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I noticed that a lot of commonly-used mathematical constants that can't be expressed in closed-form can be expressed by integrals, such as $$\pi=\int_{-\infty}^\infty \frac{dx}{x^2+1}$$ and $$\frac{1}{1+\Omega}=\int_{-\infty}^\infty \frac{dx}{(e^x-x)^2+\pi^2}$$ I was wondering if anyone knows how to express the Dottie Number $\omega$, or the unique solution to the equation $$\cos(\omega)=\omega$$ using an integral.

In general, what are some strategies for expressing constants as integrals? I'm also struggling to express the reciprocal fibonacci constant as an integral (but don't tell me how to do that one).

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  • $\begingroup$ The word to look up here is period. $\endgroup$ – Chappers Sep 26 '17 at 23:16
  • $\begingroup$ @Chappers What do you mean by that? $\endgroup$ – Frpzzd Sep 26 '17 at 23:16
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    $\begingroup$ en.wikipedia.org/wiki/Ring_of_periods $\endgroup$ – Chappers Sep 26 '17 at 23:19
  • $\begingroup$ @Chappers Ah, I see. Thanks! Any insights about how I might go about solving this problem? $\endgroup$ – Frpzzd Sep 26 '17 at 23:21
  • $\begingroup$ Not an area I know much about, I'm afraid. $\endgroup$ – Chappers Sep 26 '17 at 23:28
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Okay, I've figured out how to do this using the residue theorem, so I should probably post an answer to my own question in case anyone happens upon it in the future. How satisfying it is to answer a question that I've had for such a long time!

Consider the function $$f(z)=\frac{1}{(z-\cos z)^2-\pi^2}$$

SINGULARITIES: The poles of $f$ occur at the zeroes of $$(z-\cos z)^2-\pi^2=(z-\cos z+\pi)(z-\cos z-\pi)$$ Let $z_k^+$ be the poles of $f$ that are zeroes of $z-\cos z+\pi$ and $z_k^-$ be the poles of $f$ that are zeroes of $z-\cos z-\pi$. Because each $z_k^+$ satisfies $$z_k^+-\cos z_k^+ +\pi=0$$ it follows that $$(z_k^++2\pi)-\cos (z_k^+ +2\pi) -\pi=0$$ and so $z_k^+$ is a zero of $z-\cos z-\pi$. Thus we may establish the following correspondence: $$z_k^++2\pi=z_k^-$$ Now let us set $z_0^+=\pi- ա$ and $z_0^-=-\pi- ա$. Note that this is the only pair of corresponding poles lying on opposite sides of the line $\Re(z)=-\pi/2$.

RESIDUES: I shall omit the algebra: $$\text{Res}(f,z_k^+)=-\frac{1}{2\pi} \frac{1}{1+\sin z_k^+}$$ $$\text{Res}(f,z_k^-)=\frac{1}{2\pi} \frac{1}{1+\sin z_k^+}$$ Notice that $$\text{Res}(f,z_k^+)+\text{Res}(f,z_k^-)=0$$ This will be important later.

CONTOUR: Let $C_1$ be a straight line contour from $-\pi/2+ri$ to $-\pi/2-ri$, and let $C_2$ be a semicircular arc stretching counterclockwise from $-\pi/2-ri$ to $-\pi/2+ri$, and let $\gamma = C_1 \cup C_2$.

