to and then counterclockwise along a semicircle centered at 0 from to . Take to be greater than , so that the imaginary number unit is enclosed within the curve. The contour integral is $$\backslash int\_C\; \backslash frac\{e^\{itz\}\; \}\{\; z^2+1\}\backslash ,dz.$$

Since is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator is zero. Since , that happens only where or . Only one of those points is in the region bounded by this contour. The residue of at is $$\backslash lim\_\{z\backslash to\; i\}(z-i)f(z)\; =\; \backslash lim\_\{z\backslash to\; i\}(z-i)\backslash frac\{e^\{itz\}\; \}\{\; z^2+1\}\; =\; \backslash lim\_\{z\backslash to\; i\}(z-i)\backslash frac\{e^\{itz\}\; \}\{\; (z-i)(z+i)\}\; =\; \backslash lim\_\{z\backslash to\; i\}\backslash frac\{e^\{itz\}\; \}\{\; z+i\}\; =\; \backslash frac\{e^\{-t\}\}\{2i\}.$$

According to the residue theorem, then, we have $$\backslash int\_C\; f(z)\backslash ,dz=2\backslash pi\; i\; \backslash operatorname\{Res\}\_\{z=i\}f(z)=2\backslash pi\; i\backslash frac\{e^\{-t\}\; \}\{\; 2i\}=\backslash pi\; e^\{-t\}.$$

The contour may be split into a "straight" part and a curved arc, so that $$\backslash int\_\backslash text\{straight\}+\backslash int\_\backslash text\{arc\}=\backslash pi\; e^\{-t\},$$ and thus $$\backslash int\_\{-a\}^a\; =\backslash pi\; e^\{-t\}-\backslash int\_\backslash text\{arc\}.$$

According to Jordan's lemma, if then $$\backslash int\_\backslash text\{arc\}\backslash frac\{e^\{itz\}\; \}\{\; z^2+1\}\backslash ,dz\; \backslash rightarrow\; 0\; \backslash mbox\{\; as\; \}\; a\backslash rightarrow\backslash infty.$$

Therefore, if then $$\backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash frac\{e^\{itx\}\; \}\{\; x^2+1\}\backslash ,dx=\backslash pi\; e^\{-t\}.$$

A similar argument with an arc that winds around rather than shows that if then $$\backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash frac\{e^\{itx\}\; \}\{\; x^2+1\}\backslash ,dx=\backslash pi\; e^t,$$ and finally we have this: $$\backslash int\_\{-\backslash infty\}^\backslash infty\; \backslash frac\{e^\{itx\}\; \}\{\; x^2+1\}\; \backslash ,dx=\backslash pi\; e^\{-|t|\}.$$

(If then the integral yields immediately to real-valued calculus methods and its value is .)

As an example, consider $$\backslash int\_\{-\backslash pi\}^\backslash pi\; \backslash frac\{1\; \}\{\; 1\; +\; 3\; (\backslash cos\; t)^2\}\; \backslash ,dt.$$

We seek to make a substitution of . Now, recall $$\backslash cos\; t\; =\; \backslash frac12\; \backslash left(e^\{it\}+e^\{-it\}\backslash right)\; =\; \backslash frac12\; \backslash left(z+\backslash frac\{1\}\{z\}\backslash right)$$ and $$\backslash frac\{dz\}\{dt\}\; =\; iz,\backslash \; dt\; =\; \backslash frac\{dz\}\{iz\}.$$

Taking to be the unit circle, we substitute to get:

The singularities to be considered are at $\backslash tfrac\{\backslash pm\; i\}\{\backslash sqrt\{3\}\}.$ Let be a small circle about $\backslash tfrac\{i\}\{\backslash sqrt\{3\}\},$ and be a small circle about $\backslash tfrac\{-i\}\{\backslash sqrt\{3\}\}.$ Then we arrive at the following:

The trick is to use the substitution where and hence $$\backslash frac\{1\}\{iz\}\; \backslash ,dz\; =\; dt.$$

This substitution maps the interval to the unit circle. Furthermore, $$\backslash sin(k\; t)\; =\; \backslash frac\{e^\{i\; k\; t\}\; -\; e^\{-\; i\; k\; t\}\}\{2\; i\}\; =\; \backslash frac\{z^k\; -\; z^\{-k\}\}\{2i\}$$ and $$\backslash cos(k\; t)\; =\; \backslash frac\{e^\{i\; k\; t\}\; +\; e^\{-\; i\; k\; t\}\}\{2\}\; =\; \backslash frac\{z^k\; +\; z^\{-k\}\}\{2\}$$ so that a rational function in results from the substitution, and the integral becomes $$\backslash oint\_$$

