NationMaster - Encyclopedia: Methods of contour integration

In complex analysis, the evaluation of integrals of real-valued functions along intervals on the real line, is not readily found with certain integrands and methods involving only real variables. Complex analysis methods described below give means of calculating these real-valued integrals by means of contour integrals in the complex plane. Complex analysis is the branch of mathematics investigating functions of complex numbers, and is of enormous practical use in many branches of mathematics, including applied mathematics. ... In calculus, the integral of a function is an extension of the concept of a sum. ... In mathematics, the set of real numbers, denoted R, or in blackboard bold , is the set of all rational and irrational numbers. ... This article is about path integrals in the general mathematical sense, and not the path integral formulation of physics which was studied by Richard Feynman. ...

These methods include

direct integration of a complex-valued function along a curve in the complex plane (a contour)
application of the Cauchy integral formula
application of the residue theorem

One method can be used, or a combination of these methods, or various limiting processes, for the purpose of finding these integrals or sums. In mathematics, a complex number is a number of the form where a and b are real numbers, and i is the imaginary unit, with the property i 2 = âˆ’1. ... Cauchys integral formula is a central statement in complex analysis. ... The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

1 Direct methods
- 1.1 Example
2 Applications of integral theorems
3 See also
4 External links

Direct methods

Direct methods involve the calculation of the integral by means of methods similar to those in calculating line integrals in several-variable calculus. This means that we use the following method:

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.

Example

A fundamental result in complex analysis is that the integral around the contour C which is the unit circle (or any Jordan curve about 0) of z⁻¹ is 2πi. Let us evaluate the integral In topology, the Jordan curve theorem states that every non-self-intersecting loop in the plane divides the plane into an inside and an outside. It was proved by Oswald Veblen in 1905. ...

$oint_C {1 over z},dz.$

In evaluating this integral, we use the unit circle |z| = 1 as our contour, which we can parametrize by γ(t) = e^it, with t ∈ [0, 2π]. Observe that γ'(t) = ie^it. Now, substituting this for z, we have

$oint_C {1 over z},dz = int_0^{2pi} {1 over e^{it}} ie^{it},dt = iint_0^{2pi} e^{-it}e^{it},dt = iint_0^{2pi} 1 ,dt$

$= left.tright]_0^{2pi} i=(2pi-0)i = 2pi i$

which is the value of the integral.

Applications of integral theorems

Applications of integral theorems are also often used to evaluate the contour integral along a contour, which means that the real-valued integral is calculated simultaneously along with calculating the contour integral.

Integral theorems such as the Cauchy integral formula or residue theorem are generally used in the following method: Cauchys integral formula is a central statement in complex analysis. ... The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

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 the Cauchy-Goursat 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 formula 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 R), and the integral that crosses the complex plane (call it I). 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 I can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integral I as described above tends to 0, the integral along R will tend to the integral around the contour R+I.

conclusion

If we can show the above step, then we can directly calculate R, the real-valued integral.

Cauchys integral formula is a central statement in complex analysis. ... The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ... The Cauchy integral theorem in complex analysis is an important statement about path integrals for holomorphic functions in the complex plane. ... Cauchys integral formula is a central statement in complex analysis. ... The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

Example (I)

Consider

$int_{-infty}^{infty} {1 over (x^2+1)^2},dx.$

To evaluate this integral, we look at the complex-valued function Image File history File links Contour Diagram File links The following pages link to this file: Residue theorem Methods of contour integration User:MaxPower/myImages ...

$f(z)={1 over (z^2+1)^2}$

which has singularities at i and −i. However, we will want to choose a contour that will enclose the real-valued integral, so we choose a semicircle as the one to the left, which we will let expand as to contain the whole real axis (a will tend to infinity). Call this contour C.

Now, there are two ways of proceeding, using the Cauchy integral formula or by the method of residues. Cauchys integral formula is a central statement in complex analysis. ...

Using the Cauchy integral formula

Observe that

$f(z)={1 over (z^2+1)^2}={1 over (z+i)^2(z-i)^2}.$

Since the only singularity in the contour is the one at i, then we can write

$f(z)={{1 over (z+i)^2} over (z-i)^2},$

which puts the function in the form for direct application of the formula.

