Q15MP - Mathematical Methods in Physical Sciences

Question

Evaluate the integrals by contour integration. 

$\mathit{I}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}\frac{\mathbf{c}\mathbf{o}\mathbf{s}\left(\theta \right)\mathbf{d}\mathbf{\theta }}{\mathbf{5}\mathbf{-}\mathbf{4}\mathbf{c}\mathbf{o}\mathbf{s}\left(\theta \right)}$

Step-by-Step Solution

Answer

Required integral is $\frac{\mathrm{\pi }}{6}$.

Step 1: Concept of Contour integral

Contour integral: Contour integration is a method of calculating integrals along paths in the complex plane.

Step 2: Solve for integral

The integration is given by,

 $I={\int }_0^{\mathrm{\pi }}\frac{\cos \left(\theta \right)d\theta }{5-4\cos \left(\theta \right)} .....\left(1\right)$

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Convert (1) to contour integral, but first, we shall assume that:

$$\begin{gathered} \cos \theta =\frac{z+{\displaystyle \frac{1}{z}}}{2} \\ -d\theta =\frac{dz}{iz} \end{gathered}$$ 

The contour is the unit circle.

So, substitution in the above equation (1)  yields as follows:

$$\begin{gathered} I_1=\oint \frac{\left({\displaystyle \frac{z+{\displaystyle \frac{1}{z}}}{2}}\right){\displaystyle \frac{dz}{iz}}}{5-4\left(\frac{z+{\displaystyle \frac{1}{z}}}{2}\right)} \\ =\frac{-1}{i}\oint \frac{1+z^{2}dz}{z\left(4z^2-10z+4\right)} \\ =\frac{-1}{i}\oint \frac{1+z^{2}dz}{z\left(2z-4\right)\left(2z-1\right)} \\ =\frac{-1}{4i}\oint \frac{1+z^{2}dz}{z\left(z-2\right)\left(z-{\displaystyle \frac{1}{2}}\right)} ....\left(2\right) \end{gathered}$$ 

 

                                                                                           

 

Step 3: Solve further the value of integral (2)

The value of integral (2) is given as follows:

$$\begin{gathered} I_1=2\mathrm{\pi i}\sum _{}\mathrm{R}\left({\mathrm{Z}}_{\mathrm{i}}\right) \\ =2\mathrm{\pi i}\left(-\frac{1}{4\mathrm{i}}\right)\left[{\left(\frac{1+{\mathrm{z}}^2}{\mathrm{z}\left(\mathrm{z}-2\right)}\right)}_{\mathrm{z}\to 0.5}+{\left(\frac{1+{\mathrm{z}}^2}{\left(\mathrm{z}-2\right)\left(\mathrm{z}-{\displaystyle \frac{1}{2}}\right)}\right)}_{\mathrm{z}\to 0}\right] \\ =2\mathrm{\pi i}\left(-\frac{1}{4\mathrm{i}}\right)\left(1-\frac{5}{3}\right) \\ =\frac{\mathrm{\pi }}{3} \end{gathered}$$ 

Because the function $\frac{\cos \left(\theta \right)}{5-4\cos \left(\theta \right)}$  is even, the value from $0\to 2\mathrm{\pi }$  is twice the value form $$" width="9" height="19" style="max-width: none; vertical-align: -4px;" >$0\to \mathrm{\pi }$, so the value of I is given as follows:

 $$\begin{gathered} I=\left(\frac{1}{2}\right)I_1 \\ I=\left(\frac{\mathrm{\pi }}{3}\right)\left(\frac{1}{2}\right) \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$

 

Therefore, the answer is $\frac{\mathrm{\pi }}{6}$.