Therefore: $$\backslash int\_C\; \backslash frac\{\backslash sqrt\; z\}\{z^2+6z+8\}\backslash ,dz=2\backslash int\_0^\backslash infty\; \backslash frac\{\backslash sqrt\; x\}\{x^2+6x+8\}\backslash ,dx.$$

By using the residue theorem or the Cauchy integral formula (first employing the partial fractions method to derive a sum of two simple contour integrals) one obtains $$\backslash pi\; i\; \backslash left(\backslash frac\{i\}\{\backslash sqrt\; 2\}-i\backslash right)=\backslash int\_0^\backslash infty\; \backslash frac\{\backslash sqrt\; x\}\{x^2+6x+8\}\backslash ,dx\; =\; \backslash pi\backslash left(1-\backslash frac\{1\}\{\backslash sqrt\; 2\}\backslash right).\backslash quad\backslash square$$

To calculate this integral, one uses the function $$f(z)\; =\; \backslash left\; (\backslash frac\{\backslash log\; z\}\{1+z^2\}\; \backslash right\; )^2$$ and the branch of the logarithm corresponding to .

We will calculate the integral of along the keyhole contour shown at right. As it turns out this integral is a multiple of the initial integral that we wish to calculate and by the Cauchy residue theorem we have

Let be the radius of the large circle, and the radius of the small one. We will denote the upper line by , and the lower line by . As before we take the limit when and . The contributions from the two circles vanish. For example, one has the following upper bound with the ML lemma: $$\backslash left|\; \backslash int\_R\; f(z)\; \backslash ,\; dz\; \backslash right|\; \backslash le\; 2\; \backslash pi\; R\; \backslash frac\{(\backslash log\; R)^2\; +\; \backslash pi^2\}\{\backslash left(R^2-1\backslash right)^2\}\; \backslash to\; 0.$$

In order to compute the contributions of and we set on and on , with : which gives $$\backslash int\_0^\backslash infty\; \backslash frac\{\backslash log\; x\}\{\backslash left(1+x^2\backslash right)^2\}\; \backslash ,\; dx\; =\; -\; \backslash frac\{\backslash pi\}\{4\}.$$

This requires a close study of $$f(z)\; =\; z^\backslash frac34\; (3-z)^\backslash frac14.$$

We will construct so that it has a branch cut on , shown in red in the diagram. To do this, we choose two branches of the logarithm, setting and

The cut of is therefore and the cut of is . It is easy to see that the cut of the product of the two, i.e. , is , because is actually continuous across . This is because when and we approach the cut from above, has the value $$r^\backslash frac34\; e^\{\backslash frac34\; \backslash pi\; i\}\; (3+r)^\backslash frac14\; e^\{\backslash frac24\; \backslash pi\; i\}\; =\; r^\backslash frac34\; (3+r)^\backslash frac14\; e^\{\backslash frac54\; \backslash pi\; i\}.$$

When we approach from below, has the value $$r^\backslash frac34\; e^\{-\backslash frac34\; \backslash pi\; i\}\; (3+r)^\backslash frac14\; e^\{\backslash frac04\; \backslash pi\; i\}\; =\; r^\backslash frac34\; (3+r)^\backslash frac14\; e^\{-\backslash frac34\; \backslash pi\; i\}.$$

But $$e^\{-\backslash frac34\; \backslash pi\; i\}\; =\; e^\{\backslash frac54\; \backslash pi\; i\},$$

so that we have continuity across the cut. This is illustrated in the diagram, where the two black oriented circles are labelled with the corresponding value of the argument of the logarithm used in and .

We will use the contour shown in green in the diagram. To do this we must compute the value of along the line segments just above and just below the cut.

Let (in the limit, i.e. as the two green circles shrink to radius zero), where . Along the upper segment, we find that has the value $$r^\backslash frac34\; e^\{\backslash frac04\; \backslash pi\; i\}\; (3-r)^\backslash frac14\; e^\{\backslash frac24\; \backslash pi\; i\}\; =\; i\; r^\backslash frac34\; (3-r)^\backslash frac14$$ and along the lower segment, $$r^\backslash frac34\; e^\{\backslash frac04\; \backslash pi\; i\}\; (3-r)^\backslash frac14\; e^\{\backslash frac04\; \backslash pi\; i\}\; =\; r^\backslash frac34\; (3-r)^\backslash frac14.$$

It follows that the integral of along the upper segment is in the limit, and along the lower segment, .

