With the idea of using a path as in
Figure 1.23.9, we work with the branch
\(\Log_{\alpha}\text{,}\) with
\(\alpha=-\pi/2\text{.}\) This is defined on
\(\C-R_{-\pi/2}\) and agrees with
\(\ln x\) on the positive reals. Let
\(f(z)=\frac{\Log_{\alpha}(z)}{z^4+1}\text{.}\) It is easy to see that
\(f\) is analytic on
\(U=\C-R_{\alpha}\) except at the roots
\(z_k=e^{i\pi/4}e^{i\pi k/2}\text{,}\) \(0\leq k\leq 3\text{,}\) of
\(z^4+1\text{.}\) With the help of
Theorem 1.21.7, we compute
\begin{align*}
\res_f z_0 \amp = \frac{\Log_\alpha(e^{i\pi/4})}{4e^{i3\pi/4}}\\
\amp = \frac{i\pi}{16}\cdot e^{-i3\pi/4} \\
\res_f z_1 \amp = \frac{\Log_\alpha(e^{i3\pi/4})}{4e^{i9\pi/4}}\\
\amp = \frac{i3\pi}{16}\cdot e^{-9i\pi/4} \\
\amp = \frac{i 3\pi}{16}\cdot e^{-i\pi/4}\text{.}
\end{align*}
For all
\(R > 1\) and
\(0 < \delta < 1\text{,}\) let
\(\gamma\) be the path as in
Figure 1.23.9. On the one hand, since such a path only encloses the poles
\(z_0\) and
\(z_1\text{,}\) we have by the Cauchy residue theorem
\begin{align*}
\int_{\gamma}f\, dz \amp = 2\pi i(\res_f z_0+\res_f z_1) \amp \\
\amp = -\frac{\pi^2}{8}(e^{-i3\pi/4}+3e^{-i\pi/4})\\
\amp = -\frac{\pi^2}{8}(\sqrt{2}-2\sqrt{2}i)\\
\amp = \frac{\pi^2}{8}(-\sqrt{2}+2\sqrt{2}i)\text{.}
\end{align*}
On the other hand, breaking the path \(\gamma\) down into separate legs, we have
\begin{align*}
\int_{\gamma}f\, dz \amp = \int_{-R}^{-\delta}\frac{\ln\abs{x}+\pi i}{x^4+1}\, dx+\int_{\delta}^R \frac{\ln x}{x^4+1}\, dx+\int_{\gamma_\delta}f\, dz+\int_{\gamma_R}f\, dz\\
\amp =2\int_{\delta}^R\frac{\ln x}{x^4+1}+i\int_{\delta}^R\frac{\pi}{x^4+1}\, dx +\int_{\gamma_\delta}f\, dz+\int_{\gamma_R}f\, dz\text{,}
\end{align*}
where here \(\gamma_\delta(t)=\delta e^{i(\pi-t)}=i\delta e^{-it}\text{,}\) \(t\in [0,\pi]\text{,}\) and \(\gamma_R(t)=Re^{it}\text{,}\) \(t\in [0,\pi]\text{.}\) We will show that \(\lim\limits_{\delta\to 0^+}\int_{\gamma_\delta}f\, dz=0\text{,}\) and \(\lim\limits_{R\to \infty}\int_{\gamma_R}f\, dz=0\text{,}\) and hence, taking limits of the equality above,
\begin{equation*}
\frac{\pi^2}{8}(-\sqrt{2}+2\sqrt{2}i)= 2\int_{\delta}^R\frac{\ln x}{x^4+1}+i\int_{\delta}^R\frac{\pi}{x^4+1}\, dx\text{.}
\end{equation*}
Equating real and imaginary parts, we then conclude that
\begin{align*}
\int_0^\infty\frac{\ln x}{x^4+1}\, dx \amp = \frac{\sqrt{2}\pi}{16} \amp \\
\int_0^\infty \frac{1}{x^4+1}\, dx \amp = \frac{\sqrt{2}\pi}{4} \amp (\text{bonus}) \text{.}
\end{align*}
To see that \(\lim\limits_{\delta\to 0^+}\int_{\gamma_\delta}f\, dz=0\text{,}\) and \(\lim\limits_{R\to \infty}\int_{\gamma_R}f\, dz=0\text{,}\) it suffices to use ML-inequalities on both these paths. In general, for any circular path \(\phi(t)=se^{it}\text{,}\) \(t\in [0,\pi]\) we have
\begin{equation*}
\abs{\Log_\alpha( se^{it})}=\abs{\ln \abs{s}+it}\leq \ln \abs{s}+\pi\text{.}
\end{equation*}
Thus, we have
\begin{align*}
\abs{\int_{\gamma_R}f\, dz} \amp\leq \frac{\ln R+\pi}{R^4-1}\cdot \pi R \\
\amp = \pi \cdot \frac{\ln R+\pi}{R^3-1/R}\\
\abs{\int_{\gamma_\delta}f\, dz} \amp\leq \frac{\ln \delta+\pi}{1-\delta^4}\cdot \pi \delta \\
\amp = \pi \frac{\delta \ln \delta+\delta\pi}{1-\delta^4}\text{.}
\end{align*}
Some straightforward, if not altogether pleasant l’Hôpital’s rule arguments now show that
\begin{align*}
\lim\limits_{R\to \infty} \pi \cdot \frac{\ln R+\pi}{R^3-\tfrac{1}{R}} \amp=0 \\
\lim\limits_{\delta\to 0^+} \pi \frac{\delta \ln \delta+\delta\pi}{1-\delta^4}\amp =0
\end{align*}
and hence that
\begin{align*}
\lim\limits_{R\to \infty}\int_{\gamma_R}f\, dz \amp= 0 \\
\lim\limits_{\delta\to 0^+} \int_{\gamma_\delta}f\, dz \amp = 0 \text{,}
\end{align*}
as desired.
