How to Integrate Using Residue Theory

Contour Integrals

Consider the integral below. The term e 1 / z {\displaystyle e^{1/z}} e^{{1/z}} is a classic example of a function with an essential singularity - a singularity that results in the function taking every complex value in the neighborhood of the function (except, for this function, the value of 0). This is due to the fact that there are an infinite number of negative power terms in the Laurent series expansion for e 1 / z . {\displaystyle e^{1/z}.} e^{{1/z}}. Below, we consider a contour γ : z ( t ) = e i t , 0 ≤ t ≤ 2 π . {\displaystyle \gamma :z(t)=e^{it},\ 0\leq t\leq 2\pi .} \gamma :z(t)=e^{{it}},\ 0\leq t\leq 2\pi . ∫ γ z 3 e 1 / z d z {\displaystyle \int _{\gamma }z^{3}e^{1/z}\mathrm {d} z} \int _{{\gamma }}z^{{3}}e^{{1/z}}{\mathrm {d}}z

Write out the Laurent expansion for the function. We want to find the residue at the singularity in order to use the residue theorem. For essential singularities, series expansions are the only way to find them. e 1 / z = 1 + 1 z + 1 2 ! z 2 + 1 3 ! z 3 + ⋯ {\displaystyle e^{1/z}=1+{\frac {1}{z}}+{\frac {1}{2!z^{2}}}+{\frac {1}{3!z^{3}}}+\cdots } e^{{1/z}}=1+{\frac {1}{z}}+{\frac {1}{2!z^{{2}}}}+{\frac {1}{3!z^{{3}}}}+\cdots z 3 e 1 / z = z 3 + z 2 + z 2 ! + 1 3 ! + 1 4 ! z + ⋯ {\displaystyle z^{3}e^{1/z}=z^{3}+z^{2}+{\frac {z}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!z}}+\cdots } z^{{3}}e^{{1/z}}=z^{{3}}+z^{{2}}+{\frac {z}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!z}}+\cdots

Use the Laurent series to find the residue. The definition of the residue of a function is the coefficient of the z − 1 {\displaystyle z^{-1}} z^{{-1}} term of the Laurent series of that function. We see that the coefficient is 1 24 . {\displaystyle {\frac {1}{24}}.} {\frac {1}{24}}. Therefore, that will be our residue.

Use the residue theorem to evaluate the integral. ∫ γ z 3 e 1 / z d z = 2 π i ⋅ 1 24 = π i 12 {\displaystyle \int _{\gamma }z^{3}e^{1/z}\mathrm {d} z=2\pi i\cdot {\frac {1}{24}}={\frac {\pi i}{12}}} \int _{{\gamma }}z^{{3}}e^{{1/z}}{\mathrm {d}}z=2\pi i\cdot {\frac {1}{24}}={\frac {\pi i}{12}}

Consider the integral below. We give another example of an integral that can technically be done without series, but the problem is that we do not know the order of the pole. The contour is the unit circle in the counterclockwise direction. ∫ γ sin ⁡ ( 2 z 4 ) z 21 d z {\displaystyle \int _{\gamma }{\frac {\sin(2z^{4})}{z^{21}}}\mathrm {d} z} \int _{{\gamma }}{\frac {\sin(2z^{{4}})}{z^{{21}}}}{\mathrm {d}}z

Expand the integrand into its Laurent series. We know the Taylor series for the sine function, so we can put in the z − 21 {\displaystyle z^{-21}} z^{{-21}} term quite easily. sin ⁡ ( 2 z 4 ) z 21 = 1 z 21 ∑ j = 0 ∞ ( − 1 ) j ( 2 z 4 ) 2 j + 1 ( 2 j + 1 ) ! = ∑ j = 0 ∞ ( − 1 ) j ( 2 j + 1 ) ! 2 2 j + 1 z 8 j − 17 {\displaystyle {\frac {\sin(2z^{4})}{z^{21}}}={\frac {1}{z^{21}}}\sum _{j=0}^{\infty }{\frac {(-1)^{j}(2z^{4})^{2j+1}}{(2j+1)!}}=\sum _{j=0}^{\infty }{\frac {(-1)^{j}}{(2j+1)!}}2^{2j+1}z^{8j-17}} {\frac {\sin(2z^{{4}})}{z^{{21}}}}={\frac {1}{z^{{21}}}}\sum _{{j=0}}^{{\infty }}{\frac {(-1)^{{j}}(2z^{{4}})^{{2j+1}}}{(2j+1)!}}=\sum _{{j=0}}^{{\infty }}{\frac {(-1)^{{j}}}{(2j+1)!}}2^{{2j+1}}z^{{8j-17}} We see that our pole is order 17. In order to find the residue by partial fractions, we would have to differentiate 16 times and then substitute 0 into our result. Clearly, this is impractical.