If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of , by the Cauchy residue theorem. Let the radius of the green circles be , where and , and apply the Estimation lemma. For the circle on the left, we find $$\backslash left|\; \backslash int\_\{C\_\backslash mathrm\{L\}\}\; \backslash frac\{f(z)\}\{5-z\}\; dz\; \backslash right|\; \backslash le\; 2\; \backslash pi\; \backslash rho\; \backslash frac\{\backslash rho^\backslash frac34\; 3.001^\backslash frac14\}\{4.999\}\; \backslash in\; \backslash mathcal\{O\}\; \backslash left(\; \backslash rho^\backslash frac74\; \backslash right)\; \backslash to\; 0.$$

Similarly, for the circle on the right, we have $$\backslash left|\; \backslash int\_\{C\_\backslash mathrm\{R\}\}\; \backslash frac\{f(z)\}\{5-z\}\; dz\; \backslash right|\; \backslash le\; 2\; \backslash pi\; \backslash rho\; \backslash frac\{3.001^\backslash frac34\; \backslash rho^\backslash frac14\}\{1.999\}\; \backslash in\; \backslash mathcal\{O\}\; \backslash left(\; \backslash rho^\backslash frac54\; \backslash right)\; \backslash to\; 0.$$

Now using the Cauchy residue theorem, we have $$(-i\; +\; 1)\; I\; =\; -2\backslash pi\; i\; \backslash left(\; \backslash operatorname\{Res\}\_\{z=5\}\; \backslash frac\{f(z)\}\{5-z\}\; +\; \backslash operatorname\{Res\}\_\{z=\backslash infty\}\; \backslash frac\{f(z)\}\{5-z\}\; \backslash right).$$ where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly $$\backslash operatorname\{Res\}\_\{z=5\}\; \backslash frac\{f(z)\}\{5-z\}\; =\; -\; 5^\backslash frac34\; e^\{\backslash frac14\; \backslash log(-2)\}.$$

The pole is shown in blue in the diagram. The value simplifies to $$-5^\backslash frac34\; e^\{\backslash frac14(\backslash log\; 2\; +\; \backslash pi\; i)\}\; =\; -e^\{\backslash frac14\; \backslash pi\; i\}\; 5^\backslash frac34\; 2^\backslash frac14.$$

We use the following formula for the residue at infinity: $$\backslash operatorname\{Res\}\_\{z=\backslash infty\}\; h(z)\; =\; \backslash operatorname\{Res\}\_\{z=0\}\; \backslash left(-\; \backslash frac\{1\}\{z^2\}\; h\backslash left(\backslash frac\{1\}\{z\}\backslash right)\backslash right).$$

Substituting, we find $$\backslash frac\{1\}\{5-\backslash frac\{1\}\{z\}\}\; =\; -z\; \backslash left(1\; +\; 5z\; +\; 5^2\; z^2\; +\; 5^3\; z^3\; +\; \backslash cdots\backslash right)$$ and $$\backslash left(\backslash frac\{1\}\{z^3\}\backslash left\; (3-\backslash frac\{1\}\{z\}\; \backslash right\; )\backslash right)^\backslash frac14\; =\; \backslash frac\{1\}\{z\}\; (3z-1)^\backslash frac14\; =\; \backslash frac\{1\}\{z\}e^\{\backslash frac14\; \backslash pi\; i\}\; (1-3z)^\backslash frac14,$$ where we have used the fact that for the second branch of the logarithm. Next we apply the binomial expansion, obtaining $$\backslash frac\{1\}\{z\}\; e^\{\backslash frac14\; \backslash pi\; i\}\; \backslash left(\; 1\; -\; \{1/4\; \backslash choose\; 1\}\; 3z\; +\; \{1/4\; \backslash choose\; 2\}\; 3^2\; z^2\; -\; \{1/4\; \backslash choose\; 3\}\; 3^3\; z^3\; +\; \backslash cdots\; \backslash right).$$

The conclusion is that $$\backslash operatorname\{Res\}\_\{z=\backslash infty\}\; \backslash frac\{f(z)\}\{5-z\}\; =\; e^\{\backslash frac14\; \backslash pi\; i\}\; \backslash left\; (5\; -\; \backslash frac34\; \backslash right\; )\; =\; e^\{\backslash frac14\; \backslash pi\; i\}\backslash frac\{17\}\{4\}.$$

