Solution.
Let f ( z ) = ( log z ) 2 ( z + a ) 2 + b 2 f\p{z} = \frac{\p{\log{z}}^2}{\p{z + a}^2 + b^2} f ( z ) = ( z + a ) 2 + b 2 ( l o g z ) 2 0 < ε < R 0 < \epsilon < R 0 < ε < R γ ε , R \gamma_{\epsilon,R} γ ε , R ε \epsilon ε R R R C R C_R C R
∣ ∫ C R f ( z ) d z ∣ ≤ ∫ 0 2 π R ( log R ) 2 ∣ ( R − a ) 2 − b 2 ∣ d θ → R → ∞ 0. \abs{\int_{C_R} f\p{z} \,\diff{z}}
\leq \int_0^{2\pi} \frac{R\p{\log{R}}^2}{\abs{\p{R - a}^2 - b^2}} \,\diff\theta
\xrightarrow{R\to\infty} 0. ∣ ∣ ∫ C R f ( z ) d z ∣ ∣ ≤ ∫ 0 2 π ∣ ( R − a ) 2 − b 2 ∣ R ( log R ) 2 d θ R → ∞ 0. Similarly,
∣ ∫ C ε f ( z ) d z ∣ ≤ ∫ 0 2 π ε ( log ε ) 2 ∣ ( ε − a ) 2 ∣ d θ → ε → 0 0. \abs{\int_{C_\epsilon} f\p{z} \,\diff{z}}
\leq \int_0^{2\pi} \frac{\epsilon\p{\log{\epsilon}}^2}{\abs{\p{\epsilon - a}^2}} \,\diff\theta
\xrightarrow{\epsilon\to0} 0. ∣ ∣ ∫ C ε f ( z ) d z ∣ ∣ ≤ ∫ 0 2 π ∣ ( ε − a ) 2 ∣ ε ( log ε ) 2 d θ ε → 0 0. Next, observe that f f f z = − a ± i b z = -a \pm ib z = − a ± ib
Res ( f ; − a ± i b ) = lim z → − a ± i b ( z − ( − a ± i b ) ) f ( z ) = ( log ( − a ± i b ) ) 2 ± 2 i b . \begin{aligned}
\Res{f}{-a \pm ib}
&= \lim_{z\to -a\pm ib} \p{z - \p{-a \pm ib}} f\p{z} \\
&= \frac{\p{\log\,\p{-a \pm ib}}^2}{\pm 2ib}.
\end{aligned} Res ( f ; − a ± ib ) = z → − a ± ib lim ( z − ( − a ± ib ) ) f ( z ) = ± 2 ib ( log ( − a ± ib ) ) 2 . Continuing, notice that from the branch cut, we have for any x ∈ R x \in \R x ∈ R
lim θ → 2 π − ( log x e i θ ) 2 = ( log x + 2 π i ) 2 = ( log x ) 2 + 4 π i log x − 4 π 2 . \lim_{\theta\to2\pi^-} \p{\log{xe^{i\theta}}}^2
= \p{\log{x} + 2\pi i}^2
= \p{\log{x}}^2 + 4\pi i \log{x} - 4\pi^2. θ → 2 π − lim ( log x e i θ ) 2 = ( log x + 2 πi ) 2 = ( log x ) 2 + 4 πi log x − 4 π 2 . Sending ε → 0 \epsilon \to 0 ε → 0 R → ∞ R \to \infty R → ∞
∫ 0 ∞ ( log x ) 2 ( x + a ) 2 + b 2 d x − ∫ 0 ∞ ( log x ) 2 + 4 π i log x − 4 π 2 ( x + a ) 2 + b 2 d x = − 4 π i ∫ 0 ∞ log x ( x + a ) 2 + b 2 d x − 4 π 2 ∫ 0 ∞ 1 ( x + a ) 2 + b 2 d x = − 4 π i ∫ 0 ∞ log x ( x + a ) 2 + b 2 d x − 4 π 2 b ( π 2 − arctan ( a b ) ) . \begin{aligned}
\int_0^\infty \frac{\p{\log{x}}^2}{\p{x + a}^2 + b^2} \,\diff{x} - \int_0^\infty \frac{\p{\log{x}}^2 + 4\pi i \log{x} - 4\pi^2}{\p{x + a}^2 + b^2} \,\diff{x}
&= -4\pi i\int_0^\infty \frac{\log{x}}{\p{x + a}^2 + b^2} \,\diff{x} - 4\pi^2 \int_0^\infty \frac{1}{\p{x + a}^2 + b^2} \,\diff{x} \\
&= -4\pi i\int_0^\infty \frac{\log{x}}{\p{x + a}^2 + b^2} \,\diff{x} - \frac{4\pi^2}{b}\p{\frac{\pi}{2} - \arctan\p{\frac{a}{b}}}.
