= Solution
{c}
Use the <principal logarithm> $\operatorname{Log}z=\log|z|+i\operatorname{Arg}z$ with $-\pi<\operatorname{Arg}z<\pi$, whose <branch cut> is the negative <real axis>. For this choice,
$$
F(z)=\frac1{(z^2+a^2)\operatorname{Log}z}
$$
has three <simple poles> away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\operatorname{Log}1=0$, must be included.
Choose $0<\varepsilon<\min\{1,a\}$ and $R>\max\{1,a\}$, and use a positively oriented <keyhole contour> around the negative <real axis>: the large <circle> is traversed counterclockwise, the upper bank from $-R$ to $-\varepsilon$, the small <circle> clockwise, and the lower bank from $-\varepsilon$ to $-R$. Formally, use radial segments at <arguments> $\pm(\pi-\eta)$, then let $\eta\downarrow0$ with $R,\varepsilon$ fixed. The <function> has continuous limiting values on each compact bank, which justifies this limit.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2021-q5-keyhole.png]
{title=Keyhole contour around the negative real axis, including the logarithm's pole at 1}
Put $L=\log a$. The <residues> are
$$
\operatorname{Res}(F,1)=\frac1{1+a^2},\qquad
\operatorname{Res}(F,ia)=\frac1{2ia(L+i\pi/2)},\qquad
\operatorname{Res}(F,-ia)=-\frac1{2ia(L-i\pi/2)}.
$$
The first follows from $\operatorname{Log}z=(z-1)+O((z-1)^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum is
$$
\Sigma=\frac1{1+a^2}-\frac{\pi}{2a(L^2+\pi^2/4)}.
$$
These expressions are valid also for $a=1$, since $L\pm i\pi/2$ never vanish for real $L$.
On the outer <circle>, $|\operatorname{Log}z|\geq\log R$ for $R>1$ and $|z^2+a^2|\geq R^2-a^2$. On the inner <circle>, $|\operatorname{Log}z|\geq|\log\varepsilon|$ and $|z^2+a^2|\geq a^2-\varepsilon^2$. Thus the <estimation lemma> bounds the two arc <integrals> by
$$
\frac{2\pi R}{(R^2-a^2)\log R}\longrightarrow0,\qquad
\frac{2\pi\varepsilon}{(a^2-\varepsilon^2)|\log\varepsilon|}\longrightarrow0,
$$
respectively, as $R\to\infty$ and $\varepsilon\downarrow0$.
On the upper bank, $z=-x+i0$ gives $\operatorname{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined <integral> is
$$
\begin{aligned}
\int_\varepsilon^R\frac1{x^2+a^2}\left(\frac1{\log x+i\pi}-\frac1{\log x-i\pi}\right)\,dx
&=-2\pi i\int_\varepsilon^R\frac{dx}{(x^2+a^2)((\log x)^2+\pi^2)}.
\end{aligned}
$$
The real <improper integral> converges: its integrand is bounded near zero by $1/(a^2\pi^2)$ and at infinity by $1/(\pi^2x^2)$. Taking the arc limits in the <residue theorem> gives $-2\pi i I=2\pi i\Sigma$, hence
$$
\boxed{\int_0^\infty\frac{dx}{(x^2+a^2)((\log x)^2+\pi^2)}
=\frac{\pi}{2a((\log a)^2+\pi^2/4)}-\frac1{1+a^2}.}
$$
The final subtraction is exactly the contribution of the pole at $1$.
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