Expand the Laurent series to find the residue. We see that the z − 1 {\displaystyle z^{-1}} z^{{-1}} coefficient is Res ⁡ ( f ( z ) ; 0 ) = 2 5 5 ! = 4 15 . {\displaystyle \operatorname {Res} (f(z);0)={\frac {2^{5}}{5!}}={\frac {4}{15}}.} \operatorname {Res}(f(z);0)={\frac {2^{{5}}}{5!}}={\frac {4}{15}}. ∑ j = 0 ∞ ( − 1 ) j ( 2 j + 1 ) ! 2 2 j + 1 z 8 j − 17 = 2 z − 17 − 2 3 3 ! z − 9 + 2 5 5 ! z − 1 − ⋯ {\displaystyle \sum _{j=0}^{\infty }{\frac {(-1)^{j}}{(2j+1)!}}2^{2j+1}z^{8j-17}=2z^{-17}-{\frac {2^{3}}{3!}}z^{-9}+{\frac {2^{5}}{5!}}z^{-1}-\cdots } \sum _{{j=0}}^{{\infty }}{\frac {(-1)^{{j}}}{(2j+1)!}}2^{{2j+1}}z^{{8j-17}}=2z^{{-17}}-{\frac {2^{{3}}}{3!}}z^{{-9}}+{\frac {2^{{5}}}{5!}}z^{{-1}}-\cdots

Use the residue theorem to evaluate the integral. The key to our efficiency here is our recognition of using the Laurent series of known functions. From here, we simply expand. ∫ γ sin ⁡ ( 2 z 4 ) z 21 d z = 2 π i ( 4 15 ) = 8 π i 15 {\displaystyle \int _{\gamma }{\frac {\sin(2z^{4})}{z^{21}}}\mathrm {d} z=2\pi i\left({\frac {4}{15}}\right)={\frac {8\pi i}{15}}} \int _{{\gamma }}{\frac {\sin(2z^{{4}})}{z^{{21}}}}{\mathrm {d}}z=2\pi i\left({\frac {4}{15}}\right)={\frac {8\pi i}{15}}

Trigonometric Integrals

Consider the integral below. The easiest trigonometric integrals to evaluate using residues will be those integrals whose bounds are [ 0 , 2 π ] , {\displaystyle [0,2\pi ],} [0,2\pi ], or any other interval 2 π {\displaystyle 2\pi } 2\pi apart. Try to evaluate this integral using elementary techniques - the process will be long-winded and difficult. ∫ 0 2 π cos ⁡ 3 x 5 − 4 cos ⁡ x d x {\displaystyle \int _{0}^{2\pi }{\frac {\cos 3x}{5-4\cos x}}\mathrm {d} x} \int _{{0}}^{{2\pi }}{\frac {\cos 3x}{5-4\cos x}}{\mathrm {d}}x In general, we can apply this to any integral of the form below - rational, trigonometric functions. ∫ 0 2 π F ( cos ⁡ x , cos ⁡ 2 x , ⋯ , sin ⁡ x , sin ⁡ 2 x , ⋯ ) G ( cos ⁡ x , cos ⁡ 2 x , ⋯ , sin ⁡ x , sin ⁡ 2 x , ⋯ ) d x {\displaystyle \int _{0}^{2\pi }{\frac {F(\cos x,\cos 2x,\cdots ,\sin x,\sin 2x,\cdots )}{G(\cos x,\cos 2x,\cdots ,\sin x,\sin 2x,\cdots )}}\mathrm {d} x} \int _{{0}}^{{2\pi }}{\frac {F(\cos x,\cos 2x,\cdots ,\sin x,\sin 2x,\cdots )}{G(\cos x,\cos 2x,\cdots ,\sin x,\sin 2x,\cdots )}}{\mathrm {d}}x

