Section 14

Computing Integrals Over a Closed Contour

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Problem 14.1

Compute the integrals of rational functions:

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(1)

∮∣z∣=1dzz\oint_{\left|z\right| = 1} \frac{d z}{z};

(2)

∮∣z∣=2dz(z−1)2\oint_{\left|z\right| = 2} \frac{d z}{(z-1)^{2}};

(3)

∮∣z∣=2dzz2+1\oint_{\left|z\right| = 2} \frac{d z}{z^{2}+1};

(4)

∮∣z∣=3dzz2+z+1\oint_{\left|z\right| = 3} \frac{d z}{z^{2}+z+1};

(5)

∮∣z∣=3z4z2+z−2dz\oint_{\left|z\right| = 3} \frac{z^{4}}{z^{2}+z-2} d z;

(6)

∮∣z+i∣=2dzz2(z2−7z+12)\oint_{\left|z+i\right| = 2} \frac{d z}{z^{2}\left(z^{2}-7 z+12\right)}.

Problem 14.2

Compute the integrals:

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(1)

∮∣z∣=3zcos⁡πzz+2idz\oint_{\left|z\right| = 3} \frac{z \cos \frac{\pi }{z}}{z+2 i} d z;

(2)

∮∣z∣=4e1/(z−1)z−2dz\oint_{\left|z\right| = 4} \frac{e^{1 /(z-1)}}{z-2} d z;

(3)

∮∣z−12∣=1z3e1/z1−z2dz\oint_{\left|z-\frac{1}{2}\right| = 1} \frac{z^{3} e^{1 / z}}{1-z^{2}} d z;

(4)

∮∣z∣=2z2sin⁡11−zdz\oint_{\left|z\right| = 2} z^{2} \sin \frac{1}{1-z} d z;

(5)

∮∣z∣=114i+3zcos⁡5i6z+3idz\oint_{\left|z\right| = 1} \frac{1}{4 i+3 z} \cos \frac{5 i}{6 z+3 i} d z;

(6)

∮∣z∣=213+ize4i/(z−i)dz\oint_{\left|z\right| = 2} \frac{1}{3+i z} e^{4 i /(z-i)} d z;

(7)

∮∣z∣=3/2ztg⁡z(z2−1)2dz\oint_{\left|z\right| = 3 / 2} \frac{z \operatorname {tg} z}{\left(z^{2}-1\right)^{2}} d z;

(8)

∮∣z∣=5/2z2z−3sin⁡zz−2dz\oint_{\left|z\right| = 5 / 2} \frac{z^{2}}{z-3} \sin \frac{z}{z-2} d z;

(9)

∮∣z−i∣=3zz2+9ch⁡zz−2dz\oint_{\left|z-i\right| = 3} \frac{z}{z^{2}+9} \operatorname {ch} \frac{z}{z-2} d z;

(10)

∮∣z∣=6z2eiz+idz\oint_{\left|z\right| = 6} \frac{z^{2}}{e^{i z}+i} d z;

(11)

∮∣z∣=3dz(z−1)2(1−e2z)\oint_{\left|z\right| = 3} \frac{d z}{(z-1)^{2}\left(1-e^{2 z}\right)};

(12)

∮∣z∣=2z2z−1e1/(z−1)dz\oint_{\left|z\right| = 2} \frac{z^{2}}{z-1} e^{1 /(z-1)} d z;

(13)

∮∣z+13∣=163z+11−sin⁡π2zdz\oint_{\left|z+\frac{1}{3}\right| = \frac{1}{6}} \frac{3 z+1}{1-\sin \frac{\pi }{2 z}} d z;

(14)

∮∣z∣=4zdze1/z+e1/2z\oint_{\left|z\right| = 4} \frac{z d z}{e^{1 / z}+e^{1 / 2 z}};

(15)

∮∣z∣=2ez5z(1+z5)dz\oint_{\left|z\right| = 2} \frac{e^{z^{5}}}{z\left(1+z^{5}\right)} d z;

(16)

