'2024/11/09 글 목록

Integration along a branch cut-033

Mathematics 2024. 11. 9. 18:04

$I=∫∞0log(1+x2)dxx1+a=πcsc(πa/2)a(0<a<2)<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mi>I</mi><mo>=</mo><msubsup><mo data-mjx-texclass="OP">∫</mo><mn>0</mn><mi mathvariant="normal">∞</mi></msubsup><mfrac><mrow><mi>log</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>x</mi><mn>2</mn></msup><mo stretchy="false">)</mo><mi>d</mi><mi>x</mi></mrow><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mfrac><mo>=</mo><mfrac><mrow><mi>π</mi><mi>csc</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn><mo stretchy="false">)</mo></mrow><mi>a</mi></mfrac><mstyle scriptlevel="0"><mspace width="2em"></mspace></mstyle><mo stretchy="false">(</mo><mn>0</mn><mo><</mo><mi>a</mi><mo><</mo><mn>2</mn><mo stretchy="false">)</mo></math>$

그림과 같은 경로에서 $f(z)=log(1+z2)z1+a<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mi>f</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mo>=</mo><mfrac><mrow><mi>log</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>z</mi><mn>2</mn></msup><mo stretchy="false">)</mo></mrow><msup><mi>z</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mfrac></math>$ 의 적분을 고려하자.

$z = \pm i <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><mo>\pm</mo><mi>i</mi></math>$ , $z = 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><mn>0</mn></math>$ 이 branch point이므로 cutline은 그림과 같이 선택한다.

각 branch point을 감싸는 미소원호에서 적분과 $C \infty <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mi mathvariant="normal">\infty</mi></msub></math>$ 에서의 적분은 0으로 수렴함은 쉽게 보일 수 있다. 그리고 $C 1 <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>1</mn></msub></math>$ , $C 2 <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>C</mi><mn>2</mn></msub></math>$ 에서 각각 $z = x e i 0 (x : 0 \to \infty) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><mi>x</mi><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mn>0</mn></mrow></msup><mtext> </mtext><mo stretchy="false">(</mo><mi>x</mi><mo>:</mo><mn>0</mn><mo accent="false" stretchy="false">\to</mo><mi mathvariant="normal">\infty</mi><mo stretchy="false">)</mo></math>$ , $z = x e i 2 π (x : \infty \to 0) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>z</mi><mo>=</mo><mi>x</mi><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mn>2</mn><mi>π</mi></mrow></msup><mtext> </mtext><mo stretchy="false">(</mo><mi>x</mi><mo>:</mo><mi mathvariant="normal">\infty</mi><mo accent="false" stretchy="false">\to</mo><mn>0</mn><mo stretchy="false">)</mo></math>$ 이므로

$∫C1=∫∞0log(1+x2)dxx1+a=I∫C2=∫0∞log(1+x2)dxei2π(1+a)x1+a=−e−i2πaI<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>1</mn></msub></mrow></msub><mo>=</mo><msubsup><mo data-mjx-texclass="OP">∫</mo><mn>0</mn><mi mathvariant="normal">∞</mi></msubsup><mfrac><mrow><mi>log</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>x</mi><mn>2</mn></msup><mo stretchy="false">)</mo><mi>d</mi><mi>x</mi></mrow><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mfrac><mo>=</mo><mi>I</mi><mspace linebreak="newline"></mspace><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>2</mn></msub></mrow></msub><mo>=</mo><msubsup><mo data-mjx-texclass="OP">∫</mo><mi mathvariant="normal">∞</mi><mn>0</mn></msubsup><mfrac><mrow><mi>log</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><msup><mi>x</mi><mn>2</mn></msup><mo stretchy="false">)</mo><mi>d</mi><mi>x</mi></mrow><mrow><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mn>2</mn><mi>π</mi><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><mi>a</mi><mo stretchy="false">)</mo></mrow></msup><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mrow></mfrac><mo>=</mo><mo>−</mo><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mo>−</mo><mi>i</mi><mn>2</mn><mi>π</mi><mi>a</mi></mrow></msup><mi>I</mi></math>$

그리고 $arg (z - i) | C 3 = arg (z - i) | C 4 + 2 π arg (z + i) | C 5 = arg (z + i) | C 6 + 2 π <math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mi>arg</mi><mo data-mjx-texclass="NONE"></mo><mo stretchy="false">(</mo><mi>z</mi><mo>-</mo><mi>i</mi><mo stretchy="false">)</mo><msub><mrow data-mjx-texclass="ORD"><mo stretchy="false">|</mo></mrow><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>3</mn></msub></mrow></msub><mo>=</mo><mi>arg</mi><mo data-mjx-texclass="NONE"></mo><mo stretchy="false">(</mo><mi>z</mi><mo>-</mo><mi>i</mi><mo stretchy="false">)</mo><msub><mrow data-mjx-texclass="ORD"><mo stretchy="false">|</mo></mrow><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>4</mn></msub></mrow></msub><mo>+</mo><mn>2</mn><mi>π</mi><mspace linebreak="newline"></mspace><mi>arg</mi><mo data-mjx-texclass="NONE"></mo><mo stretchy="false">(</mo><mi>z</mi><mo>+</mo><mi>i</mi><mo stretchy="false">)</mo><msub><mrow data-mjx-texclass="ORD"><mo stretchy="false">|</mo></mrow><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>5</mn></msub></mrow></msub><mo>=</mo><mi>arg</mi><mo data-mjx-texclass="NONE"></mo><mo stretchy="false">(</mo><mi>z</mi><mo>+</mo><mi>i</mi><mo stretchy="false">)</mo><msub><mrow data-mjx-texclass="ORD"><mo stretchy="false">|</mo></mrow><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>6</mn></msub></mrow></msub><mo>+</mo><mn>2</mn><mi>π</mi></math>$