Parameterize the unit circle. The integral is a one-dimensional integral integrated along the real axis. However, we can convert the interval [ 0 , 2 π ] {\displaystyle [0,2\pi ]} [0,2\pi ] to one along the unit circle. We describe this below by the contour γ , {\displaystyle \gamma ,} \gamma , a positively oriented contour along the unit circle z ( x ) = e i x . {\displaystyle z(x)=e^{ix}.} z(x)=e^{{ix}}. Then d z = i e i x d x , {\displaystyle \mathrm {d} z=ie^{ix}\mathrm {d} x,} {\mathrm {d}}z=ie^{{ix}}{\mathrm {d}}x, and therefore we arrive at the important change of variables written below. d x = 1 i z d z {\displaystyle \mathrm {d} x={\frac {1}{iz}}\mathrm {d} z} {\mathrm {d}}x={\frac {1}{iz}}{\mathrm {d}}z

Rewrite the trigonometric functions in terms of complex exponentials. Recall that cos ⁡ a x = 1 2 ( e i a x + e − i a x ) . {\displaystyle \cos ax={\frac {1}{2}}(e^{iax}+e^{-iax}).} \cos ax={\frac {1}{2}}(e^{{iax}}+e^{{-iax}}). Then as a result of our previous parameterization, we can rewrite the terms cos ⁡ 3 x {\displaystyle \cos 3x} \cos 3x and cos ⁡ x {\displaystyle \cos x} \cos x like so. cos ⁡ 3 x = 1 2 ( z 3 + 1 z 3 ) {\displaystyle \cos 3x={\frac {1}{2}}\left(z^{3}+{\frac {1}{z^{3}}}\right)} \cos 3x={\frac {1}{2}}\left(z^{{3}}+{\frac {1}{z^{{3}}}}\right) cos ⁡ x = 1 2 ( z + 1 z ) {\displaystyle \cos x={\frac {1}{2}}\left(z+{\frac {1}{z}}\right)} \cos x={\frac {1}{2}}\left(z+{\frac {1}{z}}\right) ∫ 0 2 π cos ⁡ 3 x 5 − 4 cos ⁡ x d x = ∫ γ 1 2 ( z 3 + 1 z 3 ) 5 − 4 ⋅ 1 2 ( z + 1 z ) 1 i z d z {\displaystyle \int _{0}^{2\pi }{\frac {\cos 3x}{5-4\cos x}}\mathrm {d} x=\int _{\gamma }{\frac {{\frac {1}{2}}\left(z^{3}+{\frac {1}{z^{3}}}\right)}{5-4\cdot {\frac {1}{2}}\left(z+{\frac {1}{z}}\right)}}{\frac {1}{iz}}\mathrm {d} z} \int _{{0}}^{{2\pi }}{\frac {\cos 3x}{5-4\cos x}}{\mathrm {d}}x=\int _{{\gamma }}{\frac {{\frac {1}{2}}\left(z^{{3}}+{\frac {1}{z^{{3}}}}\right)}{5-4\cdot {\frac {1}{2}}\left(z+{\frac {1}{z}}\right)}}{\frac {1}{iz}}{\mathrm {d}}z

Simplify the integral. We bring out factors and multiply the top and bottom by z 3 . {\displaystyle z^{3}.} z^{{3}}. Then we factor to identify the singularities. We recall that our contour is the unit circle e i x . {\displaystyle e^{ix}.} e^{{ix}}. As such, only the poles at z 0 = 0 {\displaystyle z_{0}=0} z_{{0}}=0 and z 1 = 1 2 {\displaystyle z_{1}={\frac {1}{2}}} z_{{1}}={\frac {1}{2}} will contribute to the integral. ∫ γ f ( z ) d z = 1 2 i ∫ γ z 3 + 1 z 3 5 − 2 ( z + 1 z ) 1 z d z = 1 2 i ∫ γ z 6 + 1 5 z 4 − 2 z 5 − 2 z 3 d z = − 1 2 i ∫ γ z 6 + 1 z 3 ( 2 z − 1 ) ( z − 2 ) d z = − 1 2 i ∫ γ z 6 + 1 2 z 3 ( z − 1 / 2 ) ( z − 2 ) d z {\displaystyle {\begin{aligned}\int _{\gamma }f(z)\mathrm {d} z&={\frac {1}{2i}}\int _{\gamma }{\frac {z^{3}+{\frac {1}{z^{3}}}}{5-2\left(z+{\frac {1}{z}}\right)}}{\frac {1}{z}}\mathrm {d} z\\&={\frac {1}{2i}}\int _{\gamma }{\frac {z^{6}+1}{5z^{4}-2z^{5}-2z^{3}}}\mathrm {d} z\\&=-{\frac {1}{2i}}\int _{\gamma }{\frac {z^{6}+1}{z^{3}(2z-1)(z-2)}}\mathrm {d} z\\&=-{\frac {1}{2i}}\int _{\gamma }{\frac {z^{6}+1}{2z^{3}(z-1/2)(z-2)}}\mathrm {d} z\end{aligned}}} {\begin{aligned}\int _{{\gamma }}f(z){\mathrm {d}}z&={\frac {1}{2i}}\int _{{\gamma }}{\frac {z^{{3}}+{\frac {1}{z^{{3}}}}}{5-2\left(z+{\frac {1}{z}}\right)}}{\frac {1}{z}}{\mathrm {d}}z\\&={\frac {1}{2i}}\int _{{\gamma }}{\frac {z^{{6}}+1}{5z^{{4}}-2z^{{5}}-2z^{{3}}}}{\mathrm {d}}z\\&=-{\frac {1}{2i}}\int _{{\gamma }}{\frac {z^{{6}}+1}{z^{{3}}(2z-1)(z-2)}}{\mathrm {d}}z\\&=-{\frac {1}{2i}}\int _{{\gamma }}{\frac {z^{{6}}+1}{2z^{{3}}(z-1/2)(z-2)}}{\mathrm {d}}z\end{aligned}}

