# 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!

• Comments are not for extended discussion; this conversation has been moved to chat. – quid Jan 1 at 1:47
• Numerical approximation suggests your integral (and the more general pattern) is true. Do you have a proof of it? – Cheerful Parsnip Jan 1 at 1:59
• I'm guessing residue analysis is the right way to prove it. – Cheerful Parsnip Jan 1 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. – Zacky Jan 1 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. – Jyrki Lahtonen Jan 3 at 12:54

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.

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

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

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

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

• Beauties! Maybe for the first one, since the integrand is an even function we can write: $$\int_{-\pi}^{\pi } \frac{ \sin ({2018} x) \sin (x)}{1-2 e \cos (x)+e^2} dx=\frac{\pi}{e^{{2019}}}$$ – Zacky Dec 31 '18 at 22:45
• @Zacky. Thanks. Happy new year! – Mariusz Iwaniuk Dec 31 '18 at 23:17
• A beginner's one $\int 2019x^{2018}dx=x^{2019}$ ☺️🎄 – user376343 Dec 31 '18 at 23:44
• $+ 2019\cdot \text{constant}$ $\quad \smile \quad$ – Zacky Jan 1 at 0:27
• Another simple one $\mathscr{L}\left[t^{2018}\right] = 2018! / s^{2019}$ Happy new year! – user150203 Jan 2 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. – user21820 Jan 1 at 14:52

\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}

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

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