∮∣z−π2(1−i)∣=πzdzcos⁡z−ch⁡z\oint_{\left|z-\frac{\pi }{2}(1-i)\right| = \pi } \frac{z d z}{\cos z-\operatorname {ch} z};

(17)

∮∣z∣=1(z−i)sin⁡1iz(z−3i)2dz\oint_{\left|z\right| = 1} \frac{(z-i) \sin \frac{1}{i z}}{(z-3 i)^{2}} d z;

(18)

∮∣z∣=1(z+1)cos⁡iz(2i−z)2dz\oint_{\left|z\right| = 1} \frac{(z+1) \cos \frac{i}{z}}{(2 i-z)^{2}} d z;

(19)

∮∣z∣=1/2z+2π(2sin⁡13z−1)dz\oint_{\left|z\right| = 1 / 2} \frac{z+\frac{2}{\pi }}{\left(2 \sin \frac{1}{3 z}-1\right)} d z;

(20)

∮∣z+i∣=3/22(z+3i)ch⁡12zz4+10z2+9dz\oint_{\left|z+i\right| = 3 / 2} \frac{2(z+3 i) \operatorname {ch} \frac{1}{2 z}}{z^{4}+10 z^{2}+9} d z;

(21)

∮∣z∣=1z−1ln⁡2e1/z−2dz\oint_{\left|z\right| = 1} \frac{z-\frac{1}{\ln 2}}{e^{1 / z}-2} d z;

(22)

∮∣z−π2∣=32z−iln⁡2tg⁡z−i3dz\oint_{\left|z-\frac{\pi }{2}\right| = 3} \frac{2 z-i \ln 2}{\operatorname {tg} z-\frac{i}{3}} d z;

(23)

∮∣z+1−i∣=2z+i(z−i)sh⁡12zdz\oint_{\left|z+1-i\right| = 2} \frac{z+i}{(z-i) \operatorname {sh} \frac{1}{2 z}} d z;

(24)

∮∣z−1∣=1zdz(π−3z)(1+cos⁡3z)\oint_{\left|z-1\right| = 1} \frac{z d z}{(\pi -3 z)(1+\cos 3 z)}.

Problem 14.3

Making an appropriate change of variable, reduce the given integral to an integral over a closed contour in C\mathbb {C} and compute it:

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(1)

∫02πdφa+cos⁡φ(a>1)\int_{0}^{2 \pi } \frac{d \varphi }{a+\cos \varphi }(a > 1);

(2)

∫02πdφ(a+bcos⁡φ)2(a>b>0)\int_{0}^{2 \pi } \frac{d \varphi }{(a+b \cos \varphi )^{2}} \quad (a > b > 0);

(3)

∫02πdφ(a+bcos⁡2φ)2\int_{0}^{2 \pi } \frac{d \varphi }{\left(a+b \cos^{2} \varphi \right)^{2}};

(4)

∫02πecos⁡φcos⁡(nφ−sin⁡φ)dφ,n∈Z\int_{0}^{2 \pi } e^{\cos \varphi } \cos (n \varphi -\sin \varphi ) d \varphi , \quad n \in \mathbb {Z}.

Remark.
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In Problems 4--6 it is assumed that, while moving along the boundary Γ\Gamma of the domain DD in the direction of integration, the domain DD itself remains on the left.

Problem 14.4

Prove that the integral

∫Γeiz(z−i)2dz \int _{\Gamma } \frac{e^{i z}}{(z-i)^{2}} d z

taken over the boundary Γ\Gamma of the half-plane {z:Im(z)>0}\left\{ z: \mathrm{Im}\left(z\right) > 0\right\}, equals the sum of the residues of the integrand in this half-plane, and find its value.

