Residual theorem

The residual theorem is an important theorem of function theory , a branch of mathematics . It represents a generalization of Cauchy's integral theorem and Cauchy's integral formula . Its importance lies not only in the far-reaching consequences within function theory, but also in the calculation of integrals over real functions .

It says that the curve integral along a closed curve over a holomorphic function except for isolated singularities depends only on the residual in the singularities inside the curve and the number of revolutions of the curve around these singularities. Instead of a curve integral, you only have to calculate residuals and circulating numbers, which is easier in many cases.

sentence

Let be an elementary domain , i.e. a simply connected domain in the complex number plane . Furthermore, let be a discrete subset of , a holomorphic function , a real interval and a closed path in . Then applies to the complex path integral ${\displaystyle D\subseteq \mathbb {C} }$ ${\displaystyle D_{f}}$ ${\displaystyle D}$ ${\displaystyle f\colon D\setminus D_{f}\to \mathbb {C} }$ ${\displaystyle I}$ ${\displaystyle \Gamma \colon I\to D\setminus D_{f}}$ ${\displaystyle D\setminus D_{f}}$

${\displaystyle {\frac {1}{2\pi \mathrm {i} }}\int _{\Gamma }f=\sum _{a\in D_{f}}\operatorname {ind} _{\Gamma }(a)\operatorname {Res} _{a}f}$,

where the rotation number of is with respect to and the residual of in . ${\displaystyle \operatorname {ind_{\Gamma }} (a)}$ ${\displaystyle \Gamma }$ ${\displaystyle a}$ ${\displaystyle \operatorname {Res} _{a}f}$ ${\displaystyle f}$ ${\displaystyle a}$

Remarks

• The sum on the right-hand side is always finite, because the area enclosed by (simply connected) is relatively compact in and therefore restricted. Because discrete is in, is finite, and only these are the points that contribute to the sum, because for all others the number of turns or the residual vanishes. ${\displaystyle \Gamma }$ ${\displaystyle \operatorname {Int} \,\Gamma }$ ${\displaystyle D}$ ${\displaystyle D_{f}}$ ${\displaystyle D}$ ${\displaystyle \operatorname {Int} \,\Gamma \cap D_{f}}$
• If the points are liftable singularities, if the residual vanishes in these points, then one obtains Cauchy's integral theorem ${\displaystyle D_{f}}$

${\displaystyle \int _{\Gamma }\,f=0}$.

• Is on holomorphic and has a first order pole in with a residual then one gets the integral formula of Cauchy ${\displaystyle f}$ ${\displaystyle D}$ ${\displaystyle z\in D}$ ${\displaystyle \textstyle \zeta \mapsto {\frac {f(\zeta )}{\zeta -z}}}$ ${\displaystyle z}$ ${\displaystyle f(z)}$

${\displaystyle {\frac {1}{2\pi \mathrm {i} }}\int _{\Gamma }{\frac {f(\zeta )}{\zeta -z}}\mathrm {d} \zeta =\operatorname {ind} _{\Gamma }(z)f(z).}$

Integral counting zero and poles

Is on meromorphic with the set of zeros , the set of poles and , then with the residue theorem it follows: ${\displaystyle f\not \equiv 0}$ ${\displaystyle D}$ ${\displaystyle N}$ ${\displaystyle P}$ ${\displaystyle \operatorname {Spur} \,\Gamma \cap \left(N\cup P\right)=\emptyset }$

${\displaystyle {\frac {1}{2\pi \mathrm {i} }}\int _{\Gamma }{\frac {f'}{f}}=\sum \limits _{a\in N\cup P}\operatorname {ind} _{\Gamma }(a)\operatorname {ord} _{a}f}$

Here designated

${\displaystyle \operatorname {ord} _{a}f:={\begin{cases}k,&{\mbox{falls }}f{\mbox{ in }}a{\mbox{ eine Nullstelle }}k{\mbox{-ter Ordnung hat}}\\-k,&{\mbox{falls }}f{\mbox{ in }}a{\mbox{ eine Polstelle }}k{\mbox{-ter Ordnung hat}}\\0,&{\mbox{sonst}}\end{cases}}}$

the zero or pole order of in . With the calculation rule of the residual for the logarithmic derivative, the following applies ${\displaystyle f}$ ${\displaystyle a}$

${\displaystyle \operatorname {ord} _{a}f=\operatorname {Res} _{a}{\frac {f'(z)}{f(z)}}}$.