By the formula, then,

$oint_C f(z),dz = oint_C {1 over (z^2+1)^2},dz = oint_C {{1 over (z+i)^2} over (z-i)^2},dz = 2pi i frac{d}{dz} left(left.{1 over (z+i)^2}right)right|_{z=i}$

$=2 pi i left.left({-2 over (z+i)^3}right)right|_{z = i} =2 pi i (-i/4)={piover 2}$

If we call the arc of the semicircle A, we need to show that the integral over A tends to zero as a tends to infinity — using the estimation lemma In mathematics, the estimation lemma gives an upper bound for a contour integral. ...

$left|oint_A f(z),dzright| le ML$

where M is an upper bound on |f(z)| and L the length of A. Now,

$int_A f(z),dz le {api over (a^2+1)^2} rightarrow 0 mathrm{as} a rightarrow infty$

$int_{-infty}^infty f(z),dz = {piover 2}.quadsquare$

Using the method of residues

Consider the Laurent series of f(z) about i, the only singularity we need to consider. We then have A Laurent series is defined with respect to a particular point c and a path of integration Î³. The path of integration must lie in an annulus (shown here in red) inside of which f(z) is holomorphic. ...

$f(z) = {1 over 4(z-i)^2} + {-i over 4(z-i)} + {3 over 16} + {i over 8}(z-i) + {-5 over 64}(z-i)^2 + cdots$

It is clear by inspection that the residue is -i/4, so, by the residue theorem, we have

$oint_C f(z),dz = oint_C {1 over (z^2+1)^2},dz = 2 pi i mathrm{Res}_{z=i} f = 2 pi i (-i/4)={piover 2}quadsquare$

$left |oint_A f(z),dzright | le ML$

where M is an upper bound on |f(z)| and L the length of A. Now,

$oint_A f(z),dz le {api over (a^2+1)^2} rightarrow 0 quad mathrm{as} a rightarrow infty.$

$int_{-infty}^{infty} f(z),dz = {piover 2}.quadsquare$

Thus we get the same result as before.

Contour note

As an aside, a question can arise whether we do not take the semicircle to include the other singularity, enclosing −i. To have the integral along the real axis moving in the correct direction, the contour must travel clockwise, ie., in a negative direction, reversing the sign of the integral overall.

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

Example (II) – Cauchy distribution

The integral

$int_{-infty}^infty {e^{itx} over x^2+1},dx$

(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 C that goes along the real line from −a to a and then counterclockwise along a semicircle centered at 0 from a to −a. Take a to be greater than 1, so that the imaginary unit i is enclosed within the curve. The contour integral is Image File history File links Contour Diagram File links The following pages link to this file: Residue theorem Methods of contour integration User:MaxPower/myImages ... Probability theory is the mathematical study of phenomena characterized by randomness or uncertainty. ... In probability theory, the characteristic function of any random variable completely defines its probability distribution. ... The Cauchy-Lorentz distribution, named after Augustin Cauchy, is a continuous probability distribution with probability density function where x0 is the location parameter, specifying the location of the peak of the distribution, and Î³ is the scale parameter which specifies the half-width at half-maximum (HWHM). ... Calculus is a central branch of mathematics. ... In mathematics, the set of real numbers, denoted R, or in blackboard bold , is the set of all rational and irrational numbers. ...

$int_C {e^{itz} over z^2+1},dz.$

Since e^itz is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z² + 1 is zero. Since z² + 1 = (z + i)(z − i), that happens only where z = i or z = −i. Only one of those points is in the region bounded by this contour. The residue of f(z) at z = i is In complex analysis, an entire function is a function that is holomorphic everywhere (ie complex-differentiable at every point) on the whole complex plane. ... In mathematics, a singularity is in general a point at which a given mathematical object is not defined, or a point of an exceptional set where it fails to be well-behaved in some particular way, such as differentiability. ... In complex analysis, the residue is a complex number which describes the behavior of path integrals of a meromorphic function around a singularity. ...