$∫C3+C4=−2πi∫C4dzz1+a⟶z=eiπ/2y−2πi∫∞1idyeiπ(1+a)/2y1+a=−i2πae−iπa/2<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>3</mn></msub><mo>+</mo><msub><mi>C</mi><mn>4</mn></msub></mrow></msub><mo>=</mo><mo>−</mo><mn>2</mn><mi>π</mi><mi>i</mi><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>4</mn></msub></mrow></msub><mfrac><mrow><mi>d</mi><mi>z</mi></mrow><msup><mi>z</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mfrac><munder><mo stretchy="false">⟶</mo><mrow><mi>z</mi><mo>=</mo><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mi>π</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><mi>y</mi></mrow></munder><mo>−</mo><mn>2</mn><mi>π</mi><mi>i</mi><msubsup><mo data-mjx-texclass="OP">∫</mo><mn>1</mn><mi mathvariant="normal">∞</mi></msubsup><mfrac><mrow><mi>i</mi><mi>d</mi><mi>y</mi></mrow><mrow><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mi>π</mi><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><mi>a</mi><mo stretchy="false">)</mo><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><msup><mi>y</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mrow></mfrac><mo>=</mo><mo>−</mo><mi>i</mi><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mi>a</mi></mfrac><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mo>−</mo><mi>i</mi><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup></math>$

$∫C5+C6=−2πi∫C6dzz1+a⟶z=ei3π/2y−2πi∫∞1−idye3π(1+a)/2y1+a=−i2πae−i3πa/2<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>5</mn></msub><mo>+</mo><msub><mi>C</mi><mn>6</mn></msub></mrow></msub><mo>=</mo><mo>−</mo><mn>2</mn><mi>π</mi><mi>i</mi><msub><mo data-mjx-texclass="OP">∫</mo><mrow data-mjx-texclass="ORD"><msub><mi>C</mi><mn>6</mn></msub></mrow></msub><mfrac><mrow><mi>d</mi><mi>z</mi></mrow><msup><mi>z</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mfrac><munder><mo stretchy="false">⟶</mo><mrow><mi>z</mi><mo>=</mo><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mn>3</mn><mi>π</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><mi>y</mi></mrow></munder><mo>−</mo><mn>2</mn><mi>π</mi><mi>i</mi><msubsup><mo data-mjx-texclass="OP">∫</mo><mn>1</mn><mi mathvariant="normal">∞</mi></msubsup><mfrac><mrow><mo>−</mo><mi>i</mi><mi>d</mi><mi>y</mi></mrow><mrow><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mn>3</mn><mi>π</mi><mo stretchy="false">(</mo><mn>1</mn><mo>+</mo><mi>a</mi><mo stretchy="false">)</mo><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><msup><mi>y</mi><mrow data-mjx-texclass="ORD"><mn>1</mn><mo>+</mo><mi>a</mi></mrow></msup></mrow></mfrac><mo>=</mo><mo>−</mo><mi>i</mi><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mi>a</mi></mfrac><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mo>−</mo><mi>i</mi><mn>3</mn><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup></math>$

Residue 정리에서 $\oint f (z) d z = 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo data-mjx-texclass="OP">\oint</mo><mi>f</mi><mo stretchy="false">(</mo><mi>z</mi><mo stretchy="false">)</mo><mi>d</mi><mi>z</mi><mo>=</mo><mn>0</mn></math>$ 이므로

$−i2πa(e−iπa/2+e−3πa/2)+2ieiπasin(πa)I=0→I=πcsc(πa/2)a<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><mo>−</mo><mi>i</mi><mfrac><mrow><mn>2</mn><mi>π</mi></mrow><mi>a</mi></mfrac><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mo>−</mo><mi>i</mi><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mo>−</mo><mn>3</mn><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn></mrow></msup><mo data-mjx-texclass="CLOSE">)</mo></mrow><mo>+</mo><mn>2</mn><mi>i</mi><msup><mi>e</mi><mrow data-mjx-texclass="ORD"><mi>i</mi><mi>π</mi><mi>a</mi></mrow></msup><mi>sin</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mi>π</mi><mi>a</mi><mo stretchy="false">)</mo><mi>I</mi><mo>=</mo><mn>0</mn><mspace linebreak="newline"></mspace><mo accent="false" stretchy="false">→</mo><mi>I</mi><mo>=</mo><mfrac><mrow><mi>π</mi><mi>csc</mi><mo data-mjx-texclass="NONE">⁡</mo><mo stretchy="false">(</mo><mi>π</mi><mi>a</mi><mrow data-mjx-texclass="ORD"><mo>/</mo></mrow><mn>2</mn><mo stretchy="false">)</mo></mrow><mi>a</mi></mfrac></math>$

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