$\int_{-\infty}^\infty \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx$ via complex analysis I'm a little new to complex analysis, so bear with me:
First, I defined a function $f(x)= \frac{\sin(x-\frac 1x)}{x+\frac 1x}$, so that I could define a new function $f(z)= \frac{e^{i(z-\frac 1z)}}{z+\frac 1z}$, which I then simplified to be $f(z)= \frac{ze^{i(z-\frac 1z)}}{z^2+1}$
Second, I defined a contour $C$ that contains the interval $[-R, -\epsilon]$, the contour $\gamma$ which a a semicircle with radius $\epsilon$ centered at the origin, the interval $[\epsilon, R]$, and the semicircular contour $\Gamma$ with radius $R$ centered at the origin. I want to take the limit as $R\rightarrow\infty$ and $\epsilon\rightarrow0$
Thus, 
$$\int_C f(z)dz=\int_{-R}^{-\epsilon} f(z)dz+\int_\gamma f(z)dz+\int_\epsilon^R f(z)dz+\int_\Gamma f(z)dz$$
For the leftmost integral, I can simply use the residue theorem, as $C$ is a closed contour with a simple pole at $i$. Thus,
$$\int_C f(z)dz=2\pi i Res(f, i)$$
$$=2\pi i\frac{1}{2e^2}=\frac{\pi i}{e^2}$$
Next, looking at the first and third integrals on the right,
substitute $z=-u$ and $dz=-du$ into the first integral, so then
$$\int_{-R}^{-\epsilon} f(z)dz+\int_\epsilon^R f(z)dz$$
$$\int_{\epsilon}^{R} \frac{-ue^{i(\frac 1u-u)}}{u^2+1}du+\int_\epsilon^R \frac{ze^{i(z-\frac 1z)}}{z^2+1}dz$$
which can then be combined to get
$$\int_{\epsilon}^{R} \frac{z(e^{i(z-\frac 1z)}-e^{-i(z-\frac 1z)})}{z^2+1}dz$$
Next, multiplying by $\frac{2i}{2i}$, we get
$$2i\int_{\epsilon}^{R} \frac{z\sin(z-\frac 1z)}{z^2+1}dz$$
And then, letting $R\rightarrow\infty$ and $\epsilon\rightarrow0$ we get
$$2i\int_{0}^{\infty} \frac{z\sin(z-\frac 1z)}{z^2+1}dz$$
Also, since this integral is only on the real line, we can exchange the $z$ for an $x$, and also simplify it so that it looks like the original
$$2i\int_{0}^{\infty} \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx$$
which then leaves us with
$$\frac{\pi i}{e^2}=2i\int_{0}^{\infty} \frac{\sin(x-\frac 1x)}{x+\frac 1x}dx+\int_\gamma f(z)dz+\int_\Gamma f(z)dz$$
The problem is that I'm not quite sure how to deal with the other two integrals in the equation. I pretty sure that the integral over $\Gamma$ tends toward 0 just by using the M-L inequality, but I'm not so sure about how to evaluate the integral over $\gamma$ 
 A: for the contour $\Gamma$ we parametrize: $\eta : [0,1] \to \mathbb{C} : t \mapsto R e^{ti \pi}$. Then $$\int_\Gamma f(z) ~dz= \int_0^1 \frac{\exp\big({i (R e^{ti \pi} - e^{-ti \pi}/R)}\big)}{Re^{ti \pi}+ e^{-ti \pi}/R}\cdot i \pi R e^{ti \pi}~dz$$Now for the enumerator we have $$\exp\big({i (R e^{ti \pi} - e^{-ti \pi}/R)}\big)=\\\exp\big( i R \cos (t \pi)\big) \cdot \exp\big(- R \sin (t \pi)\big)\cdot\exp\big(-i\cos(-t \pi)/R\big)\cdot \exp\big( \sin(-t\pi)/R\big).$$The second  term hast is the important one. The ones with purely imaginary argument in the $\exp$ are of absolute value one. The last one tends to $1$ for $R\to \infty$ and as $-R \sin(t\pi)$ tends to $- \infty$ (because on $[0,\pi]$ we have $\sin \ge 0$) the second term goes to zero exponentially, overweighing the $R$ which we have from the derivative.The denominator is bounded by $2R$ hence we get that the limit of the integral is zero. The contour $\gamma$ can be handled the same way, the trick again is writing $\exp\big(e^{it}\big)=\exp\big( i \cos(t) - \sin(t) \big)$.