Evaluate the residue Res ⁡ ( f ; z 1 ) {\displaystyle \operatorname {Res} (f;z_{1})} \operatorname {Res}(f;z_{{1}}). Because z 1 = 1 2 {\displaystyle z_{1}={\frac {1}{2}}} z_{{1}}={\frac {1}{2}} is a simple pole (pole of order 1), we can use the method of partial fractions. Res ⁡ ( f ; z 1 ) = lim z → 1 / 2 z 6 + 1 2 z 3 ( z − 2 ) = − 65 24 {\displaystyle \operatorname {Res} (f;z_{1})=\lim _{z\to 1/2}{\frac {z^{6}+1}{2z^{3}(z-2)}}=-{\frac {65}{24}}} \operatorname {Res}(f;z_{{1}})=\lim _{{z\to 1/2}}{\frac {z^{{6}}+1}{2z^{{3}}(z-2)}}=-{\frac {65}{24}}

Evaluate the residue at the other singularity. The singularity at z 0 = 0 {\displaystyle z_{0}=0} z_{{0}}=0 is a pole of order 3. This means that we will need to do a bit more work to get the residue. We can use the formula below as one method. Keep in mind that as the order increases, these calculations can quickly become cumbersome. Expanding out the functions in series will be preferred. Res ⁡ ( f ; z 0 ) = 1 2 ! lim z → 0 d 2 d z 2 z 3 ⋅ z 6 + 1 z 3 ( 2 z 2 − 5 z + 2 ) = 21 8 {\displaystyle \operatorname {Res} (f;z_{0})={\frac {1}{2!}}\lim _{z\to 0}{\frac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}z^{3}\cdot {\frac {z^{6}+1}{z^{3}(2z^{2}-5z+2)}}={\frac {21}{8}}} \operatorname {Res}(f;z_{{0}})={\frac {1}{2!}}\lim _{{z\to 0}}{\frac {{\mathrm {d}}^{{2}}}{{\mathrm {d}}z^{{2}}}}z^{{3}}\cdot {\frac {z^{{6}}+1}{z^{{3}}(2z^{{2}}-5z+2)}}={\frac {21}{8}} In general, we use the formula below, where n {\displaystyle n} n denotes the order of the pole. Res ⁡ ( f ; z n ) = 1 ( n − 1 ) ! lim z → z n d n − 1 d z n − 1 ( z − z n ) n f ( z ) {\displaystyle \operatorname {Res} (f;z_{n})={\frac {1}{(n-1)!}}\lim _{z\to z_{n}}{\frac {\mathrm {d} ^{n-1}}{\mathrm {d} z^{n-1}}}(z-z_{n})^{n}f(z)} \operatorname {Res}(f;z_{{n}})={\frac {1}{(n-1)!}}\lim _{{z\to z_{{n}}}}{\frac {{\mathrm {d}}^{{n-1}}}{{\mathrm {d}}z^{{n-1}}}}(z-z_{{n}})^{{n}}f(z) We can also use series to find the residue. First, the residue of the function f {\displaystyle f} f is the coefficient of the z − 1 {\displaystyle z^{-1}} z^{{-1}} term. If we consider the function z 6 + 1 ( 2 z − 1 ) ( z − 2 ) {\displaystyle {\frac {z^{6}+1}{(2z-1)(z-2)}}} {\frac {z^{{6}}+1}{(2z-1)(z-2)}} instead, then the residue at z 0 {\displaystyle z_{0}} z_{{0}} will be the coefficient of the z 2 {\displaystyle z^{2}} z^{{2}} term. If we expand the function into two terms, we see that the first term cannot contain