Laurent expansion of a principal value integral

Question

Let $f(t)$ be a nice Hölder continuous function. Also, suppose that $f$ is even. I'm interested in evaluating integrals of the form:

$$\oint (1-z)^{k+1}\int_0^1 \frac{f(t)}{(1-zt)^{n+1}}dtdz,$$

where for the contour integral, one may assume any loop around $z=1$. Equivalently, I'm interested in calculating Laurent expansions of $F_n(z):=\int_0^1\frac{f(t)}{(1-zt)^{n+1}}dt$ about $z=1$. Note that for $z\in(1,\infty)$, $F_n(z)$ is defined as a Cauchy principle value integral when $n$ is even and as as a Hadamard principle value integral when $n$ is odd. The primary difficulty I'm encountering is that when $z>1,$ the contour integral becomes rather nontrivial. In other words, one needs to understand what's going on as $z$ approaches the real axis to the right of 1. This looks like a sort of Riemann Hilbert problem to me, at least for the evaluation of $F_n(z)$. To this extent, is there a generalization of the Riemann Hilbert method for such double integrals? I mention this because I would like to consider $f_n(t)$, instead of $f(t)$ and then perform asymptotics on the Laurent series coefficeints, as $n\rightarrow\infty$.

Upon swapping integrals, it seems like the problem depends on whether it's a Cauchy or Hadamard integral. Specifically, there is a sharp transition when $1/t$ enters the area bounded by the $\lambda$ contour. This gives me a nonsensical answer that depends on the contour, which is impossible.

Answer 1

define $f_{m}(t)$ for $m=1,2,\ldots$ by $\frac{d^{m}}{dt^{m}}f_{m}(t)=f(t)$,

with $\lim_{t\rightarrow 0}\frac{d^{n}}{dt^n}f_{m}(t)=0$ for $0\leq n<m$.

in particular, $f(t)=t^p$ produces $f_{m}(t)=t^{p+m}\frac{p!}{(p+m)!}$

Laurent expansion:

$$\int_0^1 \frac{f(t)}{(1-zt)^{n+1}}dt=\sum_{q=0}^\infty \frac{c_q}{(z-1)^{q+n+1}}$$

with coefficients

$$c_q=(-1)^{n+1}\frac{1}{n!q!}\sum_{s=q}^{\infty}\frac{(n+s)!s!}{(s-q)!}f_{s+1}(1)$$

not quite the simple answer one might have desired, perhaps still of some use for the large-$n$ asymptotics.

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Derivation:

insert a formal Taylor expansion $f(t)=\sum_{p=0}^{\infty}a_p t^{p}$ and integrate term by term,

$$\int_0^1 \frac{t^{p}}{(1-zt)^{n+1}}dt=\frac{1}{1+p}{}_2F_1(1+n,1+p,2+p;z)$$

$$=\frac{1}{1+p}(1-z)^{-1-n}{}_2F_1(1+n,1,2+p;1+(z-1)^{-1})$$

$$=\frac{1}{1+p}(1-z)^{-1-n}\frac{(p+1)!}{n!}\sum_{s=0}^{\infty}\frac{(n+s)!}{(p+s+1)!}\left(1+\frac{1}{z-1}\right)^s$$

$$=(1-z)^{-1-n}\frac{p!}{n!}\sum_{s=0}^{\infty}\frac{(n+s)!}{(p+s+1)!}\sum_{q=0}^s\frac{s!}{q!(s-q)!}\frac{1}{(z-1)^q}$$

$$=(1-z)^{-1-n}\frac{p!}{n!}\sum_{q=0}^{\infty}\frac{1}{(z-1)^q}\sum_{s=q}^{\infty}\frac{(n+s)!s!}{(p+s+1)!q!(s-q)!}$$

return to the integral of $f(t)$,

$$\int_0^1 \frac{f(t)}{(1-zt)^{n+1}}dt=(1-z)^{-1-n}\frac{1}{n!}\sum_{q=0}^{\infty}\frac{1}{(z-1)^q}\sum_{s=q}^{\infty}\frac{(n+s)!s!}{q!(s-q)!}\sum_{p=0}^{\infty}a_p\frac{p!}{(p+s+1)!}$$

$$=(1-z)^{-1-n}\frac{1}{n!}\sum_{q=0}^{\infty}\frac{1}{(z-1)^q}\sum_{s=q}^{\infty}\frac{(n+s)!s!}{q!(s-q)!}f_{s+1}(1)$$

$$=\sum_{q=0}^\infty \frac{c_q}{(z-1)^{q+n+1}}$$

with the coefficients $c_q$ given above.