Does the following transition between $2018$ and $2019$ hold true?$$\large\bbox[10pt,#000,border:5px solid green]{\color{#58A}{\color{#A0A}\int_{\color{#0F5}{-\infty}}^{\color{#0F5}{+\infty}} \frac{\color{yellow}\sin\left(\color{#0AF}x\color{violet}-\frac{\color{tomato}{2018}}{\color{#0AF}x}\right)}{\color{#0AF}x\color{violet}+\frac{\color{aqua}1}{\color{#0AF}x}} \color{#A0A}{\mathrm d}\color{#0AF}x\color{aqua}=\frac{\color{magenta}\pi}{\color{magenta}e^{\color{red}{2019}}}}}$$
$$\large\color{red}{\text{Happy new year!}}$$
I must say that I got lucky arriving at this integral.
Earlier this year I have encountered the following integral:$$\int_0^\infty \frac{\sqrt{x^4+3x^2+1}\cos\left[x-\frac{1}{x} +\arctan\left(x+\frac{1}{x}\right)\right]}{x(x^2+1)^2}dx=\frac34\cdot \frac{\pi}{e^2}$$
Which at the first sight looks quite scary, but after some manipulations it breaks up into two integrals, one of which is:$$\int_{-\infty}^\infty \frac{\sin\left(x-\frac{1}{x}\right)}{x+\frac{1}{x}}dx$$
And while trying to solve it I also noticed a pattern on an integral of this type.
Also today when I saw this combinatorics problem I tried to make something similar and remembered about the older integral. $\ddot \smile$
If you have other integral of the same type feel free to add!
Answer
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.
The integral around the arc near the origin limits to zero by elementary estimates, concluding the proof.
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