$lim_{zto i}(z-i)f(z)=lim_{zto i}(z-i){e^{itz} over z^2+1}=lim_{zto i}(z-i){e^{itz} over (z-i)(z+i)}$

$=lim_{zto i}{e^{itz} over z+i}={e^{iti} over i+i}={e^{-t}over 2i}.$

According to the residue theorem, then, we have The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

$int_C f(z),dz=2pi icdotoperatorname{Res}_{z=i}f(z)=2pi i{e^{-t} over 2i}=pi e^{-t}.$

The contour C may be split into a "straight" part and a curved arc, so that

∫	+	∫	= πe ^{− t},
straight		arc

and thus

$int_{-a}^a =pi e^{-t}-int_{mbox{arc}}.$

It can be shown that if t > 0 then

$int_{mbox{arc}}{e^{itz} over z^2+1},dz rightarrow 0 mbox{as} arightarrowinfty.$

Therefore if t > 0 then

$int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^{-t}.$

A similar argument with an arc that winds around −i rather than i shows that if t < 0 then

$int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^t,$

and finally we have

$int_{-infty}^infty{e^{itz} over z^2+1},dz=pi e^{-left|tright|}.quadsquare$

(If t = 0 then the integral yields immediately to real-valued calculus methods and its value is π.)

Example (III) – trigonometric integrals

Certain substitutions can be made to integrals involving trigonometric functions, so the integral is transformed into a rational function of a complex variable and then the above methods can be used in order to evaluate the integral. In mathematics, the trigonometric functions are functions of an angle, important when studying triangles and modeling periodic phenomena. ...

As an example, consider

$int_{-pi}^{pi} {1 over 1 + 3 (cos{t})^2} ,dt.$

We seek to make a substitution of z = e^it.

Now, recall

$cos{t} = {1 over 2} left(e^{it}+e^{-it}right) = {1 over 2} left(z+{1 over z}right)$

and

${dz over dt} = iz, dt = {dz over iz}.$

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

$oint_C {1 over 1 + 3 ({1 over 2} (z+{1 over z}))^2} ,{dzover iz}$

$= oint_C {1 over 1 + {3 over 4} (z+{1 over z})^2}{1 over iz} ,dz = oint_C {-i over z+{3over 4}z(z+{1over z})^2},dz = -i oint_C { 1 over z+{3over 4}z(z^2+2+{1over z^2})} ,dz$

$= -i oint_C {1over z+{3over 4}(z^3+2z+{1 over z})} ,dz = -i oint_C {1 over {3over 4 }z^3+{5 over 2}z+{3 over 4z}} ,dz$

$= -i oint_C {4 over 3z^3+10z+{3over z}},dz = -4i oint_C {1 over 3z^3+10z+{3over z}},dz$

$= -4i oint_C { z over 3z^4+10z^2+3 } ,dz.$

We use the Cauchy integral formula. Factorize the denominator:

$= -4i oint_C { z over 3z^4+10z^2+3 } ,dz = -4i oint_C {z over 3(z^2+3)(z^2+1/3)},dz$

$= -4i oint_C {z over 3(z+sqrt{3}i)left(z-sqrt{3}iright)left(z+sqrt{1over 3}iright)left(z-sqrt{1over 3}iright)},dz$

$= -{4over 3}i oint_C {z over (z+sqrt{3}i)(z-sqrt{3}i)left(z+sqrt{1over 3}iright)left(z-sqrt{1over 3}iright)},dz.$

The singularities then to be considered are at 3^−1/2i, −3^−1/2i. We can now reduce the integral:

$= -{4over 3}i oint_{C_1} {,{z over (z+sqrt{3}i)(z-sqrt{3}i)left(z+sqrt{1over 3}iright)}, over left(z-sqrt{1over 3}iright)},dz + -{4over 3}i oint_{C_2} {,{z over (z+sqrt{3}i)(z-sqrt{3}i)left(z-sqrt{1over 3}iright)}, over left(z+sqrt{1over 3}iright)}$

where C₁ is a small circle about 3^−1/2i, and C₂ is a small circle about −3^−1/2i. We can now apply the formula:

$= -{4over 3}i left( 2pi i left.left({z over (z+sqrt{3}i)(z-sqrt{3}i)(z+sqrt{1over 3}i)}right)right|_{z=sqrt{1 over 3}i} right.$