the residue, because the smallest non-zero coefficient lies with a term with a degree greater than 2. Then, we simply rewrite the denominator in terms of power series, multiply them out, and check the coefficient of the z 2 {\displaystyle z^{2}} z^{{2}} term. Note that we can be lazy with the multiplication for the other coefficients, because we don't care about them. 1 ( 2 z − 1 ) ( z − 2 ) = 1 2 1 ( 1 − 2 z ) ( 1 − z / 2 ) = 1 2 ( 1 + 2 z + ( 2 z ) 2 + ⋯ ) ( 1 + z 2 + ( z 2 ) 2 + ⋯ ) = 1 2 ( ⋯ + ( z 2 ) 2 + ( 2 z ) ( z 2 ) + ( 2 z ) 2 + ⋯ ) = 1 2 21 4 z 2 {\displaystyle {\begin{aligned}{\frac {1}{(2z-1)(z-2)}}&={\frac {1}{2}}{\frac {1}{(1-2z)(1-z/2)}}\\&={\frac {1}{2}}\left(1+2z+(2z)^{2}+\cdots \right)\left(1+{\frac {z}{2}}+\left({\frac {z}{2}}\right)^{2}+\cdots \right)\\&={\frac {1}{2}}\left(\cdots +\left({\frac {z}{2}}\right)^{2}+(2z)\left({\frac {z}{2}}\right)+(2z)^{2}+\cdots \right)\\&={\frac {1}{2}}{\frac {21}{4}}z^{2}\end{aligned}}} {\begin{aligned}{\frac {1}{(2z-1)(z-2)}}&={\frac {1}{2}}{\frac {1}{(1-2z)(1-z/2)}}\\&={\frac {1}{2}}\left(1+2z+(2z)^{{2}}+\cdots \right)\left(1+{\frac {z}{2}}+\left({\frac {z}{2}}\right)^{{2}}+\cdots \right)\\&={\frac {1}{2}}\left(\cdots +\left({\frac {z}{2}}\right)^{{2}}+(2z)\left({\frac {z}{2}}\right)+(2z)^{{2}}+\cdots \right)\\&={\frac {1}{2}}{\frac {21}{4}}z^{{2}}\end{aligned}} We see that our residue is 21 8 , {\displaystyle {\frac {21}{8}},} {\frac {21}{8}}, as found from before.

Use the residue theorem to evaluate the integral. Summing everything up, we can finally evaluate the original integral. ∫ 0 2 π cos ⁡ 3 x 5 − 4 cos ⁡ x d x = − 1 2 i ( 2 π i ) ( 21 8 − 65 24 ) = π 12 {\displaystyle \int _{0}^{2\pi }{\frac {\cos 3x}{5-4\cos x}}\mathrm {d} x=-{\frac {1}{2i}}(2\pi i)\left({\frac {21}{8}}-{\frac {65}{24}}\right)={\frac {\pi }{12}}} \int _{{0}}^{{2\pi }}{\frac {\cos 3x}{5-4\cos x}}{\mathrm {d}}x=-{\frac {1}{2i}}(2\pi i)\left({\frac {21}{8}}-{\frac {65}{24}}\right)={\frac {\pi }{12}}

Consider the integral below. As before, we will convert this integral into a contour integral, find its residues, and evaluate using the residue theorem. Below, a {\displaystyle a} a and b {\displaystyle b} b are real numbers such that a 2 > b 2 . {\displaystyle a^{2}>b^{2}.} a^{{2}}>b^{{2}}. ∫ 0 2 π 1 a + b cos ⁡ θ d θ {\displaystyle \int _{0}^{2\pi }{\frac {1}{a+b\cos \theta }}\mathrm {d} \theta } \int _{{0}}^{{2\pi }}{\frac {1}{a+b\cos \theta }}{\mathrm {d}}\theta