Finally, it follows that the value of is $$I\; =\; 2\; \backslash pi\; i\; \backslash frac\{e^\{\backslash frac14\; \backslash pi\; i\}\}\{-1+i\}\; \backslash left(\backslash frac\{17\}\{4\}\; -\; 5^\backslash frac34\; 2^\backslash frac14\; \backslash right)\; =\; 2\; \backslash pi\; 2^\{-\backslash frac12\}\; \backslash left(\backslash frac\{17\}\{4\}\; -\; 5^\backslash frac34\; 2^\backslash frac14\; \backslash right)$$ which yields $$I\; =\; \backslash frac\{\backslash pi\}\{2\backslash sqrt\; 2\}\; \backslash left(17\; -\; 5^\backslash frac34\; 2^\backslash frac94\; \backslash right)\; =\; \backslash frac\{\backslash pi\}\{2\backslash sqrt\; 2\}\; \backslash left(17\; -\; 40^\backslash frac34\; \backslash right).$$

Using the residue theorem, let us evaluate this contour integral. $$\backslash oint\_C\; \backslash frac\{e^z\}\{z^3\}\backslash ,dz$$

Recall that the residue theorem states $$\backslash oint\_\{C\}\; f(z)\; dz=2\backslash pi\; i\backslash cdot\; \backslash sum\backslash operatorname\{Res\}(f,a\_k)\; ,$$ where $\backslash operatorname\{Res\}$ is the residue of $f(z)$, and the $a\_k$ are the singularities of $f(z)$ lying inside the contour $C$ (with none of them lying directly on $C$).

$f(z)$ has only one pole, $0$. From that, we determine that the residue of $f(z)$ to be $\backslash tfrac\{1\}\{2\}$

Thus, using the residue theorem, we can determine: $$\backslash oint\_C\; \backslash frac\{e^z\}\{z^3\}\; dz\; =\; \backslash pi\; i.$$

In addition, we also need to evaluate $\backslash nabla\backslash cdot\; \backslash mathbf\{F\}$ where $\backslash nabla\; \backslash cdot\; \backslash mathbf\{F\}$ is an alternate notation of $\backslash operatorname\{div\}\; (\backslash mathbf\{F\})$. The divergence of any dimension can be described as

The corresponding double contour integral would be set up as such:

We now evaluate $\backslash nabla\backslash cdot\backslash mathbf\{F\}$. Meanwhile, set up the corresponding triple integral:

To evaluate this, we must utilize the divergence theorem as stated before, and we must evaluate $\backslash nabla\backslash cdot\backslash mathbf\{F\}$. Let $dV\; =\; dx\; \backslash ,\; dy\; \backslash ,\; dz\; \backslash ,\; du$

Thus, we can evaluate a contour integral with $n=4$. We can use the same method to evaluate contour integrals for any vector field with $n>4$ as well.

One of the most important examples is Cauchy's integral formula, which provides a way to reconstruct an analytic function from its values on a surrounding contour:

$f(z)\; =\; \backslash frac\{1\}\{2\backslash pi\; i\}\; \backslash oint\_\backslash gamma\; \backslash frac\{f(\backslash zeta)\}\{\backslash zeta\; -\; z\}\; \backslash ,\; d\backslash zeta$

Where $f$ is a function holomorphic on and inside the simple closed contour $\backslash gamma$, $z$ is a point inside $\backslash gamma$, and $\backslash zeta$ is the variable of integration. This formula shows that the values of $f$ inside the contour are determined by its values along the contour.

This integral expresses a function $f(t)$ in terms of its Laplace transform $F(s)$.

$\backslash frac\{\backslash sin\; x\}\{x\}\; =\; \backslash frac\{1\}\{\backslash pi\}\; \backslash int\_0^\backslash infty\; \backslash cos(xt)\; \backslash ,\; dt$

Although this is a real integral, methods from contour integration are often used in its derivation or evaluation.

$\backslash Gamma(z)\; =\; \backslash int\_0^\backslash infty\; t^\{z-1\}\; e^\{-t\}\; \backslash ,\; dt\; \backslash quad\; \backslash text\{for\; \}\; \backslash operatorname\{Re\}(z)\; >\; 0$

Extensions of this definition involve contour integrals in the complex plane.

• Residue (complex analysis)
• Cauchy principal value
• Poisson integral
• Pochhammer contour

• ; reprinted, 1968,
• Marko Riedel et al., Problème d'intégrale, Les-Mathematiques.net, in French.
• Marko Riedel et al., Integral by residue, math.stackexchange.com.
• W W L Chen, Introduction to Complex Analysis
• Various authors, sin límites ni cotas, es.ciencia.matematicas, in Spanish.