You can actually see directly that there is no elementary expression for the integral, using contour integration. For your integral is the same as
$$\mathrm{Im}\int_0^\infty {e^{iax} \over x^2 + b^2}\mathrm dx$$
Note that
$${1 \over x^2 + b^2} = {1 \over 2ib}\left({1 \over x - ib} - {1 \over x + ib}\right)$$
So it suffices to evaluate the integrals
$$\int_0^\infty {e^{iax} \over x \pm bi}\,\mathrm dx$$
These integrals converge despite the denominator having only degree one because the exponential factor modulates it, similar to for $\dfrac{\sin(x)}{x}$. You can use your quarter circle idea on each of these two terms (without even having to worry about indentations)... the upper quarter circle for $+bi$ and the bottom quarter circle for $-bi$. For example you get that the first one is equal to
$$\int_0^\infty {e^{-ax} \over x + b}\,\mathrm dx$$
Letting $x = z - b$ this becomes
$$e^{ab}\int_b^\infty {e^{-az} \over z}\,\mathrm dz$$
Then letting $z = \dfrac{w}{a}$ this turns into
$$e^{ab}\int_{ab}^\infty {e^{-w} \over w}\,\mathrm dw$$
$$= e^{ab}\mathrm{Ei}(-ab)$$
Here $\mathrm{Ei}$ is the exponential function which is defined in terms of the above integral. So unless the exponential functions from the two quarter circles somehow cancel out (which I seriously doubt in view of Sasha's answer), you won't be able to get an elementary expression for the integral.
Call $\displaystyle K=\int_0^\infty \frac{\sqrt{t}}{1+t^2} \mathrm dt$ and $\displaystyle J_\gamma=\int_\gamma\frac{\sqrt{t}}{1+t^2} \mathrm dt$.
Basically, you made the same mistake twice.
(1) Evaluating $J_\gamma$ thanks to the residue theorem yields
$$
J_\gamma=2\pi\mathrm i\cdot\left.\left(\frac{\mathrm e^{(\log t)/2}}{t+\mathrm i}\right)\right|_{t=\mathrm i}=\pi\mathrm e^{(\log \mathrm i)/2}=\pi\mathrm e^{\mathrm i\pi/4}=\pi(1+\mathrm i)/\sqrt2.
$$
Here the mistake is to believe that (the principal value of) $\log\mathrm i$ is $\pi/2$ instead of $\mathrm i\pi/2$: consider that $\mathrm e^{\mathrm i\pi/2}=\cos(\pi/2)+\mathrm i\sin(\pi/2)=\mathrm i$ and that $\mathrm e^{\pi/2}$ is... well, a real number close to $4.81$.
(2) Evaluating $J_\gamma$ through an integral on the real line yields
$$
\int_{-\infty}^\infty \frac{\sqrt{t}}{1+t^2} \mathrm dt=K+\int_{-\infty}^0 \frac{\sqrt{t}}{1+t^2} \mathrm dt=K+\int_0^{+\infty} \frac{\mathrm i\sqrt{t}}{1+t^2} \mathrm dt=(1+\mathrm i)\,K.
$$
Here, the mistake is to believe that, for every positive real number $t$, $\sqrt{-t}$ is $\mathrm e^{\pi/2}\sqrt{t}$ instead of $\mathrm e^{\mathrm i\pi/2}\sqrt{t}=\mathrm i\sqrt{t}$.
Correcting steps (1) and (2), your reasoning yields the relation $$
(1+\mathrm i)K=\lim\limits_{R\to+\infty}J_{\gamma}=\pi(1+\mathrm i)/\sqrt2,
$$
hence the (exact) value $K=\pi/\sqrt2$.
Note: A "purely real" road is available to compute $K$. To do that, first use in succession the changes of variables $u=\sqrt{t}$, $v=\pm u$, and $w=1/v$ to get
$$
K=\int_0^\infty\frac{2u^2}{1+u^4}\mathrm du=\int_{-\infty}^\infty\frac{v^2}{1+v^4}\mathrm dv=\int_{-\infty}^\infty\frac{\mathrm dw}{1+w^4}.
$$
Thus,
$$
K=\frac12\int_{-\infty}^\infty\frac{x^2+1}{x^4+1}\mathrm dx.
$$
The factorisation $x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$ implies that
$$
\frac{x^2+1}{x^4+1}=\frac1{(\sqrt2x+1)^2+1}+\frac1{(\sqrt2x-1)^2+1},
$$
hence $K=\frac12(L_1+L_{-1})$ where, for every real number $a$,
$$
L_a=\int_{-\infty}^\infty\frac{\mathrm dx}{(\sqrt2x+a)^2+1}.
$$
Using the change of variable $u=\sqrt2x+a$, one sees that, for every real number $a$,
$$
L_a=\frac1{\sqrt2}\int_{-\infty}^\infty\frac{\mathrm du}{u^2+1}=\frac{\pi}{\sqrt2},
$$
where the last equality is direct if one recognizes the density of a standard Cauchy distribution. This proves finally that $K=\pi/\sqrt2$.
Best Answer
I'm afraid there is no a closed form for this integral. $$I=\int_0^\infty \frac{4\pi}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{e^{-x}-1}) dx=4\pi\int_0^\infty \frac{dx}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{2}\coth\frac{x}{2}-\frac{1}{2}) dx$$ $$=-\frac{\pi}{4}+\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\Big(\frac{1}{t}-\coth t\Big)=-\frac{\pi}{4}-\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\frac{d}{dt}\Big(\ln\frac{\sinh t}{t}\Big)$$ Using the representation $\,\,\ln\frac{\sinh t}{t}=\sum_{k=1}^\infty\ln\Big(1+\frac{t^2}{\pi^2k^2}\Big)$ and making the change $x=t^2$ $$I=-\frac{\pi}{4}-\pi\int_0^\infty\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^N\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}$$ $$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\sum_{k=1}^\infty\int_0^N\frac{dx}{4\pi^2+x}\frac{1}{\pi^2k^2+x}$$ $$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^Ndx\bigg(\frac{1}{3\pi^2}\Big(\frac{1}{\pi^2+x}-\frac{1}{4\pi^2+x}\Big)+\frac{1}{(4\pi^2+x)^2}\bigg)$$ $$-\pi\lim_{N\to\infty}\sum_{k=3}^\infty\int_0^N\frac{dx}{\pi^2(k^2-4)}\Big(\frac{1}{4\pi^2+x}-\frac{1}{\pi^2k^2+x}\Big)$$ Integrating and taking the limit $$=-\frac{\pi}{4}-\frac{2\ln 2}{3\pi}-\frac{1}{4\pi}-\pi \sum_{k=3}^\infty\frac{1}{\pi^2(k^2-4)}\ln\frac{k^2}{4}$$ $$I(n)=-\frac{\pi}{4}+\frac{3\ln 2}{8\pi}-\frac{1}{4\pi}-\frac{2}{\pi}\sum_{k=3}^\infty\frac{\ln k}{k^2-4}\approx-1.36824...$$ It is not clear whether a closed form for the sum $\sum_{k=3}^\infty\frac{\ln k}{k^2-4}$ exists.