$left. + 2pi i left.left({z over (z+sqrt{3}i)(z-sqrt{3}i)(z-sqrt{1over 3}i)}right)right|_{z=-sqrt{1 over 3}i} right)$

$= -{4over 3}i left( 2pi i left( { sqrt{1over 3}i over (sqrt{1over 3}i+sqrt{3}i)(sqrt{1over 3}i-sqrt{3}i)(sqrt{1over 3}i+sqrt{1over 3}i)} right) right.$

$left. +2pi i left( { -sqrt{1over 3}i over (-sqrt{1over 3}i+sqrt{3}i)(-sqrt{1over 3}i-sqrt{3}i)(-sqrt{1over 3}i-sqrt{1over 3}i } right)right)$

$= -{4over 3}i left( 2pi i left( {sqrt{1over 3}i over ({4 over sqrt{3}}i)(-{2 over sqrt{3}}i)({2 over sqrt{3}}i)} right) + 2 pi i left( {-sqrt{1over 3}i over ({2 over sqrt{3}}i)(-{4 over sqrt{3}}i)(-{2 over sqrt{3}}i)} right) right)$

$= -{4over 3}i left( 2pi i left( {sqrt{1over 3}i over i({4 over sqrt{3}})({2 over sqrt{3}})({2 over sqrt{3}})} right) + 2pi i left( {-sqrt{1over 3}i over -i({2 over sqrt{3}})({4 over sqrt{3}})({2 over sqrt{3}})} right) right)$

$= -{4over 3}i left( 2pi i left( {sqrt{1over 3} over ({4 over sqrt{3}})({2 over sqrt{3}})({2 over sqrt{3}})} right) + 2pi i left( {sqrt{1over 3} over ({2 over sqrt{3}})({4 over sqrt{3}})({2 over sqrt{3}})} right) right)$

$= -{4over 3}i left( 2pi i left( { ,sqrt{1over 3} ,over {16 over 3sqrt{3}} } right) + 2pi i left( {, sqrt{1over 3} ,over {16 over 3sqrt{3}} } right) right)$

$= -{4over 3}i left(2 pi i left({3over 16}right) + 2 pi i left({3over 16}right)right) = -{4over 3}i left(pi i left({3 over 8}+{3 over 8}right)right) = {4over 3}left({3 over 4}right)pi = pi.quadsquare$

Example (IV) – branch cuts

Consider

$int_0^infty {sqrt{x} over x^2+6x+8},dx.$

We can begin by formulating the complex integral

$int_C {sqrt{z} over z^2+6z+8},dz=I.$

We can use the Cauchy integral formula or residue theorem again to obtain the relevant residues. However, the important thing to note is that z^1/2=e^{1/2 Log(z)}, so z^1/2 has a branch cut. This affects our choice of the contour C. Normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculation of the integral slightly more complex, so we define it to be the positive real axis. Image File history File links Typical keyhole contour, drawn by myself. ... In complex analysis, a branch point may be thought of informally as a point z0 at which a multiple_valued function changes values when one winds once around z0. ...

Then, we use the so-called keyhole contour, which consists of a small circle about the origin of radius ε say, extending to a line segment parallel and close to the positive real axis but not touching it, to an almost full circle, returning to a line segment parallel, close, and below the positive real axis in the negative sense, returning to the small circle in the middle.

Let γ be the small circle of radius ε, Γ the larger, with radius r, then

$begin{align} int_C {sqrt{z} over z^2+6z+8},dz & = & int_epsilon^R {sqrt{z} over z^2+6z+8},dz & + & int_Gamma {sqrt{z} over z^2+6z+8},dz & + & int_R^epsilon {sqrt{z} over z^2+6z+8},dz & + & int_gamma {sqrt{z} over z^2+6z+8},dz. end{align}$

Since z^1/2=e^{1/2 Log(z)}, along the contour below the branch cut, we have gained 2π in argument along Γ, so

$int_R^epsilon {sqrt{z} over z^2+6z+8},dz=int_R^epsilon {e^{{1over 2} mathrm{Log}(z)} over z^2+6z+8},dz=int_R^epsilon {e^{{1over 2}(log{|z|}+i arg{z})} over z^2+6z+8},dz$

