I want to do following integrals \begin{align} \int_{-\pi}^{\pi}\frac{1}{ai+b\cos(x)}dx \end{align} where $a,b$ are real

Frist my trial was using the ideas of complex analysis, but here I don't know whether the poles are inside $|z|<1$ or not. (Since i didn't fix the magnitude of a and b)

any ideas? To this integral be finite, do i have to restrict the magnitude of $|a|$ and $|b|$ (i.e. |a|<|b|)

For the simple case, via mathematica i can obtain some results, for example setting a=1 or b=1 case. I want to know how to calculate such integrals.

For $a+b\cos(\theta)$ case, introducting complex variables or parametrizing cos(x) into functions of $tan^2(x/2)$ i can do the integral without any problem, but i want to do it in more general

  • $\begingroup$ Unless $a=0$ you have that $ai+b\cos(x)$ is never zero for $x\in(-\pi,\pi)$, so no extra assumption on $a,b$ is needed. On the other hand, if $|b|<|a|$ you may simply tackle such integral by expanding $\frac{1}{ai+b\cos x}$ as a geometric series and by recalling what $$\int_{-\pi}^{\pi}\left(\cos x\right)^n\,dx $$ is. Or you may just exploit parity and the tangent half-angle substitution. $\endgroup$ Apr 3, 2018 at 4:10

3 Answers 3


Let $t=\tan(x/2)$ then $dx=\frac{2dt}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}.$ So you want:

$$\begin{align}\int_{-\infty}^{\infty}\frac{2\,dt}{ai(1+t^2)+b(1-t^2)}&=\int_{-\infty}^{\infty}\frac{2dt}{(b+ai)+(-b+ai)t^2}\\ &=2(b+ai)\int_{-\infty}^{\infty} \frac{dt}{(b+ai)^2-(a^2+b^2)t^2} \end{align}$$

Letting $w=b+ai$ then this is:


Now, $$\frac{2w}{w^2-|w|^2t^2}=\frac{1}{w-|w|t} +\frac{1}{w+|w|t}$$

So the indefinite integral is $$\frac{1}{|w|} \left(\log\frac{w+|w|t}{w-|w|t}\right)$$

Now, $\dfrac{w+|w|t}{w-|w|t}\to -1$ as $t\to\infty$ and $t\to-\infty,$ but it appraches $-1$ from different directs.

So the integral is $\pm\dfrac{2\pi i}{|w|},$ where the sign is the opposite of the sign of $a.$

The sign is because when $t$ is positive, and $a>0,$ the angle to get from $w-|w|t$ to $w+|w|t$ is clockwise rotation, while when $a<0$ this rotation is counter-clockwise.

The answer can be rewritten as:


From this, we see that we should try to show:

$$\int_{-\pi}^{\pi}\frac{dx}{1+ci\cos x}=\frac{2\pi}{\sqrt{1+c^2}}$$

Then if $c=\frac{-b}{a}$ then we'd get (almost) our original integral.

If $f(x)=\frac{1}{1+ci\cos x}=\frac{1-ci\cos x}{1+c^2\cos^2 x}$, then $f(x+\pi)=\overline{f(x)}$, so the imaginary part of the integral is zero, and thus we are reduced to showing:

$$\int_{-\pi}^{\pi}\frac{dx}{1+c^2\cos^2 x}=\frac{2\pi}{\sqrt{1+c^2}}$$

Now, we're done if you can show:

$$\int_{-\pi/2}^{\pi/2}\frac{dx}{1+c^2\cos^2 x}=\frac{\pi}{\sqrt{1+c^2}}$$

Letting $t=\tan x$ then $dx=\frac{dt}{1+t^2}, \cos^2 x=\frac{1}{1+t^2}$ amd you need to show:

$$\int_{-\infty}^{\infty} \frac{dt}{1+c^2+t^2}=\frac{\pi}{\sqrt{1+c^2}}$$

which is a fairly standard integral.


Recall that $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$, so we have

$$\int_{-\pi}^\pi\frac{dx}{ai+b\cos x}=\int_{-\pi}^\pi\frac{dx}{ai+\frac{b}{2}(e^{ix}+e^{-ix})}=\frac{1}{i}\int_{-\pi}^\pi\frac{ie^{ix}dx}{\frac{b}{2}e^{2ix}+iae^{ix}+\frac{b}{2}}.$$

Use change of variable $z=e^{ix}$ and let $\gamma$ be the unit circle with counter clockwise orientation. Then we have

$$\int_{-\pi}^\pi\frac{dx}{ai+b\cos x}=-i\int_\gamma\frac{dz}{\frac{b}{2}z^2+iaz+\frac{b}{2}}.$$

Notice that $\frac{b}{2}z^2+iaz+\frac{b}{2}=\frac{b}{2}(z-z_+)(z-z_-)$, where $z_\pm=\frac{i}{b}\left[-a\pm\sqrt{a^2+b^2}\right]$. We will split into cases.

Case 1: assume $a>0$.