As we let $r\to\infty$, this contour will enclose all poles of $f$ to the right of the line $\Re(z)=-\pi/2$. Because all such poles can be put in equal and opposite pairs except for $z_0^+=\pi- ա$, the sum of residues inside of $\gamma$ as $r\to\infty$ will approach $$\text{Res}(f,\pi- ա)$$ which is equal to $$\frac{1}{2\pi} \frac{1}{1+\sqrt{1-ա^2}}$$

EVALUATION: By the residue theorem, $$\oint_\gamma f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$ or $$\int_{C_1} f(z)dz+\int_{C_2} f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$ Since $\int_{C_2} f(z)dz$ vanishes as $r\to\infty$, we have $$\int_{C_1} f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$ or $$\int_{-\pi/2+i\infty}^{-\pi/2-i\infty} \frac{dz}{(z-\cos z)^2-\pi^2}=\frac{i}{1+\sqrt{1-ա^2}}$$

Now I'm going to omit a lot of nasty algebra. After simplifying this and equating real and imaginary parts, one ends up with the result $$\int_{-\infty}^\infty \frac{12\pi^2+16(z-\sinh z)^2}{(3\pi^2+4(z-\sinh z)^2)^2+16\pi^2(z-\sinh z)^2}dz=\frac{1}{1+\sqrt{1-ա^2}}$$ which is equivalent to $$\color{green}{\int_{0}^\infty \frac{3\pi^2+4(z-\sinh z)^2}{(3\pi^2+4(z-\sinh z)^2)^2+16\pi^2(z-\sinh z)^2}dz=\frac{1}{8+8\sqrt{1-ա^2}}}$$ Wolfram Alpha agrees with this to a lot of digits. Whoopee!

Of course, this is a nasty integral and no-one would ever want to try and evaluate it directly, so I'm still looking for something a little nicer.

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You seem to have provided an answer to your own question. Thank you. I will bookmark this post; it may help me in some of my own work. I'd like to add a little more information; it may suggest other ways to tackle your goal, perhaps direct you to a more concise solution.

The Dottie Number (D) also happens to be the solution of Kepler's Equation of Elliptical Motion (it satisfies the "equal area swept out in equal time" condition) at the quarter period for e = 1. i.e.,

$$ E - e \sin(E) = M $$

$$ E - \sin(E) = \frac\pi 2 $$

(Aside #1: When e = 1, the conic is a parabola, suggesting D can be expressed in the form

$$ D = a b^2 $$ where a and be are constants yet to be determined.)

Kepler's Equation reduces to $$ \cos(\sin(E)) = \sin(E) $$

This is the definition of the Dottie Number, where $\sin(E) = D$.

I bring up this connection because Kepler's Equation can be expressed in terms of a Bessel Function of the First Kind ($J_k$): $$ E = M + 2 \sum_{k=1}^\infty \frac{1}{k} J_k (ke) \sin(kM) $$

It's an infinite series, not an integral, but perhaps it helps you.

(Aside #2: The Dottie Number is just a special case of a more general equation: $$ \cos(k x) = x $$ where 0 $\le$ k $\le$ 1. The Dottie Number is the solution to the equation for k = 1. A plot of the solutions to this equation for 0 $\le$ k $\le$ 1 is tantalizingly close to a graph of $\text{sech}(0.817326346581 k)$. Since $E$ is periodic and a function of itself, I suppose a solution should include sine and exponential terms, but I haven't found an exact relationship. Perhaps it involves fractional derivatives.)

(Aside #3: There used to be a blog solely about the Dottie Number. Have you come across it? It has lot of good information about the Dottie Number, for example, Bertrand's Semicircle.)

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  • $\begingroup$ Interesting. Do you have a link to this blog? Or to any other relevant documents online? $\endgroup$ – Frpzzd Dec 25 '17 at 20:29
  • $\begingroup$ The blogger seems to have let the original blog die. However, I found it backed up on the Wayback Machine: The Dottie Number Blog. The post about Betrand's Semicircle is from February 2011. $\endgroup$ – DavidB2013 Dec 25 '17 at 23:27
  • $\begingroup$ Okay, do you have a link? $\endgroup$ – Frpzzd Dec 25 '17 at 23:28
  • $\begingroup$ This answer seems similar to my answer (posted in 2015). math.stackexchange.com/a/1174724/215560 $\endgroup$ – giorgiomugnaini Jul 5 '18 at 16:51

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