Question

Use Residue theorem to compute
the integral from 0 to 2 pi
[ Sin theta ] / [ 4+ sin (theta) + cos (theta) ] dtheta


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Answer 1

Answer on Question#50256 – Math – Other

$$\pi \left(1 - \frac {2 \sqrt {1 4}}{7}\right)$$

Solution

$$\int_ {0} ^ {2 \pi} \frac {\sin \theta}{4 + \sin \theta + \cos \theta} d \theta$$

General formula:

$$\int_ {0} ^ {2 \pi} R (\cos \theta , \sin \theta) d \theta = \oint_ {U (0, 1)} R \left(\frac {1}{2} \left(z + \frac {1}{z}\right), \frac {1}{2 i} \left(z - \frac {1}{z}\right)\right) \frac {d z}{i z}$$

, where $U(0,1)$ – unit circle centered at origin.

$$\int_ {0} ^ {2 \pi} \frac {\sin \theta}{4 + \sin \theta + \cos \theta} d \theta = \oint_ {U (0, 1)} \frac {\frac {1}{2 i} \left(z - \frac {1}{z}\right)}{4 + \frac {1}{2 i} \left(z - \frac {1}{z}\right) + \frac {1}{2} \left(z + \frac {1}{z}\right)} \frac {d z}{i z} =$$ $$\oint_ {U (0, 1)} \frac {1 - z ^ {2}}{z ((1 - i) z ^ {2} + 8 z + (1 + i))} d z =$$ 2 \pi i \sum_ {U (0, 1)} \frac {\operatorname {R e s} _ {U (0 , 1)} \frac {1 - z ^ {2}}{z ((1 - i) z ^ {2} + 8 z + (1 + i))}

, summation over all singularities within unit circle.

Let us determine singularities.

Examine zeros of numerator.

$$1 - z ^ {2} = 0$$ $$z ^ {2} = 1$$ $$z _ {n 1} = 1$$ $$z _ {n 2} = - 1$$

Examine zeros of denominator.

$$z = 0$$ $$z _ {1} = 0 - \text{simple pole}$$ $$(1 - i) z ^ {2} + 8 z + (1 + i) = 0$$ $$z _ {2} = \left(- 2 - \sqrt {\frac {7}{2}}\right) (1 + i) - \text{simple pole}$$ $$z _ {3} = \left(- 2 + \sqrt {\frac {7}{2}}\right) (1 + i) - \text{simple pole}$$

Look, which singularities are inside the unit circle.

$$z _ {1} = 0 - \text{inside}$$ $$z _ {2} = \left(- 2 - \sqrt {\frac {7}{2}}\right) (1 + i) \approx - 3.9 + 3.9i - \text{outside}$$ $$z _ {3} = \left(- 2 + \sqrt {\frac {7}{2}}\right) (1 + i) \approx - 0.13 + 0.13i - \text{inside}$$

Calculate related residues.

As soon as we work with simple poles, we can use formula:

$$\operatorname{Res}_{z = a} \frac{\phi(z)}{\psi(z)} = \frac{\phi(a)}{\psi'(a)}$$

, where $\phi(a) \neq 0; \psi(a) = 0; \psi'(a) \neq 0$.

To use it effectively rearrange expression.

$$\frac {1 - z ^ {2}}{z \left((1 - i) z ^ {2} + 8 z + (1 + i)\right)} = \frac {\left(\frac {1 + i}{2}\right) - \left(\frac {1 + i}{2}\right) z ^ {2}}{z \left(z ^ {2} + 4 (1 + i) z + i\right)} = \frac {A}{z} + \frac {B z + C}{z ^ {2} + 4 (1 + i) z + i}$$ $$z ^ {0}: i A = \frac {1 + i}{2} \rightarrow A = \frac {1 - i}{2}$$ $$z ^ {1}: 4 (1 + i) A + C = 0 \rightarrow C = - 4 A (1 + i) = - \frac {4}{2} (1 + i) (1 - i) = - 4$$ $$z ^ {2}: A + B = - \left(\frac {1 + i}{2}\right) \rightarrow B = - \left(\frac {1 + i}{2}\right) - A = - \left(\frac {1 + i}{2}\right) - \left(\frac {1 - i}{2}\right) = - 1$$ $$\frac {1 - z ^ {2}}{z \left((1 - i) z ^ {2} + 8 z + (1 + i)\right)} = \frac {\frac {1 - i}{2}}{z} + \frac {- z - 4}{z ^ {2} + 4 (1 + i) z + i}$$

Recall that we shall use formula only to addenda, which contain singularity.

$$\operatorname{Res}_{z = 0} \frac {\frac {1 - i}{2}}{z} = \frac {1 - i}{2}$$ $$\operatorname{Res}_{z = z _ {3}} \frac {- z - 4}{z ^ {2} + 4 (1 + i) z + i} = \frac {- z _ {3} - 4}{2 z _ {3} + 4 (1 + i)} =$$ $$\frac {- \left(- 2 + \sqrt {\frac {7}{2}}\right) (1 + i) - 4}{2 \left(- 2 + \sqrt {\frac {7}{2}}\right) (1 + i) + 4 (1 + i)} =$$ $$\frac {2 + 2 i - \sqrt {\frac {7}{2}} - i \sqrt {\frac {7}{2}} - 4}{- 4 - 4 i + \sqrt {14} + i \sqrt {14} + 4 + 4 i} = \frac {- \left(2 + \sqrt {\frac {7}{2}}\right) + i \left(2 - \sqrt {\frac {7}{2}}\right)}{\sqrt {14} (1 + i)} = \frac {- \left(2 + \sqrt {\frac {7}{2}}\right) + i \left(2 - \sqrt {\frac {7}{2}}\right) + i \left(2 + \sqrt {\frac {7}{2}}\right) + \left(2 - \sqrt {\frac {7}{2}}\right)}{2 \sqrt {14}} = \frac {- 2 \sqrt {\frac {7}{2}} + 4 i}{2 \sqrt {14}} = \frac {- 7 + 2 \sqrt {14} i}{14} = - \frac {1}{2} + \frac {\sqrt {14}}{7} i$$

Finally:

$$2\pi i \sum \underset {U (0, 1)} {\operatorname{Res}} \frac {1 - z ^ {2}}{z ((1 - i) z ^ {2} + 8 z + (1 + i))} = 2\pi i \left(\underset {z = 0} {\operatorname{Res}} \frac {\frac {1 - i}{2}}{z} + \underset {z = z _ {3}} {\operatorname{Res}} \frac {- z - 4}{z ^ {2} + 4 (1 + i) z + i}\right) =$$ $$2\pi i \left(\left(\frac {1 - i}{2}\right) + \left(- \frac {1}{2} + \frac {\sqrt {14}}{7} i\right)\right) = 2\pi i \left(- \frac {i}{2} + \frac {\sqrt {14}}{7} i\right) = \pi \left(1 - \frac {2 \sqrt {14}}{7}\right)$$

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