10.3: Trigonometric Integrals
- Page ID
- 6532
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The trick here is to put together some elementary properties of \(z = e^{i \theta}\) on the unit circle.
- \(e^{-i \theta} = 1/z.\)
- \(\cos (\theta) = \dfrac{e^{i \theta} + e^{-i \theta}}{2} = \dfrac{z + 1/z}{2}.\)
- \(\sin (\theta) = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} = \dfrac{z - 1/z}{2i}.\)
We start with an example. After that we’ll state a more general theorem.
Compute
\[\int_{0}^{2\pi} \dfrac{d \theta}{1 + a^2 - 2a \cos (\theta)}. \nonumber \]
Assume that \(|a| \ne 1\).
Solution
Notice that \([0, 2\pi]\) is the interval used to parametrize the unit circle as \(z = e^{i \theta}\). We need to make two substitutions:
\[\begin{array} {rcl} {\cos (\theta)} & = & {\dfrac{z + 1/z}{2}} \\ {dz} & = & {i e^{i \theta} \ d\theta \ \ \ \ \Leftrightarrow \ \ \ \ d \theta = \dfrac{dz}{iz}} \end{array} \nonumber \]
Making these substitutions we get
\[\begin{array} {rcl} {I} & = & {\int_{0}^{2\pi} \dfrac{d \theta}{1 + a^2 - 2a \cos (\theta)}} \\ {} & = & {\int_{|z| = 1} \dfrac{1}{1 + a^2 - 2a (z + 1/z)/2} \cdot \dfrac{dz}{iz}} \\ {} & = & {\int_{|z| = 1} \dfrac{1}{i((1 + a^2) z - a(z^2 +1))} \ dz.} \end{array} \nonumber \]
So, let
\[f(z) = \dfrac{1}{i((1 + a^2) z - a(z^2 + 1))}. \nonumber \]
The residue theorem implies
\[I = 2\pi i \sum \text{ residues of } f \text{ inside the unit circle.} \nonumber \]
We can factor the denominator:
\[f(z) = \dfrac{-1}{ia (z - a) (z - 1/a)}. \nonumber \]
The poles are at \(a\), \(1/a\). One is inside the unit circle and one is outside.
If \(|a| > 1\) then \(1/a\) is inside the unit circle and \(\text{Res} (f, 1/a) = \dfrac{1}{i(a^2 - 1)}\)
If \(|a| < 1\) then \(a\) is inside the unit circle and \(\text{Res} (f, a) = \dfrac{1}{i(1 - a^2)}\)
We have
\[I = \begin{cases} \dfrac{2 \pi}{a^2 - 1} & \text{if } |a| > 1 \\ \dfrac{2 \pi}{1 - a^2} & \text{if } |a| < 1 \end{cases} \nonumber \]
The example illustrates a general technique which we state now.
Suppose \(R(x, y)\) is a rational function with no poles on the circle
\[x^2 + y^2 = 1 \nonumber \]
then for
\[f(z) = \dfrac{1}{iz} R (\dfrac{z + 1/z}{2}, \dfrac{z - 1/z}{2i}) \nonumber \]
we have
\[\int_{0}^{2\pi} R(\cos (\theta), \sin (\theta)) \ d \theta = 2 \pi i \sum \text{ residues of } f \text{ inside } |z| = 1. \nonumber \]
- Proof
-
We make the same substitutions as in Example 10.4.1. So,
\[\int_{0}^{2\pi} R(\cos (\theta), \sin (\theta)) \ d \theta = \int_{|z| = 1} R (\dfrac{z + 1/z}{2}, \dfrac{z - 1/z}{2i}) \dfrac{dz}{iz} \nonumber \]
The assumption about poles means that \(f\) has no poles on the contour \(|z| = 1\). The residue theorem now implies the theorem.