$=int_R^epsilon { e^{{1over 2}log{|z|}}e^{1/2(2pi i)} over z^2+6z+8},dz=int_R^epsilon { e^{{1over 2}log{|z|}}e^{pi i} over z^2+6z+8},dz$

$=int_R^epsilon {-sqrt{x} over x^2+6x+8},dx=-int_epsilon^R {-sqrt{x} over x^2+6x+8},dx$

simplifying,

$=int_epsilon^R {sqrt{x} over x^2+6x+8},dx,$

and then

$begin{align} int_C {sqrt{z} over z^2+6z+8},dz & = & int_epsilon^R {sqrt{z} over z^2+6z+8},dz & + & int_Gamma {sqrt{z} over z^2+6z+8},dz & + & int_epsilon^R {sqrt{z} over z^2+6z+8},dz & + & int_gamma {sqrt{z} over z^2+6z+8},dz. end{align}$

It can be shown that the integrals over Γ and γ both tend to zero as ε tends to zero and R tends to infinity, by an estimation argument above. Thus, then,

$int_C {sqrt{z} over z^2+6z+8},dz=2int_0^infty {sqrt{z} over z^2+6z+8},dz.$

By using the residue theorem or the Cauchy integral formula one obtains

$pi i left({iover sqrt{2}}-iright)=int_0^infty {sqrt{z} over z^2+6z+8},dz = pileft(1-{1oversqrt{2}}right).quadsquare$

Example (V) – logarithms and the residue at infinity

We seek to evaluate Image File history File links ContourLogs. ...

$I = int_0^3 {left(x^3 (3-x)right)^{1/4} over 5-x},dx.$

This requires a close study of

$f(z) = left(z^3 (3-z)right)^{1/4}.$

We will construct $f (z)$ so that it has a branch cut on $[0,3]$ , shown in red in the diagram. To do this, we choose two branches of the logarithm, setting

$(z^3)^{1/4} = z^{3/4} = exp(3/4 log(z)) quad mbox{where} quad -pi le arg(log(z)) < pi$

and

$(3-z)^{1/4} = exp(1/4 log(3-z)) quad mbox{where} quad 0 le arg(log(z)) < 2pi.$

The cut of $z 3 / 4$ is therefore $(-infty, 0]$ and the cut of $(3 - z) 1 / 4$ is $(-infty, 3]$ . It is easy to see that the cut of the product of the two, i.e. $f (z)$ , is $[0,3]$ , because $f (z)$ is actually continuous across $(-infty, 0).$ This is because when $z = - r < 0$ and we approach the cut from above, $f (z)$ has the value

$r^{3/4} exp(3/4 pi i) (3+r)^{1/4} exp(2/4 pi i) = r^{3/4} (3+r)^{1/4} exp(5/4 pi i).,$

When we approach from below, $f (z)$ has the value

$r^{3/4} exp(-3/4 pi i) (3+r)^{1/4} exp(0/4 pi i) = r^{3/4} (3+r)^{1/4} exp(-3/4 pi i).,$

But $exp( - 3 / 4π i) = exp(5 / 4π i),$ 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 $z 3 / 4$ and $(3 - z) 1 / 4 .$

We will use the contour shown in green in the diagram. To do this we must compute the value of $f (z)$ along the line segments just above and just below the cut. Let $z = r$ (in the limit, i.e. as the two green circles shrink to radius zero), where $0 le r le 3.$ Along the upper segment, we find that $f (z)$ has the value

$r^{3/4} exp(3/4 , 0 , pi i) (3-r)^{1/4} exp(2/4 pi i) = i , r^{3/4} (3-r)^{1/4},$

and along the lower segment,

$r^{3/4} exp(3/4 , 0 , pi i) (3-r)^{1/4} exp(0/4 pi i) = r^{3/4} (3-r)^{1/4},$

It follows that the integral of $f(z)/(5-z),$ along the upper segment is $-i , I$ in the limit, and along the lower segment, $I.,$

If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of $I,$ , by the Cauchy residue theorem. Let the radius of the green circles be $ρ$ , where $ρ < 1 / 1000$ and $rho rightarrow 0,$ and apply the ML-inequality. For the circle $C L$ on the left, we find The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

