Finding $\int_0^{2\pi}\,d\theta\,\,\frac{a-i\cdot{b}\cos(\pi+\theta)}{c-i\cdot{d}\cos(\pi+\theta)}$

Question

I would like to calculate the following definite integral:

$$\int_0^{2\pi}\,d\theta\,\,\frac{a-i\cdot{b}\cos(\pi+\theta)}{c-i\cdot{d}\cos(\pi+\theta)}$$

Anybody has any suggestion? thanks

So according to the hints, I should proceed in this way:

$$\int_C\,\,\frac{2az+ib(z^2+1)}{iz(2cz+id(z^2+1))}\,\,dz$$

To apply the residue theorem I should know the roots of the denominator:

$$iz(2cz+id(z^2+1))=-d\,z\left(z-i\frac{c+\sqrt{c^2+d^2}}{d}\right)\left(z-i\frac{c-\sqrt{c^2+d^2}}{d}\right)$$

Now, if $c>0,d>0,c/d=k>0$

$$|\frac{c+\sqrt{c^2+d^2}}{d}|=|\frac{kd+\sqrt{(kd)^2+d^2}}{d}|=|k+\sqrt{k^2+1}|>1$$

and

$$|\frac{c-\sqrt{c^2+d^2}}{d}|=|\frac{kd-\sqrt{(kd)^2+d^2}}{d}|=|k-\sqrt{k^2+1}|<1$$

Therefore, the only roots inside the unit circle are $z=0$ and $z=i\frac{c-\sqrt{c^2+d^2}}{d}$.

$$\int_C\,\,\frac{2az+ib(z^2+1)}{iz(2cz+id(z^2+1))}\,\,dz=-\frac{1}{d}\int_C\,\,\frac{2az+ib(z^2+1)}{z\left(z-i\frac{c+\sqrt{c^2+d^2}}{d}\right)\left(z-i\frac{c-\sqrt{c^2+d^2}}{d}\right)}\,\,dz=\\=-\frac{2\pi i}{d}\cdot\frac{ib}{\left(-i\frac{c+\sqrt{c^2+d^2}}{d}\right)\left(-i\frac{c-\sqrt{c^2+d^2}}{d}\right)}-\frac{2\pi i}{d}\cdot\frac{2ai\frac{c-\sqrt{c^2+d^2}}{d}+ib((i\frac{c-\sqrt{c^2+d^2}}{d})^2+1)}{i\frac{c-\sqrt{c^2+d^2}}{d}\left(i\frac{c-\sqrt{c^2+d^2}}{d}-i\frac{c+\sqrt{c^2+d^2}}{d}\right)}=\\=2\pi\,\,\frac{ac+bd+a\sqrt{c^2+d^2}}{\sqrt{c^2+d^2}(\sqrt{c^2+d^2}+c)}$$

Answer 1

Hint 1: use $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$

Hint 2: let $z=e^{i\theta}$, so that $d\theta=\frac{dz}{iz}$, $\cos(\theta)=\frac{1}{2}(z+1/z)$ and $\sin(\theta)=\frac{1}{2i}(z-1/z)$

Compute the contour integral around the unit circle using the residue theorem.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{2\pi}{a - \ic\,b\cos\pars{\pi + \theta} \over c - \ic\,d\cos\pars{\pi + \theta}}\,\dd\theta:\ {\large ?}}$

\begin{align} &\color{#c00000}{\int_{0}^{2\pi}{a - \ic\,b\cos\pars{\pi + \theta} \over c - \ic\,d\cos\pars{\pi + \theta}}\,\dd\theta} =\int_{0}^{2\pi}{a + \ic\,b\cos\pars{\theta} \over c + \ic\,d\cos\pars{\theta}}\,\dd\theta \\[3mm]&=\int_{0}^{2\pi}{a +\pars{b/d}\braces{\bracks{\ic\,d\cos\pars{\theta} + c} - c} \over c + \ic\,d\cos\pars{\theta}}\,\dd\theta =2\pi\,{b \over d} + \pars{a - {bc \over d}} \int_{0}^{2\pi}{\dd\theta \over c + \ic\,d\cos\pars{\theta}} \\[3mm]&=2\pi\,{b \over d} + 2\pars{a - {bc \over d}} \color{#04f}{\int_{0}^{\pi}{\dd\theta \over c - \ic\,d\cos\pars{\theta}}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{1} \end{align}

We'll evaluate $\ds{\color{#04f}{\int_{0}^{\pi}{\dd\theta \over c - \ic\,d\cos\pars{\theta}}}}$ with the Weierstrass Substitution $\ds{t \equiv \tan\pars{\theta \over 2}}$: \begin{align} &\color{#04f}{\int_{0}^{\pi}{\dd\theta \over c - \ic\,d\cos\pars{\theta}}} =\int_{0}^{\infty}{2\,\dd t\,/\pars{1 + t^{2}} \over c - \ic\,d\pars{1 - t^{2}}/\pars{1 + t^{2}}} =2\int_{0}^{\infty}{\dd t \over \pars{c + \ic\,d}t^{2} + c - \ic\,d} \\[3mm]&={2 \over c - \ic\,d}\int_{0}^{\infty} {\dd t \over \pars{\root{\pars{c + \ic\,d}/\pars{c - \ic d}}\ t}^{2} + 1} ={2 \over c - \ic\,d}\int_{0}^{\infty} {\dd t \over \pars{\expo{\ic\phi}\ t}^{2} + 1}\quad\pars{2} \end{align} where $\ds{\phi \equiv \arctan\pars{d \over c}}$. When $\ds{\phi = \pm\,{\pi \over 2} + 2n\pi}$ $\pars{~n \in {\mathbb Z}~}$, which occurs when $\ds{c \to 0}$, the integral in expression $\pars{2}$ diverges. So, we'll assume $\ds{c \not= 0}$.

With $\ds{c \not= 0}$, expression $\pars{2}$ is reduced to: \begin{align} &\color{#04f}{\int_{0}^{\pi}{\dd\theta \over c - \ic\,d\cos\pars{\theta}}} ={2 \over \root{c^{2} + d^{2}}}\int_{0}^{\infty\expo{\ic\phi}} {\dd t \over t^{2} + 1} \end{align}

\begin{align} &\color{#00f}{\large\int_{0}^{2\pi}{a - \ic\,b\cos\pars{\pi + \theta} \over c - \ic\,d\cos\pars{\pi + \theta}}\,\dd\theta =2\pi\,{b \over d} + 4\,{a - bc/d \over \root{c^{2} + d^{2}}}\int_{0}^{\infty\expo{\ic\phi}} \!\!\!\!\!\!\!\!{\dd t \over t^{2} + 1}}\,,\qquad c \not= 0 \end{align} The last integral $\ds{\pars{~\mbox{over}\ t~}}$ depends critically on the particular values of $\ds{c}$ and $\ds{d}$.