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Problem 14.5

Verify the applicability of the residue theorem to the following integrals, taken over the boundaries Γ\Gamma of unbounded domains DD, and compute these integrals:

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(1)

∫Γze−zz2−1dz(D={z:Re(z)>0})\int_{\Gamma } \frac{z e^{-z}}{z^{2}-1} d z \quad (D = \left\{ z: \mathrm{Re}\left(z\right) > 0\right\} );

(2)

∫Γezsh⁡2zdz(D={z:−π4<Im(z)<π4})\int_{\Gamma } \frac{e^{z}}{\operatorname {sh} 2 z} d z \quad \left(D = \left\{ z:-\frac{\pi }{4} < \mathrm{Im}\left(z\right) < \frac{\pi }{4}\right\} \right);

(3)

∫Γz3(z−1)2e−z3dz(D={z:−π6<arg⁡z<π6})\int_{\Gamma } \frac{z^{3}}{(z-1)^{2}} e^{-z^{3}} d z \quad \left(D = \left\{ z:-\frac{\pi }{6} < \arg z < \frac{\pi }{6}\right\} \right).

Problem 14.6

Verify that the residue theorem is not applicable to the integrals taken over the boundaries Γ\Gamma of the indicated domains DD:

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(1)

I=∫Γe−z2dz(D={z:Im(z)>0})I = \int_{\Gamma } e^{-z^{2}} d z \quad (D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} );

(2)

I=∫Γsin⁡z1+z2dz(D={z:Im(z)>0})I = \int_{\Gamma } \frac{\sin z}{1+z^{2}} d z \quad (D = \left\{ z: \mathrm{Im}\left(z\right) > 0\right\} ).

Problem 14.7

Let the functions f(z)f(z) and g(z)g(z) be regular in the closure Dˉ\bar{D} of a bounded domain DD, where the function f(z)f(z) has, in the domain DD, zeros a1,…,ana_{1}, \ldots , a_{n} (each zero written as many times as its order) and has no other zeros either in the domain DD or on its boundary Γ\Gamma. Prove that

∮Γg(z)f′(z)f(z)dz=2πi∑k=1ng(ak) \oint _{\Gamma } g(z) \frac{f^{\prime }(z)}{f(z)} d z = 2 \pi i \sum _{k = 1}^{n} g\left(a_{k}\right)
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Problem 14.8

Let the function f(z)f(z) of Problem 7 also have, in the domain DD, poles b1,…,bmb_{1}, \ldots , b_{m} (each pole also written as many times as its order). Prove that

∮Γg(z)f′(z)f(z)dz=2πi∑k=1ng(ak)−2πi∑k=1mg(bk) \oint _{\Gamma } g(z) \frac{f^{\prime }(z)}{f(z)} d z = 2 \pi i \sum _{k = 1}^{n} g\left(a_{k}\right)-2 \pi i \sum _{k = 1}^{m} g\left(b_{k}\right)
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Problem 14.9

Let the function f(z)f(z) be regular in the closure Dˉ\bar{D} of a bounded domain DD with boundary Γ\Gamma, and let the points a1,…,ana_{1}, \ldots , a_{n} lie in the domain DD and be pairwise distinct. Denote

P(z)=(z−a1)…(z−an) P(z) = \left(z-a_{1}\right) \ldots \left(z-a_{n}\right)

and

Φ(z)=−P(z)2πi∮Γf(ξ)P(ξ)dξξ−z(z∉Dˉ) \Phi (z) = -\frac{P(z)}{2 \pi i} \oint _{\Gamma } \frac{f(\xi )}{P(\xi )} \frac{d \xi }{\xi -z} \quad (z \notin \bar{D})

Prove that the function Φ(z)\Phi (z) extends analytically to the entire plane and is a polynomial of degree n−1n-1 satisfying the conditions

Φ(ak)=f(ak)(k=1,2,…,n). \Phi \left(a_{k}\right) = f\left(a_{k}\right) \quad (k = 1,2, \ldots , n).

(The polynomial Φ(z)\Phi (z) is called the Lagrange interpolation polynomial.)

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Problem 14.10

Prove the formula

12πi∮∣z∣=1ex2(z+1z)dz=∑n=0∞(x2)2n+11n!(n+1)!. \frac{1}{2 \pi i} \oint _{\left|z\right| = 1} e^{\frac{x}{2}\left(z+\frac{1}{z}\right)} d z = \sum _{n = 0}^{\infty }\left(\frac{x}{2}\right)^{2 n+1} \frac{1}{n!(n+1)!}.
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