Application examples

With the residual theorem one can calculate real integrals with infinite integration limits. To do this, a closed curve is introduced in the complex plane, which covers the real integration limits; the integral over the rest of the curve is usually constructed in such a way that it disappears after the limit has been crossed. The complex level is supplemented by a point at infinity ( Riemann number ball ). This method of calculating improper real integrals is often referred to in theoretical physics as the “method of residuals”.

Fractional rational functions

The integral over the semicircle disappears for , it remains the integral over the real axis. ${\displaystyle R\to \infty }$

Is the quotient of two polynomials with and for all , is ${\displaystyle f={\tfrac {p}{q}}}$ ${\displaystyle \operatorname {deg} \,p+2\leq \operatorname {deg} \,q}$ ${\displaystyle q(z)\neq 0}$ ${\displaystyle z\in \mathbb {R} }$

${\displaystyle \int _{-\infty }^{\infty }f(z)\mathrm {d} z=2\pi \mathrm {i} \sum _{a\in \mathbb {H} }\operatorname {Res} _{a}f(z)}$,

where is the upper half-level, because you can integrate with , for a large one , over the closed semicircle and cross the border . Because of large and constants follows with the standard estimate for curve integrals ${\displaystyle \mathbb {H} :=\{z\in \mathbb {C} :\operatorname {Im} z>0\}}$ ${\displaystyle \alpha \colon [0,\pi ]\to \mathbb {C} }$ ${\displaystyle t\mapsto Re^{\mathrm {i} t}}$ ${\displaystyle R\in \mathbb {R} }$ ${\displaystyle \Gamma :=[-R,R]\oplus \alpha }$ ${\displaystyle R\rightarrow \infty }$ ${\displaystyle \left|{\tfrac {p(z)}{q(z)}}\right|\leq {\tfrac {c_{p}|z|^{\operatorname {deg} \,p}}{c_{q}|z|^{\operatorname {deg} \,q}}}\leq {\tfrac {c}{|z|^{2}}}}$ ${\displaystyle |z|}$ ${\displaystyle c,c_{p},c_{q}\in \mathbb {R} }$

${\displaystyle \left|\int _{\alpha }f\right|\leq L(\alpha )\cdot \max _{\zeta \in \operatorname {im} \alpha }\left|f(\zeta )\right|\leq \pi R\cdot {\frac {c}{R^{2}}}\rightarrow 0\,(R\rightarrow \infty )}$,

so it holds and because of the above estimate the latter integral also exists. The calculation formula follows with the residual rate. ${\displaystyle \textstyle \int _{\Gamma }f\rightarrow \int _{-\infty }^{\infty }f(z)\mathrm {d} z\,(R\rightarrow \infty )}$

Example: Let , with Poland 1st order in . Then is , and with it . ${\displaystyle f\colon \mathbb {C} \setminus \{\pm \mathrm {i} \}\to \mathbb {C} }$ ${\displaystyle z\mapsto {\tfrac {1}{z^{2}+1}}}$ ${\displaystyle \pm \mathrm {i} }$ ${\displaystyle \operatorname {Res} _{\mathrm {i} }f(z)={\tfrac {1}{2\mathrm {i} }}}$ ${\displaystyle \textstyle \int _{-\infty }^{\infty }f(z)\mathrm {d} z=2\pi \mathrm {i} \cdot {\tfrac {1}{2\mathrm {i} }}=\pi }$

Fractional rational functions with exponential function

The integral over the three upper sides of the rectangle vanishes for , it remains the integral over the real axis. ${\displaystyle r\to \infty }$

${\displaystyle P}$and let polynomials with , the polynomial has no real zeros and the zeros in the upper complex half-plane. Then applies to each ${\displaystyle Q}$ ${\displaystyle deg(P)+1\leq deg(Q)}$ ${\displaystyle Q}$ ${\displaystyle a_{1},\ldots ,a_{k}}$ ${\displaystyle \alpha >0}$