$left| int_{C_L} frac{f(z)}{5-z} dz right| le 2 pi rho frac{rho^{3/4} (3+1/1000)^{1/4}}{5-1/1000} in mathcal{O} left( rho^{7/4} right) rightarrow 0.$

Similarly, for the circle $C R$ on the right, we have

$left| int_{C_R} frac{f(z)}{5-z} dz right| le 2 pi rho frac{(3+1/1000)^{3/4} rho^{1/4}}{2-1/1000} in mathcal{O} left( rho^{5/4} right) rightarrow 0.$

Now using the Cauchy residue theorem, we have The residue theorem in complex analysis is a powerful tool to evaluate path integrals of meromorphic functions over closed curves and can often be used to compute real integrals as well. ...

$(1-i) I = - 2pi i left( mathrm{Res}_{z=5} frac{f(z)}{5-z} + mathrm{Res}_{z=infty} frac{f(z)}{5-z} right).$

Using the branch of the logarithm from before, clearly

$mathrm{Res}_{z=5} frac{f(z)}{5-z} = - 5^{3/4} exp(1/4 log(-2)).$

The pole is shown in blue in the diagram. The value simplifies to

$- 5^{3/4} exp(1/4 (log(2) + pi i) ) = - exp(1/4 pi i) 5^{3/4} 2^{1/4}.,$

We use the following formula for the residue at infinity:

$mathrm{Res}_{z=infty} h(z) = mathrm{Res}_{z=0} - frac{1}{z^2} hleft(frac{1}{z}right).$

Substituting, we find

$1/(5-1/z) = -z left(1 + 5z + 5^2 z^2 + 5^3 z^3 + cdotsright)$

and

$left(1/z^3 (3-1/z)right)^{1/4} = 1/z , (3z-1)^{1/4} = 1/z exp(1/4 pi i) (1-3z)^{1/4},$

where we have used the fact that $- 1 = exp(π i)$ for the second branch of the logarithm. Next we apply the binomial expansion, obtaining

$1/z exp(1/4 pi i) left( 1 - {1/4 choose 1} 3z + {1/4 choose 2} 3^2 z^2 - {1/4 choose 3} 3^3 z^3 + cdots right).$

The conclusion is that

$mathrm{Res}_{z=infty} frac{f(z)}{5-z} = exp(1/4 pi i) (5 - 3/4) = exp(1/4 pi i) 17/4.$

Finally, it follows that the value of $I,$ is

$I = - 2 pi i , frac{exp(1/4 pi i)}{1-i} left( 17/4 - 5^{3/4} 2^{1/4} right) = - 2 pi i , 2^{-1/2} i left( 17/4 - 5^{3/4} 2^{1/4} right)$

which yields

$I = frac{pi}{2sqrt{2}} left(17 - 5^{3/4} 2^{9/4} right) = frac{pi}{2sqrt{2}} left(17 - 40^{3/4} right).$

The integral was discussed on Les-Mathematiques.net, where an additional residue computation is shown. The link is here. In complex analysis, the evaluation of integrals of real-valued functions along intervals on the real line, is not readily found with certain integrands and methods involving only real variables. ...

External links

Jean Jacquelin, Marko Riedel, Branche univalente, Les-Mathematiques.net, in French.
A collection of examples

Category: Complex analysis

Results from FactBites:

Methods of contour integration - Wikipedia, the free encyclopedia (1602 words)
	Direct methods involve the calculation of the integral by means of methods similar to those in calculating line integrals in several-variable calculus.
	The integral is evaluated in a method akin to a real-variable integral.
	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

Springer Online Reference Works (748 words)
	The well-known modern applications of the method of complex integration, which use the Cauchy theorem on residues, the Phragmén–Lindelöf theorem on Dirichlet series, the saddle point method, etc., are extremely varied in their form and content.
	A classical example of the method of complex integration is illustrated by the proof of the analytic continuation and the derivation of the functional equation of the Riemann zeta-function (see [2], [3]).
	A novel approach to the prime number theorem using the method of contour integration was given in [a1].

More results at FactBites »

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Direct methods GA_googleFillSlot("encyclopedia_square");