Rewrite the integral in terms of a contour integral. We parameterize using z ( θ ) = e i θ , {\displaystyle z(\theta )=e^{i\theta },} z(\theta )=e^{{i\theta }}, the unit circle, recognize the important relation d θ = 1 i z d z , {\displaystyle \mathrm {d} \theta ={\frac {1}{iz}}\mathrm {d} z,} {\mathrm {d}}\theta ={\frac {1}{iz}}{\mathrm {d}}z, and rewrite cos ⁡ θ {\displaystyle \cos \theta } \cos \theta in terms of exponentials. We simplify by bringing out constants and a b {\displaystyle b} b factor. ∫ 0 2 π 1 a + b cos ⁡ θ d θ = ∫ γ 1 a + b 2 ( z + 1 / z 2 ) 1 i z d z = 2 i b ∫ γ 1 z 2 + 2 a b z + 1 d z {\displaystyle {\begin{aligned}\int _{0}^{2\pi }{\frac {1}{a+b\cos \theta }}\mathrm {d} \theta &=\int _{\gamma }{\frac {1}{a+{\frac {b}{2}}\left({\frac {z+1/z}{2}}\right)}}{\frac {1}{iz}}\mathrm {d} z\\&={\frac {2}{ib}}\int _{\gamma }{\frac {1}{z^{2}+{\frac {2a}{b}}z+1}}\mathrm {d} z\end{aligned}}} {\begin{aligned}\int _{{0}}^{{2\pi }}{\frac {1}{a+b\cos \theta }}{\mathrm {d}}\theta &=\int _{{\gamma }}{\frac {1}{a+{\frac {b}{2}}\left({\frac {z+1/z}{2}}\right)}}{\frac {1}{iz}}{\mathrm {d}}z\\&={\frac {2}{ib}}\int _{{\gamma }}{\frac {1}{z^{{2}}+{\frac {2a}{b}}z+1}}{\mathrm {d}}z\end{aligned}}

Find the residues. The residues are easily found because the expression in the denominator is quadratic, so both poles are simple poles. We label the larger residue as z + {\displaystyle z_{+}} z_{{+}} and the smaller one as z − . {\displaystyle z_{-}.} z_{{-}}. z ± = − 2 a b ± 4 a 2 b 2 − 4 2 = − a b ± a 2 b 2 − 1 {\displaystyle z_{\pm }={\frac {-{\frac {2a}{b}}\pm {\sqrt {{\frac {4a^{2}}{b^{2}}}-4}}}{2}}=-{\frac {a}{b}}\pm {\sqrt {{\frac {a^{2}}{b^{2}}}-1}}} z_{{\pm }}={\frac {-{\frac {2a}{b}}\pm {\sqrt {{\frac {4a^{{2}}}{b^{{2}}}}-4}}}{2}}=-{\frac {a}{b}}\pm {\sqrt {{\frac {a^{{2}}}{b^{{2}}}}-1}} The function has two poles at these locations. However, only one of them lies within the contour - the other lies outside and will not contribute to the integral. With the constraint a 2 > b 2 , {\displaystyle a^{2}>b^{2},} a^{{2}}>b^{{2}}, we see that − a b < − 1 {\displaystyle -{\frac {a}{b}}<-1} -{\frac {a}{b}} and a 2 b 2 > 1 , {\displaystyle {\frac {a^{2}}{b^{2}}}>1,} {\frac {a^{{2}}}{b^{{2}}}}>1, making the square root term positive. That means that z − < − 1 , {\displaystyle z_{-}<-1,} z_{{-}} and therefore it must lie outside the contour, the unit circle. Now that we know that z + {\displaystyle z_{+}} z_{{+}} is the only pole within the contour, we can find the residue there. We can use the residue formula to do this. Res ⁡ ( f ; z + ) = 1 z + − z − = 1 2 a 2 b 2 − 1 {\displaystyle \operatorname {Res} (f;z_{+})={\frac {1}{z_{+}-z_{-}}}={\frac {1}{2{\sqrt {{\frac {a^{2}}{b^{2}}}-1}}}}} \operatorname {Res}(f;z_{{+}})={\frac {1}{z_{{+}}-z_{{-}}}}={\frac {1}{2{\sqrt {{\frac {a^{{2}}}{b^{{2}}}}-1}}}}