${\displaystyle \int _{-\infty }^{\infty }{\frac {P(x)}{Q(x)}}\exp(i\alpha x)\mathrm {d} x=2\pi \mathrm {i} \sum _{i=1}^{k}\operatorname {Res} _{a_{i}}f(z)}$

with . As above, a closed path is also defined here , which consists of the straight path from to , but instead of the semicircle one uses the rectangle built above it with a height , which is traversed counterclockwise. The function can by assumption to a constant times are estimated. The integrals over the vertical lines are then by means of a standard estimate , which approaches zero. For the top horizontal side is and so . The integral over this side of the rectangle is then using a standard estimate . It follows that the integral over the entire upper part of the rectangle converges to zero and one obtains the assertion. ${\displaystyle f(z):={\frac {P(z)}{Q(z)}}\exp(i\alpha z)}$ ${\displaystyle \Gamma }$ ${\displaystyle -r}$ ${\displaystyle r}$ ${\displaystyle {\sqrt {r}}}$ ${\displaystyle z\mapsto {\tfrac {P(z)}{Q(z)}}}$ ${\displaystyle C}$ ${\displaystyle {\tfrac {1}{|z|}}}$ ${\displaystyle \leq C{\tfrac {1}{\sqrt {r}}}}$ ${\displaystyle Im(z)={\sqrt {r}}}$ ${\displaystyle |\exp(i\alpha z)|=\exp(-\alpha {\sqrt {r}})}$ ${\displaystyle \leq 2C\exp(-\alpha {\sqrt {r}})}$ ${\displaystyle r\to \infty }$

Example: consider the function . It fulfills all of the above conditions: The polynomial in the denominator only has zeros and therefore none on the real axis. Hence: ${\displaystyle {\frac {x\exp {(2ix)}}{x^{2}+1}}}$ ${\displaystyle \pm i}$

${\displaystyle \int _{-\infty }^{\infty }{\frac {x\exp {2ix}}{(x+i)(x-i)}}\mathrm {d} x=2\pi \mathrm {i} \operatorname {Res} _{i}f(z)=i\pi \exp {(-2)}}$

Fractional rational function with a non-integer term

Are and polynomials for which applies, where applies, have no zeros in and no zeros in zero. Then applies ${\displaystyle P}$ ${\displaystyle Q}$ ${\displaystyle \operatorname {deg} \,Q>\operatorname {deg} \,P+\lambda }$ ${\displaystyle \lambda \in \mathbb {R} ^{+}\backslash \mathbb {Z} }$ ${\displaystyle Q}$ ${\displaystyle \mathbb {R} ^{+}}$ ${\displaystyle P/Q}$

${\displaystyle \int _{0}^{\infty }x^{\lambda -1}{\frac {P(x)}{Q(x)}}\mathrm {d} x={\frac {\pi }{\sin {(\lambda \pi )}}}\sum _{p\in \mathbb {C} \backslash \mathbb {R^{+}} }\operatorname {Res} _{p}(-z)^{\lambda -1}{\frac {P(z)}{Q(z)}}}$

Example: Is , so is , the function has the poles and all other requirements are also met. It is accordingly . Thus applies ${\displaystyle f(x)={\frac {x^{3/2-1}}{x^{2}+1}}}$ ${\displaystyle \lambda =3/2}$ ${\displaystyle \pm i}$ ${\displaystyle \operatorname {Res} _{\pm i}f(z)={\frac {(\mp i)^{1/2}}{\pm 2i}}}$

${\displaystyle \int _{0}^{\infty }{\frac {\sqrt {x}}{1+x^{2}}}\mathrm {d} x=-\pi \left({\frac {\sqrt {-i}}{2i}}+{\frac {\sqrt {i}}{-2i}}\right)={\frac {\pi }{\sqrt {2}}}}$

Trigonometric functions

Is the quotient of two polynomials with for all with . Then applies ${\displaystyle r={\tfrac {p}{q}}}$ ${\displaystyle q(x,y)\neq 0}$ ${\displaystyle x,y\in \mathbb {R} }$ ${\displaystyle x^{2}+y^{2}=1}$

${\displaystyle {\begin{aligned}\int _{0}^{2\pi }r(\cos t,\sin t)\mathrm {d} t&=\int _{0}^{2\pi }r\left({\frac {e^{\mathrm {i} t}+e^{-\mathrm {i} t}}{2}},{\frac {e^{\mathrm {i} t}-e^{-\mathrm {i} t}}{2\mathrm {i} }}\right)\mathrm {d} t\\&=\int _{\partial \mathbb {E} }{\frac {1}{\mathrm {i} z}}\cdot r\left({\frac {z+{\frac {1}{z}}}{2}},{\frac {z-{\frac {1}{z}}}{2\mathrm {i} }}\right)\mathrm {d} {z}\\&=2\pi \sum _{a\in \mathbb {E} }\operatorname {Res} _{a}\left({\frac {1}{z}}\cdot r\left({\frac {z+{\frac {1}{z}}}{2}},{\frac {z-{\frac {1}{z}}}{2\mathrm {i} }}\right)\right),\end{aligned}}}$

