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If asymptotics is enough, an accurate large-$s$ approximation is $\phi(s)\approx 2.8-\ln |s|$, see the plot above (the error is of order $1/s$, so almost invisible on the scale of the plot for the l …

I don't think there is much hope for an exact expression; for small $x$ a series expansion gives closed-form results, the first three terms are $$\sum_{n=0}^{\infty} \sum_{k=0}^{3^n-1} \frac1 { \left …

the equality is not explicit in Nakajima's lecture notes (which you can download from here); proposition 5.7 on page 60 comes closest.

I'll take your first integral, to give you an indication of what types of closed-form expressions you can expect: $${\cal I}(x)={\cal P}\int_{-1}^{1}\frac{\sqrt{1-t^2}\sin kt}{t-x}dt$$ For $x=0$ thi …

I would close the contour in the upper half of the complex plane, the principal value picks up $i\pi$ times the residue$^\ast$ at $t=0$, which is $u/(1-u)$. There are no other poles.$^{\ast\ast}$ $^\a …

Proof of the existence of a complex root in the general case, expanding Oleg's remark: Think of the $n$ complex numbers $\sin(\lambda_j z)$ as vectors in the 2D plane; for $n\geq 3$ and fixed $z$ add …

Let me first remove the $Bx$ term by completing the square, $$I=\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2+iBx}}{x - a}\,dx=e^{-iB^2/4A}\int\limits_{-\infty}^{\infty} \frac{e^{i A x^2}}{x - a-B/2 …

A 2010 status report is given by D. Khavinson et al. in Borcea's variance conjectures on the critical points of polynomials. A 2019 update is in A note on a recent attempt to prove Sendov's conjecture …

How about $t^{-1}|e^{int}-1|=t^{-1}\sqrt{2-2\cos nt}$? Here is the comparison plot for $n=80$. Orange: $|1+z+z^2\cdots +z^{n-1}|$ with $z=1+it$ and $n=80$, as a function of $t$; Blue $t^{-1}|e^{int}- …

Here is a derivation using the Fourier series of $\cos ax$: $$\cos ax=\frac{\sin \pi a}{\pi a}+\frac{2a\sin\pi a}{\pi}\sum_{n=1}^\infty\frac{(-1)^n\cos nx}{a^2-n^2}.$$ Hence for $x=0$ we have the iden …

Cartan's theorem (1931) is the extension of Schwarz's lemma to multivariable functions. See The Schwarz lemma at the boundary, page 9. H. Cartan, Les fonctions de deux variables complexes et le probl …

You can use the residue theorem, given the series expansion of $$h(z)=e^{b/(z-c)}g(z)=\sum_{n=0}^\infty h_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates t …

U.R Freiberg and M.R. Lancia, Energy Form on a Closed Fractal Curve (2004): The Koch snow flake is the union of three Koch curves of Hausdorff dimension $D=\ln 4/\ln 3$ and Hölder exponent $\beta=\lo …

For $m=1$, this is the residue: $$\operatorname{Res} \left( z^{kn-1} \left( az+1 \right)^{-k}; -1/a \right)=\frac{(-1)^{k n-k} }{a^{kn}(k-1)!}\prod _{p=1}^{k-1} (k n-p).$$

Linearization in $\phi_k(r)$ gives the desired approximation: $$S=\left|\int_0^\infty dr\, A(r)e^{-i[\phi_0(r)+\sum_{k=1}^n \phi_k(r)]} dr \right|^2$$ $$=\int_0^\infty dr\,\int_0^\infty dr'\, A(r)A(r' …