I will show that $$\int_{-\infty}^{\infty} \frac{\sin(x-nx^{-1})}{x+x^{-1}}\,dx=\frac{\pi}{e^{n+1}}.$$ I will do this using residue theory. We consider the function $$F(z)=\frac{z\exp(i(z-nz^{-1}))}{z^2+1}.$$ On the real axis, this has imaginary part equal to our integrand. We integrate around a contour that goes from $$-R$$ to $$R$$, with a short half circle detour around the pole at $$0$$. Then we enclose it by a circular arc through the upper half plane, $$C_R$$. The integral around this contour is $$2\pi i$$ times the residue of the pole at $$z=+i$$. Using the formula (see Wikipedia, the formula under "simple poles") for the residue of the quotient of two functions which are holomorphic near a pole, we see that the residue is $$Res(F,i)=\frac{i\exp(i(i-i^{-1}n)}{2i}=\frac{1}{2}e^{-(n+1)}.$$ Thus the value of the integral is $$2\pi iRes(F,i)=i\frac{\pi}{e^{n+1}}$$. This is the answer we want up to a constant of $$i$$, which comes from the fact that our original integrand is the imaginary part of the function $$F(z)$$. We are therefore done if we can show that the integral around $$C_R$$ approaches $$0$$ as $$R\to \infty$$ as well as the integral around the little arc detour at the origin going to $$0$$ as its radius gets smaller. The fact that the $$C_R$$ integral approaches $$0$$ follows from Theorem 9.2(a) in these notes. This is because we can take $$f(z)=\frac{z e^{-inz^{-1}}}{z^2+1}$$ in that theorem to get $$F(z)=f(z)e^{iz}$$. The modulus $$|e^{-inz^{-1}}|=|e^{-inR^{-1}(\cos\theta-i\sin\theta)}|=e^{-\frac{n}{R}\sin\theta}.$$ Note that $$\sin\theta \geq 0$$ in the upper half plane, so we can bound this modulus by $$1$$. So we get that $$|f(z)|\leq |z|/|z^2+1|$$ and moreover $$z/(z^2+1)$$ behaves like $$1/z$$ as $$R$$ increases, so the hypotheses of Theorem 9.2a are satisfied.