Use the residue theorem to evaluate the integral. It is not difficult to show that we would obtain the negative of this result if a < 0. {\displaystyle a<0.} a This result is remarkable in its simplicity, and after computing this integral, one starts to see the true potential of residue theory in evaluating real integrals. ∫ 0 2 π 1 a + b cos ⁡ θ d θ = 2 i b ( 2 π i ) 1 2 a 2 b 2 − 1 = 2 π a 2 − b 2 {\displaystyle \int _{0}^{2\pi }{\frac {1}{a+b\cos \theta }}\mathrm {d} \theta ={\frac {2}{ib}}(2\pi i){\frac {1}{2{\sqrt {{\frac {a^{2}}{b^{2}}}-1}}}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}} \int _{{0}}^{{2\pi }}{\frac {1}{a+b\cos \theta }}{\mathrm {d}}\theta ={\frac {2}{ib}}(2\pi i){\frac {1}{2{\sqrt {{\frac {a^{{2}}}{b^{{2}}}}-1}}}}={\frac {2\pi }{{\sqrt {a^{{2}}-b^{{2}}}}}}

Improper Integrals

Consider the integral below. This is an integral evaluated over the entire real axis. The easiest integrals will have such bounds. Note that this integral should be finite, because the 1 x 4 {\displaystyle {\frac {1}{x^{4}}}} {\frac {1}{x^{{4}}}} term dominates as x → ∞ . {\displaystyle x\to \infty .} x\to \infty . Therefore, this integral will be equal to its principal value. ∫ − ∞ ∞ 1 ( x 2 + 1 ) 2 d x {\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(x^{2}+1)^{2}}}\mathrm {d} x} \int _{{-\infty }}^{{\infty }}{\frac {1}{(x^{{2}}+1)^{{2}}}}{\mathrm {d}}x

Consider the contour integral. We switch all the x {\displaystyle x} x's to z {\displaystyle z} z's. Then we define a closed contour Γ {\displaystyle \Gamma } \Gamma that goes from − a {\displaystyle -a} -a to a . {\displaystyle a.} a. Then the contour traces a semicircle and loops back to − a {\displaystyle -a} -a in the counterclockwise direction. This part of the contour will have the parameterization γ : z ( t ) = a e i t . {\displaystyle \gamma :z(t)=ae^{it}.} \gamma :z(t)=ae^{{it}}. ∫ Γ 1 ( z 2 + 1 ) 2 d z = ∫ − ∞ ∞ 1 ( x 2 + 1 ) 2 d x + ∫ γ 1 ( z 2 + 1 ) 2 d z {\displaystyle \int _{\Gamma }{\frac {1}{(z^{2}+1)^{2}}}\mathrm {d} z=\int _{-\infty }^{\infty }{\frac {1}{(x^{2}+1)^{2}}}\mathrm {d} x+\int _{\gamma }{\frac {1}{(z^{2}+1)^{2}}}\mathrm {d} z} \int _{{\Gamma }}{\frac {1}{(z^{{2}}+1)^{{2}}}}{\mathrm {d}}z=\int _{{-\infty }}^{{\infty }}{\frac {1}{(x^{{2}}+1)^{{2}}}}{\mathrm {d}}x+\int _{{\gamma }}{\frac {1}{(z^{{2}}+1)^{{2}}}}{\mathrm {d}}z There will be two things to note here. First, we will find the residues of the integral on the left. Second, we will need to show that the second integral on the right goes to zero. Once we do both of these things, we will have completed the evaluation.

Find the residues of the integral on the left. First, we factor the denominator. 1 ( z 2 + 1 ) 2 = 1 ( z − i ) 2 ( z + i ) 2 {\displaystyle {\frac {1}{(z^{2}+1)^{2}}}={\frac {1}{(z-i)^{2}(z+i)^{2}}}} {\frac {1}{(z^{{2}}+1)^{{2}}}}={\frac {1}{(z-i)^{{2}}(z+i)^{{2}}}} We recognize that the only pole that contributes to the integral will be the pole at z = i , {\displaystyle z=i,} z=i, a pole of order 2. The other pole lies outside the contour. Equivalently, we could've chosen γ {\displaystyle \gamma } \gamma such that it made a clockwise loop and encircled the pole at z = − i . {\displaystyle z=-i.} z=-i. Next, we use partial fractions. Remember that out of four fractions in the expansion, only the term 1 z − i {\displaystyle {\frac {1}{z-i}}} {\frac {1}{z-i}} will contribute to the integral. The coefficient of this term will be the residue. lim z → i d d z ( z − i ) 2 ⋅ 1 ( z − i ) 2 ( z + i ) 2 = − i 4 {\displaystyle \lim _{z\to i}{\frac {\mathrm {d} }{\mathrm {d} z}}(z-i)^{2}\cdot {\frac {1}{(z-i)^{2}(z+i)^{2}}}=-{\frac {i}{4}}} \lim _{{z\to i}}{\frac {{\mathrm {d}}}{{\mathrm {d}}z}}(z-i)^{{2}}\cdot {\frac {1}{(z-i)^{{2}}(z+i)^{{2}}}}=-{\frac {i}{4}} Notice that this residue is imaginary - it must, if it is to cancel out the 2 π i , {\displaystyle 2\pi i,} 2\pi i, so that our final result will be real.

