# 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!

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– quid
Jan 1, 2019 at 1:47
• Numerical approximation suggests your integral (and the more general pattern) is true. Do you have a proof of it? Jan 1, 2019 at 1:59
• I'm guessing residue analysis is the right way to prove it. Jan 1, 2019 at 2:00
• Yes, I have some sort of proof, otherwise I could not make the assumption in the quesiton. My residue analysis skill is inexistent, so I can't speak of that. Jan 1, 2019 at 2:15
• Keep it simple. $$\int_0^1x^{2018}\,dx=\frac1{2019}.$$ The changes necessary for this to work at the next new year are left to the reader as an exercise. Jan 3, 2019 at 12:54

$$\int_0^{\pi } \frac{2 \sin (2018 x) \sin (x)}{1-2 e \cos (x)+e^2} \, dx=\frac{\pi }{e^{2019}}$$

$$\int_0^1 (-\ln (x))^{2018} \, dx=\Gamma (2019)$$

$$\int_0^1 \frac{\frac{1-x^{2018}}{1-x}-2018}{\ln (x)} \, dx=\ln (\Gamma (2019))$$

$$\int_0^{\infty } \frac{\tan ^{-1}(2018 x)}{x \left(1+x^2\right)} \, dx=\frac{1}{2} \pi \ln (2019)$$

• Beauties! Maybe for the first one, since the integrand is an even function we can write: $$\int_{-\pi}^{\pi } \frac{ \sin ({2018} x) \sin (x)}{1-2 e \cos (x)+e^2} dx=\frac{\pi}{e^{{2019}}}$$ Dec 31, 2018 at 22:45
• @Zacky. Thanks. Happy new year! Dec 31, 2018 at 23:17
• A beginner's one $\int 2019x^{2018}dx=x^{2019}$ ☺️🎄 Dec 31, 2018 at 23:44
• $+ 2019\cdot \text{constant}$ $\quad \smile \quad$ Jan 1, 2019 at 0:27
• Another simple one $\mathscr{L}\left[t^{2018}\right] = 2018! / s^{2019}$ Happy new year!
– user150203
Jan 2, 2019 at 10:26

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!

• Actually this is a rather good one! The description is only one line, and 2019 is the first positive integer to satisfy it. Jan 1, 2019 at 14:52

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.

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 $$\log((1+\sqrt{5})/2)<\log(2)$$.

\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}

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 $$\int_{-\infty}^{\infty} \frac{\cos \left(x-\frac{1}{x} \right)}{1+x^{2}} \ dx$$

• Indeed, a possible approach. Small typo: $$\mathfrak{I}(b)=\int\limits_0^{\infty}\mathrm dx\,\frac {\sin\left(x-\frac {\color{red} b} {x} \right)}{x+\frac 1x}$$ Jan 1, 2019 at 2:45
• @Zacky Yes you are correct. Jan 1, 2019 at 2:56