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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.

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  • $\begingroup$ The "principal value" modification won't do anything if $n$ is odd and $0<z<1$, since $\lim \int_{|t-z|>\delta} f(t)/(t-z)^{2k} = \int f(t)/(t-z)^{2k}$ by MC. But you don't need to worry about this anyway, since $F_n$ is obviously holomorphic on $|z|>1$, which is enough to give you a Laurent expansion. $\endgroup$ Jul 5 '14 at 1:37
  • $\begingroup$ An obvious try would be to change the order of integration, evaluate the $z$ integral with the calculus of residues and see what happens. $\endgroup$ Jul 5 '14 at 2:33
  • $\begingroup$ @ChristianRemling: seeing as how it's a principle value integral for $z<1$, I'm not entirely sure how to justify the swap of integrals. When one factors out $t$ to get a power series in $z/t$, there's something going wrong at $t=0$. $\endgroup$
    – Alex R.
    Jul 5 '14 at 5:37
  • $\begingroup$ You don't need to expand, just use Fubini (for $|z|>1$). $\endgroup$ Jul 5 '14 at 23:10
  • $\begingroup$ @ChristianRemling: in my last comment I meant to say that swapping the integrals and doing contour integral first gives (via residues) something proportional to $1/t^{n+1+k}$, which is not integrable on $[0,1]$. Perhaps I'm being dense? $\endgroup$
    – Alex R.
    Jul 6 '14 at 0:12
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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.


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.

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  • $\begingroup$ Thanks, this looks like a great approach! There seems to be a bug though. A toy example of $f=1,n=1$ gives: $\int_0^11/(1-zt)^2dt=1/(1-z)$, whereas the expansion starts at $1/(1-z)^2$ (also the $c_q$ coefficients blow up). Hopefully this is just a minor index issue. $\endgroup$
    – Alex R.
    Jul 9 '14 at 18:06
  • $\begingroup$ Ok here's the issue, something I've been hit by again and again thinking about this problem. Just before the "return to the integral of $f(t)$", the summand $\sum_{s=q}^\infty $ is divergent when $p+1-q-n<0$. This issue arises because when you do the binomial expansion, you are effectively calculating ${}_2F_1(n+1,1,p+2;1)$, which is actually divergent when $p+2-1-n-1\leq 0$. $\endgroup$
    – Alex R.
    Jul 9 '14 at 19:27
  • $\begingroup$ you're right, it fails; I'll leave the answer for the record, but have made it CW to avoid more rep. $\endgroup$ Jul 10 '14 at 7:08
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    $\begingroup$ The basic issue might be that in the new version of the question, it isn't even clear that $F_n$ has a Laurent expansion (the integral defining it [or not] is now singular for all real $z>1$). $\endgroup$ Jul 12 '14 at 19:29

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