Solution.
Consider
f ( z ) = e i z ( 1 + z 2 ) 2 , f\p{z} = \frac{e^{iz}}{\p{1 + z^2}^2}, f ( z ) = ( 1 + z 2 ) 2 e i z , which is meromorphic on C \C C z = − i z = -i z = − i z = i z = i z = i γ R \gamma_R γ R [ − R , R ] \br{-R, R} [ − R , R ] R R R
Observe that on the circular section of γ R \gamma_R γ R z = R e i θ z = Re^{i\theta} z = R e i θ ∣ e z ∣ = e Re z \abs{e^z} = e^{\Re{z}} ∣ e z ∣ = e Re z ∣ z ∣ 2 − 1 ≤ ∣ z 2 + 1 ∣ \abs{z}^2 - 1 \leq \abs{z^2 + 1} ∣ z ∣ 2 − 1 ≤ ∣ ∣ z 2 + 1 ∣ ∣ R > 1 R > 1 R > 1
∣ ∫ 0 π i R e i θ f ( R e i θ ) d θ ∣ ≤ R ∫ 0 π e Re ( i R e i θ ) ( R 2 − 1 ) 2 d θ = R ( R 2 − 1 ) 2 ∫ 0 π e − R sin θ d θ . \begin{aligned}
\abs{\int_0^\pi iRe^{i\theta}f\p{Re^{i\theta}} \,\diff\theta}
&\leq R \int_0^\pi \frac{e^{\Re\p{iRe^{i\theta}}}}{\p{R^2 - 1}^2} \,\diff\theta \\
&= \frac{R}{\p{R^2 - 1}^2} \int_0^\pi e^{-R\sin\theta} \,\diff\theta.
\end{aligned} ∣ ∣ ∫ 0 π i R e i θ f ( R e i θ ) d θ ∣ ∣ ≤ R ∫ 0 π ( R 2 − 1 ) 2 e Re ( i R e i θ ) d θ = ( R 2 − 1 ) 2 R ∫ 0 π e − R s i n θ d θ . Notice that sin x \sin{x} sin x [ 0 , π 2 ] \br{0, \frac{\pi}{2}} [ 0 , 2 π ] t ∈ [ 0 , 1 ] t \in \br{0, 1} t ∈ [ 0 , 1 ]
sin t π 2 ≥ t sin π 2 = t . \sin\frac{t\pi}{2}
\geq t\sin\frac{\pi}{2} = t. sin 2 t π ≥ t sin 2 π = t . Thus, if we set θ = t π 2 \theta = \frac{t\pi}{2} θ = 2 t π sin θ ≥ 2 θ π \sin\theta \geq \frac{2\theta}{\pi} sin θ ≥ π 2 θ θ ∈ [ 0 , π 2 ] \theta \in \br{0, \frac{\pi}{2}} θ ∈ [ 0 , 2 π ]
R ( R 2 − 1 ) 2 ∫ 0 π e − R sin θ d θ ≤ 2 R ( R 2 − 1 ) 2 ∫ 0 π / 2 e − 2 R θ / π d θ = 2 R ( R 2 − 1 ) 2 π 2 R ( 1 − e − R ) → R → ∞ 0. \begin{aligned}
\frac{R}{\p{R^2 - 1}^2} \int_0^\pi e^{-R\sin\theta} \,\diff\theta
&\leq \frac{2R}{\p{R^2 - 1}^2} \int_0^{\pi/2} e^{-2R\theta/\pi} \,\diff\theta \\
&= \frac{2R}{\p{R^2 - 1}^2} \frac{\pi}{2R}\p{1 - e^{-R}}
\xrightarrow{R\to\infty} 0.
\end{aligned} ( R 2 − 1 ) 2 R ∫ 0 π e − R s i n θ d θ ≤ ( R 2 − 1 ) 2 2 R ∫ 0 π /2 e − 2 Rθ / π d θ = ( R 2 − 1 ) 2 2 R 2 R π ( 1 − e − R ) R → ∞ 0. Thus, the integral over the top arc of γ R \gamma_R γ R
Next, notice that f f f z = i z = i z = i f f f z = i z = i z = i
Res ( f ; i ) = lim z → i ( z − i ) f ( z ) = lim z → i d d z ( z − i ) 2 e i z ( z − i ) 2 ( z + i ) 2 = lim z → i i e i z ( z + i ) 2 − 2 e i z ( z + i ) ( z + i ) 4 = − 4 i e − 1 − 4 i e − 1 16 = − i 2 e \begin{aligned}
\Res{f}{i}
&= \lim_{z \to i}\,\p{z - i}f\p{z} \\
&= \lim_{z \to i}\, \deriv{}{z} \p{z - i}^2 \frac{e^{iz}}{\p{z - i}^2\p{z + i}^2} \\
&= \lim_{z \to i} \frac{ie^{iz}\p{z + i}^2 - 2e^{iz}\p{z + i}}{\p{z + i}^4} \\
&= \frac{-4ie^{-1} - 4ie^{-1}}{16} \\
&= -\frac{i}{2e}
\end{aligned} Res ( f ; i ) = z → i lim ( z − i ) f ( z ) = z → i lim d z d ( z − i ) 2 ( z − i ) 2 ( z + i ) 2 e i z = z → i lim ( z + i ) 4 i e i z ( z + i ) 2 − 2 e i z ( z + i ) = 16 − 4 i e − 1 − 4 i e − 1 = − 2 e i For R > 1 R > 1 R > 1 i i i γ R \gamma_R γ R
1 2 π i ∫ γ R f ( z ) d z = − i 2 e ⟹ ∫ γ R f ( z ) d z = π e . \frac{1}{2\pi i} \int_{\gamma_R} f\p{z} \,\diff{z}
= -\frac{i}{2e}
\implies \int_{\gamma_R} f\p{z} \,\diff{z} = \frac{\pi}{e}. 2 πi 1 ∫ γ R f ( z ) d z = − 2 e i ⟹ ∫ γ R f ( z ) d z = e π . Taking R → ∞ R \to \infty R → ∞ x ∈ R x \in \R x ∈ R f ( x ) f\p{x} f ( x )
∫ 0 ∞ cos x ( 1 + x 2 ) 2 d x = 1 2 ∫ γ R f ( z ) d z = π 2 e . \int_0^\infty \frac{\cos{x}}{\p{1 + x^2}^2} \,\diff{x}
= \frac{1}{2} \int_{\gamma_R} f\p{z} \,\diff{z}
= \frac{\pi}{2e}. ∫ 0 ∞ ( 1 + x 2 ) 2 cos x d x = 2 1 ∫ γ R f ( z ) d z = 2 e π .