Jtdylktuy

Consider the catalan numbers $C_n=frac{1}{n+1}binom{2n}{n}$, which have the following generating function:

$$C(x)=sum_{n=0}^{infty}C_nx^n=frac{1-sqrt{1-4x}}{2x}.$$

I am well aware of the usual proofs of this identity: for example, one can write $C(x)$ as the root of a quadratic equation via combinatorial identities, or one can Taylor-expand $sqrt{1-4x}$.

I am wondering if there is a way to start from $C(x)=frac{1-sqrt{1-4x}}{2x}$ and extract the Taylor coefficients via residue calculus. That is, if there is a direct way to evaluate the integral

$$oint_{C}frac{1-sqrt{1-4z}}{2z^{n+1}}dz$$

for some appropriately chosen $C$.

The solution could have been much shorter, but I prefer details to concision.

One wants to show that
$$C_n=frac1{2pi i}oint_{C}frac{1-sqrt{1-4z}}{2z}frac{dz}{z^{n+1}}$$

It is easy to see
$$oint_{C}frac{1-sqrt{1-4z}}{2z}frac{dz}{z^{n+1}}=-frac12oint_{C}frac{sqrt{1-4z}}{z^{n+2}}dz$$

Definitions

We will take the branch cut of $sqrt{1-4z}$ on $mathbb R_{ge1/4}$, i.e.
$$sqrt{1-4z}=expbigg[frac{ln|1-4z|+iarg(1-4z)}2bigg]qquad{arg(1-4z)in[-pi,pi)}$$

I will show how to evaluate

$$I_n:=oint_Csqrt{1-4z}~~ z^{-n-2}dz$$ directly.

Decomposing the contour integral

Take $C$ to be the keyhole contour centered at $1/4$ which avoids the branch cut.

The two integrals, one around branch point and one along the infinitely large circle, vanish.

The integral above the real axis $I^+$ is (using the parametrization $z=frac14+te^{itheta}$, under the limit $thetato0^+$)

$$begin{align}
I^+
&=lim_{thetato0^+}int^infty_0frac{sqrt{1-4(1/4+te^{itheta})}} {(1/4+te^{itheta})^{n+2}}e^{itheta}dt \
&=int^infty_0frac{lim_{thetato0^+}sqrt{4te^{i(theta-pi)}}} {(1/4+t)^{n+2}}dt \
&=-2iint^infty_0frac{sqrt{t}}{(1/4+t)^{n+2}}dt \
end{align}
$$

The integral below the real axis $I^-$ is (using the parametrization $z=frac14+te^{itheta}$, under the limit $thetato2pi^-$)
$$begin{align}
I^-
&=lim_{thetato2pi^-}int_infty^0frac{sqrt{1-4(1/4+te^{itheta})}}{(1/4+te^{itheta})^{n+2}}e^{itheta}dt \
&=int_infty^0frac{lim_{thetato2pi^-}sqrt{4te^{i(theta-pi)}}}{(1/4+t)^{n+2}}dt \
&=-2iint^infty_0frac{sqrt{t}}{(1/4+t)^{n+2}}dt=I^+
end{align}
$$

Therefore, $I_nequiv I^+ + I^-=2I^+=-4idisplaystyle{int^infty_0frac{sqrt{t}}{(1/4+t)^{n+2}}dt}$.

Evaluating $I_n$ directly

By the substitution $u=frac1{1+4t}$,
$$begin{align}
frac1{-4i}I_n
&=int^infty_0frac{sqrt{t}}{(1/4+t)^{n+2}}dt \
&=2^{2n+1}int^1_0sqrt{1-u}~~u^{n-1/2}du \
&=2^{2n+1}Bleft(frac32,n+frac12right)\
&=2^{2n+1}cdotGamma(3/2)cdotGamma(n+1/2)cdotfrac1{Gamma(n+2)}\
&=2^{2n+1}cdotfrac{sqrtpi}{2}cdotfrac{(2n)!}{2^{2n}n!}sqrtpicdotfrac1{n+1}frac1{n!}\
&=picdotfrac1{n+1}binom{2n}{n}\
I_n&=frac{-4pi i}{n+1}binom{2n}{n}\
end{align}
$$

Assembling

$$oint_{C}frac{1-sqrt{1-4z}}{2z^{n+2}}dz=-frac12oint_{C}frac{sqrt{1-4z}}{z^{n+2}}dz=-frac12I_n$$
$$=-frac12cdotfrac{-4pi i}{n+1}binom{2n}{n}=2pi ifrac1{n+1}binom{2n}{n}$$

Thus,
$$color{red}{oint_{C}frac{1-sqrt{1-4z}}{2z}frac{dz}{z^{n+1}}=2pi i~C_n}$$ as expected.

With the functional equation

$$C(z) = 1 + z C(z)^2$$

and the solution

$$C(z) = frac{1-sqrt{1-4z}}{2z}$$

we may choose the branch cut of the square root to be $(1/4, +infty)$
so that we have an analytic function in the neighborhood of the
origin and the Cauchy Coefficient Formula applies.

We have

$$n C_n = [z^{n-1}] C'(z) =
frac{1}{2pi i}
int_{|z|=epsilon} frac{1}{z^n} C'(z) ; dz.$$

With the substitution $C(z) = w$ we get that $z = (w-1)/w^2$
so that

$$n C_n =
frac{1}{2pi i}
int_{|w-1|=gamma} frac{w^{2n}}{(w-1)^n} ; dw
\ = frac{1}{2pi i}
int_{|w-1|=gamma} frac{1}{(w-1)^n}
sum_{q=0}^{2n} {2nchoose q} (w-1)^q ; dw
\ = {2nchoose n-1}.$$

We thus have

$$C_n = frac{1}{n} {2nchoose n-1}
= frac{1}{n+1} {2nchoose n}$$

as required. This was for $nge 1$ and we see that this last formula
also yields the correct value for $n=0.$ Here we have made use of the
fact that $C(z) = 1 + z + cdots$ so the image contour of the circle
$|z|=epsilon$ can be deformed to a circle $|w-1|=gamma$ making a
single turn. The chosen branch cut in fact corresponds to the
principal branch of the logarithm with argument $(-pi, pi].$