\end{aligned} ∫ 0 ∞ ( x + a ) 2 + b 2 ( log x ) 2 d x − ∫ 0 ∞ ( x + a ) 2 + b 2 ( log x ) 2 + 4 πi log x − 4 π 2 d x = − 4 πi ∫ 0 ∞ ( x + a ) 2 + b 2 log x d x − 4 π 2 ∫ 0 ∞ ( x + a ) 2 + b 2 1 d x = − 4 πi ∫ 0 ∞ ( x + a ) 2 + b 2 log x d x − b 4 π 2 ( 2 π − arctan ( b a ) ) . Also, if − a + i b = r e i θ -a + ib = re^{i\theta} − a + ib = r e i θ − a − i b = r e − i θ -a - ib = re^{-i\theta} − a − ib = r e − i θ
( log ( − a + i b ) ) 2 − ( log ( − a + i b ) ) 2 = ( log r + i θ ) 2 − ( log r − i θ ) 2 = i 4 θ log r . \p{\log\p{-a + ib}}^2 - \p{\log\p{-a + ib}}^2
= \p{\log{r} + i\theta}^2 - \p{\log{r} - i\theta}^2
= i4\theta \log{r}. ( log ( − a + ib ) ) 2 − ( log ( − a + ib ) ) 2 = ( log r + i θ ) 2 − ( log r − i θ ) 2 = i 4 θ log r . By the residue theorem, we get
− 4 π i ∫ 0 ∞ log x ( x + a ) 2 + b 2 d x − 4 π 2 b ( π 2 − arctan ( a b ) ) = π b ( ( log ( − a + i b ) ) 2 − ( log ( − a + i b ) ) 2 ) . -4\pi i\int_0^\infty \frac{\log{x}}{\p{x + a}^2 + b^2} \,\diff{x} - \frac{4\pi^2}{b}\p{\frac{\pi}{2} - \arctan\p{\frac{a}{b}}}
= \frac{\pi}{b}\p{\p{\log\p{-a + ib}}^2 - \p{\log\p{-a + ib}}^2}. − 4 πi ∫ 0 ∞ ( x + a ) 2 + b 2 log x d x − b 4 π 2 ( 2 π − arctan ( b a ) ) = b π ( ( log ( − a + ib ) ) 2 − ( log ( − a + ib ) ) 2 ) . Taking imaginary parts and simplifying,
∫ 0 ∞ log x ( x + a ) 2 + b 2 d x = − θ log r b = − arctan ( − b a ) log ( a 2 + b 2 ) 2 b = arctan ( b a ) log ( a 2 + b 2 ) 2 b . \begin{aligned}
\int_0^\infty \frac{\log{x}}{\p{x + a}^2 + b^2} \,\diff{x}
&= -\frac{\theta\log{r}}{b} \\
&= -\frac{\arctan\p{-\frac{b}{a}}\log\,\p{a^2 + b^2}}{2b} \\
&= \frac{\arctan\p{\frac{b}{a}}\log\,\p{a^2 + b^2}}{2b}.
\end{aligned} ∫ 0 ∞ ( x + a ) 2 + b 2 log x d x = − b θ log r = − 2 b arctan ( − a b ) log ( a 2 + b 2 ) = 2 b arctan ( a b ) log ( a 2 + b 2 ) .