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I was attempting to go through the addendum of Chapter 21 of Donald Knuth's Selected Papers on Analysis of Algorithms. In a lengthy derivation, Knuth asymptotically expands the following integral:

$$ \frac{1}{2 \pi i} \int_{c-i \infty}^{c + i \infty } \frac{\Gamma(s-5/2)\Gamma(4-s) m^s \zeta(s)}{\Gamma(3/2)} ds \sim \frac{\Gamma(-3/2)\Gamma(3)}{\Gamma(3/2)} m + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{\Gamma(3/2+n)}{\Gamma(3/2)} m^{5/2-n} \zeta(5/2-n) $$

where $\frac{5}{2} < c < 4$. Knuth only states that the asymptotic series is obtained "by decreasing c as far as we like and adding up the residues of the poles of the integrand" - however I am having difficulty tracing the logic in how he did so.

Considering the more general integral

$$ \frac{1}{2 \pi i} \int_{c-i \infty }^{c + i \infty } \frac{\Gamma(s-\alpha) \Gamma(\alpha+\beta-s)m^s\zeta(s)}{\Gamma(\beta)} $$

where $\alpha = A+\frac{1}{2}, \beta = B-\frac{1}{2}$ and both $A,B \in \mathbb{N}$. The domain of integration is taken as the limit of the line integral from $c-iR$ to $c+iR$ as $R \to \infty$.

Note first that the integrand has poles whenever $s = k, k + \frac{1}{2}, k \in \mathbb{Z}$. This function stems from an inverse Mellin transform with a fundamental strip of $\{ z \in \mathbb{C} : \alpha < \Re(z) < \alpha + \beta \}$, so the integrand is only defined on this strip but can be analytically continued to the whole complex plane.

Consider the contour $C = L \cup S$ where $L$ is the line segment going from $s=c-iR$ to $s=c+iR$ and $S$ be the semicircular contour going from $s=c+iR$ to $s=c-iR$ extending to the negative real axis (i.e. parametrized by $S(t) = c + Re^{i \theta}$ for $\theta \in (\frac{\pi}{2}, \frac{3\pi}{2})$). By the Residue Theorem, we have that $$\int_C f(z) dz = 2 \pi i \sum_{z \in [c, c-R] \cap \left( \mathbb{Z} \cup \mathbb{Z} + \frac{1}{2} \right)} \text{Res}\left(f; z\right)$$

where the sum is taken over the residues inside the contour (i.e. all integers and half integers in the interval $[c, c-R]$). We also have:

$$ 2 \pi i \sum \text{Res} f(z) = \int_C f(z) dz = \int_{c-iR}^{c+iR} f(z) dz + \int_S f(z) dz $$

If we have that $\int_S f(z) dz \to 0$ as $R \to \infty$, then Knuth's claim follows - however this is where I am stuck. I have tried to manipulate the inner gamma function terms with Stirling's approximation and have gotten nowhere, so some assistance here would be appreciated.

Note: I am aware that there are some theorems involving Mellin transforms that I think are mentioned in the original paper that Knuth cites - however, I am trying to avoid them as it is made to seem as though this is a straightforward computation. Any assistance would be appreciated!

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Let $$f(s):=\frac{\Gamma(s-\alpha) \Gamma(\alpha+\beta-s)\zeta(s)}{\Gamma(\beta)},\ \ \ I:=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} m^s f(s) \, ds.$$ You can derive an asymptotic series for $I$ if you integrate over the boundary of a rectangle which has a right edge at $\hbox{Re } z=c$, a left edge at $\hbox{Re } z = K < 0$, and top and bottom edges, $\cal T$ and $\cal B$, at $\hbox{Im } z = \pm R$. To do this, you can use the uniform bounds $$ |\Gamma(x+iy)|\sim \sqrt{2 \pi}|y|^{x-(1/2)} \exp (-\pi |y|/2), \qquad \hbox{$|y|\to\infty$, $x$ in any bounded interval,} $$ from [1] (derivable from Stirling's approximation. Added: To derive this, you can start with Stirling's approximation in the form $$ \log \Gamma(z)=(z - \frac 12)\log z - z + \log \sqrt{2 \pi} + O(\frac{1}{|z|}), \qquad |z|\to\infty, $$ which is uniformly valid in $\{z\mid |\arg z|\le \pi-\delta\}$, for any fixed $\delta>0$.