where is the unit disk. Because the number of turns of the unit circle is inside the unit circle , and according to the assumption there are no singularities on the unit circle line. Theoretically, such integrals can also be solved using the Weierstrasse substitution , but this is usually more complex. If the interval limits of the integral to be calculated are not exact and zero, this can be achieved by means of a linear substitution or by means of symmetry arguments. ${\displaystyle \mathbb {E} :=\{z\in \mathbb {C} :|z|<1\}}$ ${\displaystyle 1}$ ${\displaystyle 2\pi }$

Example: It applies

${\displaystyle \int _{0}^{2\pi }{\frac {\mathrm {d} t}{2+\sin t}}=2\int _{\partial \mathbb {E} }{\frac {\mathrm {d} z}{z^{2}+4\mathrm {i} z-1}}=4\pi \mathrm {i} \cdot {\frac {1}{2{\sqrt {3}}\mathrm {i} }}={\frac {2\pi }{\sqrt {3}}}}$,

because in pole has 1st order, but only the pole at is in , and there has the residual . ${\displaystyle f\colon \mathbb {C} \to \mathbb {C} ,\ z\mapsto {\tfrac {1}{z^{2}+4\mathrm {i} z-1}}}$ ${\displaystyle \left(-2\pm {\sqrt {3}}\right)\mathrm {i} }$ ${\displaystyle \left(-2+{\sqrt {3}}\right)\mathrm {i} }$ ${\displaystyle \mathbb {E} }$ ${\displaystyle f}$ ${\displaystyle {\tfrac {1}{2{\sqrt {3}}\mathrm {i} }}}$

Fourier transform

Given a function . Furthermore, there are points with , where is. Are there then two numbers with for large , then for all the formula ${\displaystyle f\in C^{\infty }(\mathbb {R} )\cap L^{1}(\mathbb {R} )}$ ${\displaystyle a_{1},\dotsc ,a_{k}\in \mathbb {H} }$ ${\displaystyle f\in \mathbb {O} (\{z,\operatorname {Im} z>-\varepsilon \}\cap \{a_{1},\dotsc ,a_{k}\})}$ ${\displaystyle \varepsilon >0}$ ${\displaystyle C,\delta >0}$ ${\displaystyle \left|f(z)\right|\leq C\left|z\right|^{-1-\delta }}$ ${\displaystyle \left|z\right|}$ ${\displaystyle x>0}$

${\displaystyle \int _{-\infty }^{\infty }f(y)e^{\mathrm {i} xy}\mathrm {d} y=2\pi \mathrm {i} \sum _{a\in \mathbb {H} }\operatorname {Res} _{a}(f(z)e^{\mathrm {i} xz}).}$

The same formula applies to . With the help of this method, complicated Fourier integrals can be calculated. The proof is carried out as above by breaking down the path of integration into the part on the real axis and the part in the upper half-plane. Then the limit value is considered again and the integral over the curve in the upper half plane disappears due to Jordan's lemma . ${\displaystyle x<0}$

The residual theorem for Riemann surfaces

The residual theorem can be generalized to compact Riemann surfaces . For a meromorphic 1-form on such a surface, the sum of the residuals is zero.

The consequence is Liouville's second theorem on elliptic functions .

literature

• Kurt Endl, Wolfgang Luh : Analysis. Volume 3: Function Theory, Differential Equations. 6th revised edition. Aula-Verlag, Wiesbaden 1987, ISBN 3-89104-456-9 , p. 229.
• Wolfgang Fischer, Ingo Lieb : Function theory. 7th improved edition. Vieweg, Braunschweig a. a. 1994, ISBN 3-528-67247-1 , p. 145, sentence 4.1.
• AP Yuzhakov: Residue of an analytic function . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).

Web links

• The residual theorem. gsi.de
• Eric W. Weisstein : Residue Theorem . In: MathWorld (English).
• On the residual theorem . In: PlanetMath . (English)
• Elements of function theory. The most important sentences and aids for applications in physical field theory. (PDF; 441 kB) astrophys-neunhof.de, residual theorem and Cauchy's integral theorem.