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Trigonometric Integrals by Contour Integration

April 12th, 2009 |

Author: Merrin

We can do a certain set of trigonometric integrals with contour integration. These integrals are of the form

$\displaystyle \int _{0}^{2\pi }f(\cos \theta ,\sin \theta )d\theta$

For example, one such tricky integral would be

$\displaystyle \int _{0}^{2\pi }\cfrac{d\theta }{2-\sin \theta }$

We already found out how to do integrals like these by the universal trigonometric substitution, but we can also solve some of the definite integral versions by the contour integration method. If we pick the contour to be the unit circle in the complex plane, then we can begin to see the connection with complex integration.

$\displaystyle z=e^{i\theta } dz=ie^{i\theta } d\theta \quad \cfrac{dz}{iz} =d\theta \quad \sin \theta =\cfrac{z-z^{-1} }{2i} \quad \cos \theta =\cfrac{z+z^{-1} }{2}$

$\displaystyle {\int _{0}^{2\pi }f(\cos \theta ,\sin \theta )d\theta =\oint f\left(\cfrac{z+z^{-1} }{2} ,\cfrac{z-z^{-1} }{2i} \right)\cfrac{dz}{iz} =\oint g(z)dz }$

The contour is the unit circle going counter clockwise. The contour integral can be calculated by the residue theorem. The points $latex z_{k} &s=1$ are where the function $latex g(z)&s=1$ has singularities.

$\displaystyle \oint g(z)dz =2\pi i\sum _{k=1}^{n}\text{Res}(g(z),z_{k} )$

Even more complex integrals involving multiple angles of the trigonometric functions are possible. Substitutions for these are no problem. For example,

$\cos (3\theta )=\cfrac{e^{3i\theta } -e^{-3i\theta } }{2} =\cfrac{1}{2} (z^{3} -\cfrac{1}{z^{3} } )$

Example 1. Calculate by contour integration the integral of

$\displaystyle \int _{0}^{2\pi }\cfrac{d\theta }{2-\sin \theta }$

Solution 1. We found out how to do hard integrals like these using the universal transformation. Now we will integrate them some more in their simpler definite integral form. The trigonometric transformation provides a contour integral.

$\displaystyle \oint g(z)dz=\oint \cfrac{dz}{iz} \cfrac{1}{2-\cfrac{z+1/z}{2i} }$

Now we have to get in there and do some algebra.

$\displaystyle g(z)=\cfrac{1}{iz} \cfrac{1}{\left(2-\cfrac{1}{2i} z+\cfrac{1}{2iz}\right) } =\cfrac{2}{4iz-z^{2} +1} =\cfrac{-2}{z^{2} -4iz-1}$

We solve the quadratic equation for the denominator to locate where the singularities are. Take care with this algebra when factoring the quadratic because sign errors are possible. The complex numbers $latex z_1 \, z_2 &s=1$, in the equation have opposite signs of the roots of the quadratic equation. Check your work by multiplying out what has been factored.

$\displaystyle {z=-2i\pm \cfrac{1}{2} \sqrt{16-4} =-2i\pm i\sqrt{3} } \\ {\cfrac{-2}{z^{2} -4iz-1} =\cfrac{-2}{(z-2i+i\sqrt{3} )(z-2i-i\sqrt{3} )} =\cfrac{-2}{(z-z_{1} )(z-z_{2} )} } \\ {z_{1} =+2i-i\sqrt{3} {\rm \,\,\,\,inside}} \\ {z_{2} =+2i+i\sqrt{3} {\rm \,\,\,\,outside}}$

Because the singularities are simple poles, where one lies inside our contour we can use the residue formula for simple poles.

$\text{Res}(g(z),z_{1} )=\mathop{\lim }\limits_{z\to z_{1} } (z-z_{1} )\cfrac{-2}{(z-z_{1} )(z-z_{2} )} =\cfrac{-2}{z_{1} -z_{2} }$

$\text{Res}(g(z),z_{1} )=\cfrac{-2}{+2i-i\sqrt{3} -(+2i+i\sqrt{3} )} =\cfrac{-2}{-2i\sqrt{3} } =\cfrac{1}{i\sqrt{3} }$

Now the last step is the most fun. Multiply $latex 2\pi i &s=1$ times the sum of the residues inside the circle. We have

$\displaystyle \int _{0}^{2\pi }\cfrac{d\theta }{2-\sin \theta } =\oint g(z)dz =2\pi i\cfrac{1}{i\sqrt{3} } =\cfrac{2\pi }{\sqrt{3} }$

It is easy to see how other definite integrals can be worked out by this method. The only difference will be in the algebra of calculating the residues for a different function.

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