Let f ( z ) = z β 1 + z 2 f\p{z} = \frac{z^\beta}{1 + z^2} f ( z ) = 1 + z 2 z β Log z \Log{z} Log z
For 0 < ε < 1 < R 0 < \epsilon < 1 < R 0 < ε < 1 < R γ ε , R \gamma_{\epsilon,R} γ ε , R [ − R , − ε ] \br{-R, -\epsilon} [ − R , − ε ] [ ε , R ] \br{\epsilon, R} [ ε , R ] { R e i θ ∣ 0 ≤ θ ≤ π } \set{Re^{i\theta} \mid 0 \leq \theta \leq \pi} { R e i θ ∣ 0 ≤ θ ≤ π } { ε e i θ ∣ 0 ≤ θ π } \set{\epsilon e^{i\theta} \mid 0 \leq \theta \pi} { ε e i θ ∣ 0 ≤ θ π }
If C R C_R C R R R R
∣ ∫ C R f ( z ) d z ∣ = ∣ ∫ 0 π R β e i β θ 1 + R 2 e i 2 θ ⋅ i R e i θ d θ ∣ ≤ ∫ 0 π R β + 1 R 2 − 1 d θ = 2 π R β + 1 R 2 − 1 → R → ∞ 0 , \begin{aligned}
\abs{\int_{C_R} f\p{z} \,\diff{z}}
&= \abs{\int_0^\pi \frac{R^\beta e^{i\beta\theta}}{1 + R^2 e^{i2\theta}} \cdot iRe^{i\theta} \,\diff\theta} \\
&\leq \int_0^\pi \frac{R^{\beta+1}}{R^2 - 1} \,\diff\theta \\
&= \frac{2\pi R^{\beta+1}}{R^2 - 1}
\xrightarrow{R\to\infty} 0,
\end{aligned} ∣ ∣ ∫ C R f ( z ) d z ∣ ∣ = ∣ ∣ ∫ 0 π 1 + R 2 e i 2 θ R β e i βθ ⋅ i R e i θ d θ ∣ ∣ ≤ ∫ 0 π R 2 − 1 R β + 1 d θ = R 2 − 1 2 π R β + 1 R → ∞ 0 , since β + 1 < 2 \beta + 1 < 2 β + 1 < 2 C ε C_\epsilon C ε
∣ ∫ C ε f ( z ) d z ∣ ≤ 2 π ε β + 1 1 − ε 2 → ε → 0 0 , \abs{\int_{C_\epsilon} f\p{z} \,\diff{z}}
\leq \frac{2\pi \epsilon^{\beta+1}}{1 - \epsilon^2}
\xrightarrow{\epsilon\to0} 0, ∣ ∣ ∫ C ε f ( z ) d z ∣ ∣ ≤ 1 − ε 2 2 π ε β + 1 ε → 0 0 , since β + 1 > 0 \beta + 1 > 0 β + 1 > 0 z = i z = i z = i
Res ( f ; i ) = lim z → i ( z + i ) f ( z ) = lim z → i z β z + i = i β 2 i = e i β π / 2 2 i . \Res{f}{i}
= \lim_{z\to i} \p{z + i}f\p{z}
= \lim_{z\to i} \frac{z^\beta}{z + i}
= \frac{i^\beta}{2i}
= \frac{e^{i\beta\pi/2}}{2i}. Res ( f ; i ) = z → i lim ( z + i ) f ( z ) = z → i lim z + i z β = 2 i i β = 2 i e i β π /2 . Finally, observe that f ( − x ) = ( − 1 ) β f ( x ) = e i β π f ( x ) f\p{-x} = \p{-1}^\beta f\p{x} = e^{i\beta\pi} f\p{x} f ( − x ) = ( − 1 ) β f ( x ) = e i β π f ( x )
( 1 + e i β π ) ∫ 0 ∞ x β 1 + x 2 d x = 2 π i ⋅ e i β π 2 i ⟹ ∫ 0 ∞ x β 1 + x 2 d x = π e − i β π / 2 + e i β π / 2 = π 2 cos ( β π 2 ) . \begin{gathered}
\p{1 + e^{i\beta\pi}} \int_0^\infty \frac{x^\beta}{1 + x^2} \,\diff{x}
= 2\pi i \cdot \frac{e^{i\beta\pi}}{2i} \\
\implies \int_0^\infty \frac{x^\beta}{1 + x^2} \,\diff{x}
= \frac{\pi}{e^{-i\beta\pi/2} + e^{i\beta\pi/2}}
= \frac{\pi}{2\cos\p{\frac{\beta\pi}{2}}}.
\end{gathered} ( 1 + e i β π ) ∫ 0 ∞ 1 + x 2 x β d x = 2 πi ⋅ 2 i e i β π ⟹ ∫ 0 ∞ 1 + x 2 x β d x = e − i β π /2 + e i β π /2 π = 2 cos ( 2 β π ) π .