How to evaluate $\int _{0}^{\frac{\pi}{2}} \frac{\mathrm dx}{a+\cos x}$ using contour integration?

Question

I want to evaluate:

$\displaystyle \int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \tag*{}$

With my basic knowledge of Cauchy Residue theorem: \begin{align*} \int_0^{\frac{\pi}{2}} \frac{\mathrm dx}{a + \cos x} &= \int_0^{\frac{\pi}{2}}\frac{\mathrm dx}{a + \frac{e^{ix} + e^{-ix}}{2}} \\ &= 2\int_0^{\frac{\pi}{2}} \frac{e^{ix} \ \mathrm dx}{2ae^{ix} + e^{2ix} + 1} && \text{Let } z=e^{ix}, \text{ so } \mathrm dz = ie^{ix} \ \mathrm dx. \\ &= \frac{2}{i} \int_{|z|=1} \frac{\mathrm dz}{z^2 + 2az + 1} \\ &= \frac{2}{i} \int_{|z|=1} \frac{\mathrm dz}{(z-z_1)(z-z_2)} \end{align*} where the circle $|z|=1$

My problem is how to proceed further from here? How do I draw circle? Clockwise or Counterclockwise?

Answer 1

Your substitution gives$$\begin{align}\int_1^i\frac{-2idz}{(z+a)^2+1-a^2}&=\left[\frac{-2i}{\sqrt{1-a^2}}\arctan\frac{z+a}{\sqrt{1-a^2}}\right]_1^i\\&=\frac{2}{\sqrt{1-a^2}}\operatorname{artanh}\sqrt{\frac{1-a}{1+a}}.\end{align}$$It's technically a contour integral; it's just the contour isn't closed, so it's a "normal" integration evaluation technique, rather than one using the residue theorem. As for how the last $=$ works, I'll leave you to work through the trigonometric identities.

Answer 2

You have to parameterize the complex number. If you are integrating over a circle then you will have that if $$z=Re^{i\theta}$$ then $$R = 1$$ $$\theta = \lambda$$ $$z = e^{i\lambda}$$ $$dz = ie^{i\lambda}d\lambda$$ Then your integral becomes $$2\int_0^{\pi/2}\frac{e^{i\lambda}d\lambda}{e^{2i\lambda}+2ae^{i\lambda}+1} $$ which you can solve as a typical integral.

Answer 3

I doubt you can use contour integration to handle this since you can't obtain a closed curve. Noting $$ \cos x=\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}$$ one has, under $\tan(\frac x2)\to t$, \begin{eqnarray} &&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \\ &=&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}} \\ &=&\int _{0}^{\frac{\pi}{2}} \dfrac{(1+\tan^2(\frac x2))\mathrm dx}{a(1+\tan^2(\frac x2))+1-\tan^2(\frac x2)} \\ &=&\int _{0}^{\frac{\pi}{2}} \dfrac{(1+\tan^2(\frac x2))\mathrm dx}{(a+1)+(a-1)\tan^2(\frac x2))} \\ &=&\int_0^1\frac{2}{(a+1)+(a-1)t^2}\mathrm dt\\ &=&\frac2{\sqrt{a^2-1}}\arctan\bigg(\sqrt{\frac{a-1}{a+1}}\bigg). \end{eqnarray}