Set $z:=x+iy=x\pm i|y|$, let $x$ be bounded and $|y|\to\infty$, and set $r+i\theta:=\log z$. Then, you get \begin{eqnarray*} r&=&\log \sqrt{x^2+y^2}=\log |y| + O(|y|^{-2}),\\ \theta&=&\pm \cos^{-1} \frac{x}{\sqrt{x^2+y^2}}\\ &=&\pm(\frac\pi 2 - \sin^{-1} \frac{x}{\sqrt{x^2+y^2}})\\ &=&\pm(\frac\pi 2 - \sin^{-1} \frac{x}{|y|(1+O(|y|^{-2}))})\\ &=&\pm(\frac\pi 2 - \frac{x}{|y|}+O(|y|^{-3})), \end{eqnarray*} so \begin{eqnarray*} \log |\Gamma(z)|&=&\hbox{Re} \log \Gamma(z)\\ &=& (x - \frac 12)r-(\pm|y|)\theta-x +\log\sqrt{2\pi}+O(\frac{1}{|y|})\\ &=& (x - \frac 12)\log |y| - \frac{\pi}{2}|y|+\log\sqrt{2\pi}+o(1), \end{eqnarray*} which is what you want), and, for any fixed right half-plane $\{z\mid \hbox{Re } z \ge \sigma_0\}$, $$ \zeta(z)=O(|\hbox{Im } z|^A), \qquad \qquad \hbox{$\hbox{Im } z\to\infty$, for some $A$}, $$ from [2]. If you fix $K$ and $m$, let $R$ become large, and apply these bounds to $m^s f(s)$ on $\cal T$ and $\cal B$, you can see that the integral of $m^s f(s)$ over $\cal T$ and $\cal B$ vanishes exponentially (it is $O(R^B \exp(-\pi R))$ for some $B$). This implies that $I$ equals the sum of the residues of the poles of $m^s f(s)$ inside the rectangle plus $m^K J$, where $$J:=\frac{1}{2\pi i} \int_{K-i\infty}^{K+i\infty} m^{i \ \small \rm Im\ \it s} f(s)\, ds. $$ If you pick $K$ so as to avoid the poles of $f$ and use the same bounds as for $\cal T$ and $\cal B$, this is enough to show that $$ \int_{K-i\infty}^{K+i\infty} |f(s)|\, ds<\infty $$ so $m^K J$ is $O(m^K)$ as $m\to\infty$, giving the remainder term in the asymptotic series.

You should not expect to be able to push $K$ to $-\infty$ for a fixed $m$ since, using the reflection formulae ([3], [4]) $$ \Gamma(s) = \frac{\pi}{\sin{(\pi s)} \Gamma(1-s)},\ \ \ \ \ \zeta(s)=\zeta(1-s) 2^s \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) $$ you get $$ m^s f(s)=\frac{m^s \Gamma(\alpha+\beta-s) \zeta(1-s) 2^s \pi^{s} \sin (\pi s/2) \Gamma(1-s)}{\sin(\pi(s-\alpha)) \Gamma(1+\alpha-s) \Gamma(\beta)}. $$ Using Stirling's approximation and $\zeta(1-s)=\Theta(1)$ for $\hbox{Re }s \le -\epsilon$, for any fixed $\epsilon>0$, you can then see that $m^s f(s)$ blows up as you let $s$ approach $-\infty$ along the negative real axis.

1: NIST Digital Library of Mathematical Functions, version 1.1.11, (5.11.9).

2: $\S$5.1, The Theory of the Riemann Zeta-Function, E. C. Titchmarsh, D. R. Heath-Brown, Oxford : Clarendon Press, 2nd ed., 1986.

3: NIST Digital Library of Mathematical Functions, version 1.1.11, (5.5.3).

4: NIST Digital Library of Mathematical Functions, version 1.1.11, (25.4.2).

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  • $\begingroup$ Thanks! I think the main thing I was struggling with was the bounds on the gamma function - out of curiosity how would you derive that from just Stirling's formula? $\endgroup$ Oct 12, 2023 at 5:54
  • $\begingroup$ I added this to the answer. $\endgroup$ Oct 13, 2023 at 15:32

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