Integral for the New Year $2019$! 
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!
 A: Not an integral, but mildly interesting, is $2019=F_{17}+F_{14}+F_9+F_6+F_4$, a sum of five Fibonnaci numbers; this is fewer addends than we would need for a binary representation even though $\log((1+\sqrt{5})/2)<\log(2)$.
A: $$\begin{align}
\star \int_0^{\infty} e^{\left(-2019^2x\left(\frac {x-6}{x-2}\right)^2\right)}\frac {1}{\sqrt {x}} dx&=\frac {\sqrt {\pi}}{2019}\\
\star \int_0^{2\pi} \frac {(1+2\cos x)^{2019}\cos(2019x)}{3+2\cos x}dx&=\frac {2\pi}{\sqrt 5} (3-\sqrt 5)^{2019}\\
\star \int_0^1 \frac {\ln(1-x)}{x}\frac {4038}{\ln^2x+(4038\pi)^2}dx&= -\ln \left(\frac {2019! e^{2019}}{(2019)^{2019}\sqrt {4038\pi}}\right)\\
\star\int_{-\infty}^{\infty} \frac {\vert \cos (2019x)\vert}{1+x^2}dx&= 4\cosh (2019)\arctan e^{-2019}\\
\star\int_0^{\infty} \frac {\ln(\tan^2(2019x))}{1+x^2}dx&=\pi\ln(\tanh (2019))
\end{align}$$
A: This is a possible start. I will finish this answer when I get more paper and time.

Denote the generalized integral as$$\mathfrak{I}(b)=\int\limits_0^{\infty}\mathrm dx\,\frac {\sin\left(x-\frac bx\right)}{x+\frac 1x}$$Observe that the integral we seek is simply $2\mathfrak{I}(b)$ due to the even - ness of the integrand. Differentiate with respect to $b$ to get that$$\mathfrak{I}'(b)=-\int\limits_0^{\infty}\mathrm dx\,\frac {\cos\left(x-\frac bx\right)}{1+x^2}$$And make the substitution $z=x-\tfrac bx$ which is a type of Cauchy - Schlomilch transformation. For reference, you can visit this link: Evaluating the integral $ \int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \ dx$
A: $$\int_0^{\pi } \frac{2 \sin (2018 x) \sin (x)}{1-2 e \cos (x)+e^2} \, dx=\frac{\pi
   }{e^{2019}}$$
$$\int_0^1 (-\ln (x))^{2018} \, dx=\Gamma (2019)$$
$$\int_0^1 \frac{\frac{1-x^{2018}}{1-x}-2018}{\ln (x)} \, dx=\ln (\Gamma (2019))$$
$$\int_0^{\infty } \frac{\tan ^{-1}(2018 x)}{x \left(1+x^2\right)} \, dx=\frac{1}{2} \pi  \ln (2019)$$
A: Here $2019$ as the sum of the squares of $3$ primes in $6$ ways:
$$2019= 7^2 + 11^2 + 43^2 $$
$$2019= 7^2 + 17^2 + 41^2 $$
$$2019= 13^2 + 13^2 + 41^2 $$
$$2019= 11^2 + 23^2 + 37^2 $$
$$2019= 17^2 + 19^2 + 37^2 $$
$$2019= 23^2 + 23^2 + 31^2 $$
Actually $2019$ is the smallest integer to have such a property. Happy New Year!
A: 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. 
