Integral Representation of the Dottie Number 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).

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.