In this situation, $|z_+|<1$ and $|z_-|>1$. We can now apply Residue Theorem.

$$-i\int_\gamma\frac{dz}{\frac{b}{2}z^2+iaz+\frac{b}{2}}=\frac{-2i(2\pi i)}{b}\frac{1}{2\pi i}\int_\gamma\frac{dz}{(z-z_+)(z-z_-)}=\frac{4\pi}{b}\frac{1}{z_+-z_-}=\frac{-2\pi i}{\sqrt{a^2+b^2}}$$

Case 2: assume $a<0$.

In this situation, $|z_+|>1$ and $|z_-|<1$. We can now apply Residue Theorem as before.

$$-i\int_\gamma\frac{dz}{\frac{b}{2}z^2+iaz+\frac{b}{2}}=\frac{-2i(2\pi i)}{b}\frac{1}{2\pi i}\int_\gamma\frac{dz}{(z-z_+)(z-z_-)}=\frac{4\pi}{b}\frac{1}{z_--z_+}=\frac{2\pi i}{\sqrt{a^2+b^2}}$$

Case 3: $a=0$

The integral does not converge when $a=0$ (unless we talk about principal value integrals).

In conclusion,

$$\int_{-\pi}^\pi\frac{dx}{ai+b\cos x}=\frac{\mp 2\pi i}{\sqrt{a^2+b^2}}$$

for $\text{sgn}\left(a\right)=\pm1$ and is divergent when $a=0$.


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mc{I}\pars{a,b} \equiv \int_{-\pi}^{\pi}{\dd x \over a\ic + b\cos\pars{x}}\,,\qquad a, b \in \mathbb{R}}$.

Note that $\ds{\bbx{\left.\mc{I}\pars{a,0}\,\right\vert_{\ a\ \not=\ 0} = -\,{2 \over a}\,\pi\ic}}$. Then $\ds{\pars{~\mbox{with}\ a \not= 0\,,\ b \not= 0\ \mbox{and}\ \mu \equiv {a \over b} \in \mathbb{R}~}}$, \begin{align} \left.\rule{0pt}{5mm}\mc{I}\pars{a,b}\,\right\vert_{\ b\ \not=\ 0} & \equiv {2 \over b}\int_{0}^{\pi}{\dd x \over \mu\ic + \cos\pars{x}} = {2 \over b}\int_{-\pi/2}^{\pi/2}{\dd x \over \mu\ic - \sin\pars{x}} \\[5mm] & = {2 \over b}\int_{0}^{\pi/2}\bracks{{1 \over \mu\ic - \sin\pars{x}} + {1 \over \mu\ic + \sin\pars{x}}}\dd x = {4\mu \over b}\,\ic\int_{0}^{\pi/2}{\dd x \over -\mu^{2} - \sin^{2}\pars{x}} \\[5mm] & = -\,{4\mu \over b}\,\ic\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \mu^{2}\sec^{2}\pars{x} + \tan^{2}\pars{x}}\dd x = -\,{4\mu \over b}\,\ic\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \pars{\mu^{2} + 1} \tan^{2}\pars{x} + \mu^{2}}\dd x \\[5mm] & = -\,{4\over b\mu}\,\ic\,{\verts{\mu} \over \root{\mu^{2} + 1}}\int_{0}^{\pi/2}{\root{\mu^{2} + 1}\sec^{2}\pars{x}/\verts{\mu} \over \bracks{\root{\mu^{2} + 1}\tan\pars{x}/\verts{\mu}}^{2} + 1}\dd x\quad \pars{\begin{array}{l} \mbox{Note that this} \\ \mbox{step requires} \\ \ds{\mu \not= 0} \\ \ds{\implies a \not= 0} \end{array}} \\[5mm] & = -\,{4\,\mrm{sgn}\pars{\mu}\over b\root{\mu^{2} + 1}}\,\ic \int_{0}^{\infty}{\dd t \over t^{2} + 1}\dd x = -\,{2\,\mrm{sgn}\pars{\mu}\over b\root{\mu^{2} + 1}}\,\pi\ic = -\,{2\,\mrm{sgn}\pars{a/b} \over b\root{a^{2}/b^{2} + 1}}\,\pi\ic \\[5mm] & = -\,{2\,\mrm{sgn}\pars{a} \over \root{a^{2} + b^{2}}}\,\pi\ic\qquad a \not= 0. \end{align}

$$ \bbx{\int_{-\pi}^{\pi}{\dd x \over a\ic + b\cos\pars{x}} = \left\{\begin{array}{lcl} \ds{-\,{2 \over a}\,\pi\ic} & \ds{\mbox{if}} & \ds{a \not = 0\,,\quad b = 0} \\[2mm] \ds{-\,{2\,\mrm{sgn}\pars{a} \over \root{a^{2} + b^{2}}}\,\pi\ic} & \mbox{if} & \ds{a \not = 0\,,\quad b \not= 0} \\[2mm] \mbox{diverges} && \mbox{otherwise} \end{array}\right.} $$


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