Show that the integral with contour γ {\displaystyle \gamma } \gamma goes to 0. We do this using ML estimation, where we recognize that the length of the contour is π a . {\displaystyle \pi a.} \pi a. | ∫ γ 1 ( z 2 + 1 ) 2 d z | ≤ ∫ γ | 1 ( z 2 + 1 ) 2 | d z ≤ ∫ γ 1 ( a 2 + 1 ) 2 d z = π a ( a 2 + 1 ) 2 {\displaystyle {\begin{aligned}\left|\int _{\gamma }{\frac {1}{(z^{2}+1)^{2}}}\mathrm {d} z\right|&\leq \int _{\gamma }\left|{\frac {1}{(z^{2}+1)^{2}}}\right|\mathrm {d} z\\&\leq \int _{\gamma }{\frac {1}{(a^{2}+1)^{2}}}\mathrm {d} z={\frac {\pi a}{(a^{2}+1)^{2}}}\end{aligned}}} {\begin{aligned}\left|\int _{{\gamma }}{\frac {1}{(z^{{2}}+1)^{{2}}}}{\mathrm {d}}z\right|&\leq \int _{{\gamma }}\left|{\frac {1}{(z^{{2}}+1)^{{2}}}}\right|{\mathrm {d}}z\\&\leq \int _{{\gamma }}{\frac {1}{(a^{{2}}+1)^{{2}}}}{\mathrm {d}}z={\frac {\pi a}{(a^{{2}}+1)^{{2}}}}\end{aligned}} lim a → ∞ π a ( a 2 + 1 ) 2 = 0 {\displaystyle \lim _{a\to \infty }{\frac {\pi a}{(a^{2}+1)^{2}}}=0} \lim _{{a\to \infty }}{\frac {\pi a}{(a^{{2}}+1)^{{2}}}}=0 In general, for any polynomial functions G ( z ) {\displaystyle G(z)} G(z) and H ( z ) , {\displaystyle H(z),} H(z), ∫ γ G ( z ) H ( z ) d z {\displaystyle \int _{\gamma }{\frac {G(z)}{H(z)}}\mathrm {d} z} \int _{{\gamma }}{\frac {G(z)}{H(z)}}{\mathrm {d}}z will go to 0 whenever deg ⁡ G ≤ deg ⁡ H + 2. {\displaystyle \operatorname {deg} G\leq \operatorname {deg} H+2.} \operatorname {deg}G\leq \operatorname {deg}H+2. That is, the degree of the denominator must be at least two greater than the degree of the numerator. This is to avoid any tricky business when the behavior of the function goes as 1 / z {\displaystyle 1/z} 1/z for large radii a . {\displaystyle a.} a. (A similar phenomenon happens with the harmonic series - the limit goes to 0, but the series diverges.)

Use the residue theorem to evaluate the integral. This, and the result from the previous section, can easily be checked using a computer algebra program such as Mathematica. A TI-89 calculator can check certain simple expressions with exact answers - for others, it will evaluate numerically. ∫ − ∞ ∞ 1 ( x 2 + 1 ) 2 d x = 2 π i ⋅ − i 4 = π 2 {\displaystyle \int _{-\infty }^{\infty }{\frac {1}{(x^{2}+1)^{2}}}\mathrm {d} x=2\pi i\cdot {\frac {-i}{4}}={\frac {\pi }{2}}} \int _{{-\infty }}^{{\infty }}{\frac {1}{(x^{{2}}+1)^{{2}}}}{\mathrm {d}}x=2\pi i\cdot {\frac {-i}{4}}={\frac {\pi }{2}}