How do I show that as $z \to \infty$ that $\int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} dt = O(z^{-1} )$??

Question

How do I show that as $z \to \infty$ we have

$$ \int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} \,dt = O(z^{-1} )? $$

According to Serge Lang, the integral on the left is the error term for Stirling's asymptotic bound for $\log(\Gamma(z))$, but I read elsewhere that the integral is equivalent to the right hand side. I am trying to find a rigorous way to show that the statement above is true, that is, without resorting to looking at a graph on WolframAlpha.

Answer 1

There is a closed form of your integral.

Proposition. Let $z$ be a complex number such that $\Re z>0$. We have

$$ \int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} dt=-\log \Gamma (z+1)+(z+1/2)\log z-z+\frac12\log (2\pi) \tag1 $$

Proof. Let $\Re z>0$. We may write $$ \begin{align} &\int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} dt \\\\&= \int_0^1 \frac{t - \lfloor t \rfloor - 1/2}{z + t} dt+\int_1^\infty \frac{\left\{t\right\} - 1/2}{z + t} dt\\\\ &= \int_0^1 \frac{t- 1/2}{z + t} dt+\int_0^1 \frac{\left\{1/u\right\} - 1/2}{u(zu+1)} du\\\\ &= \int_0^1 \frac{t- 1/2}{z + t} dt+\int_0^1 \frac{\left\{1/u\right\} - 1/2}{u} du-z\int_0^1 \frac{\left\{1/u\right\} - 1/2}{zu+1} du\\\\ &= \underbrace{\int_0^1 \frac{t- 1/2}{z + t} dt+\frac{z}{2}\int_0^1 \frac{1}{zu+1} du}+\underbrace{\int_0^1 \frac{\left\{1/u\right\} - 1/2}{u} du}-\underbrace{\int_0^1 \frac{\left\{1/u\right\}}{u+1/z} du} \\\\ &= \qquad \qquad \qquad \quad I_1\qquad \qquad +\qquad \qquad \quad I_2\qquad-\qquad \qquad I_3 \end{align} $$ $I_1$ is elementary: $$ I_1=1+\left(z+1/2\right) \log z -z \log (1+z) $$

The integral $I_2$ has already been evaluated (see for example here): $$ I_2=-1+\frac12 \log (2\pi) $$ To evaluate $I_3$, we have $$ \begin{align} I_3=\int_0^1 \frac{\left\{1/u\right\}}{u+1/z} du &= \sum_{k=1}^{\infty} \displaystyle \int_{1/(k+1)}^{1/k} \frac{\left\{1/u\right\}}{u+1/z} du \\\\ & = z\sum_{k=1}^{\infty} \int_{k}^{k+1} \frac{\left\{v\right\}}{v(v+z)}dv \\\\ & = z\int_{0}^{1}\sum_{k=1}^{\infty} \frac{x}{(x+k)(x+k+z)}dx \\\\ & = \int_{0}^{1} x \left(\psi(x+1+z)-\psi(x+1)\right) dx \\\\ &=\log \Gamma (z+2)-\int_0^1 \left( \log \Gamma (x+1+z)-\log \Gamma (x+1)\right) dx\\\\ &=\log \Gamma (z+1)-z\log(1+z)+z \end{align} $$ where $\psi:=\Gamma'/\Gamma$ is the digamma function and where we have used Raabe's formula. The previous results give $(1)$.

Remark. If we combine the closed form $(1)$ with Jack D'Aurizio's answer we get a proof for the Stirling formula.

Answer 2

$f(t)=t-\lfloor t\rfloor-\frac{1}{2}$ is a $1$-periodic function with mean zero on every period, and $\frac{1}{z+t}$ is a function slowly decreasing towards zero. Then, just prove that for any $u\in\mathbb{R}^+$ $$ F(u) = \int_{0}^{u}f(t)\,dt \in\left[-\frac{1}{8},0\right] $$ and use integration by parts, or the second mean value theorem for integration.

Answer 3

It's also possible to obtain a complete asymptotic expansion for this integral (and hence obtain the asymptotic expansion for $\log\Gamma(z+1)$).

The quantity $t-\lfloor t \rfloor - 1/2$ is a sawtooth wave and has the known Fourier series

$$ t - \lfloor t \rfloor - \frac{1}{2} = - \frac{1}{\pi}\sum_{k=1}^{\infty} \frac{\sin(2\pi k t)}{k}. $$

We can substitute this series into the integral and exchange the order of integration and summation; indeed, if we set $f(t) = t-\lfloor t \rfloor - 1/2$ and

$$ f_n(t) = - \frac{1}{\pi}\sum_{k=1}^{n} \frac{\sin(2\pi k t)}{k} $$

then, using integration by parts,

$$ \begin{align} \left|\int_0^\infty [f(t) - f_n(t)] (z+t)^{-1}\,dt \right| &= \left| \int_0^\infty (z+t)^{-2} \int_0^t [f(s)-f_n(s)]\,ds\,dt \right| \\ &= \left| \int_0^\infty (z+t)^{-2} \int_0^{\{t\}} [f(s)-f_n(s)]\,ds\,dt \right| \\ &\leq \int_0^\infty (z+t)^{-2} \int_0^1 |f(s)-f_n(s)|\,ds\,dt \\ &= z^{-1} \|f-f_n\|_{L^1([0,1])} \\ &\to 0 \end{align} $$

as $n \to \infty$, where $\{t\} = t - \lfloor t \rfloor$. Thus

$$ \begin{align} \int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} \,dt &= -\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \int_0^\infty \frac{\sin(2\pi k t)}{z+t}\,dt \\ &= -\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} \int_0^\infty \left(1+t^2\right)^{-1} e^{-2\pi k z t}\,dt, \tag{1} \end{align} $$

where the second equality follows from [1] (see also [2, pp. 187-188]). For each $k$ we have

$$ \int_0^\infty \left(1+t^2\right)^{-1} e^{-2\pi k z t}\,dt \sim \sum_{n=0}^{\infty} \frac{(-1)^n (2n)!}{(2\pi k z)^{2n+1}} $$

by Watson's lemma. This asymptotic series can be interchanged with the convergent series in $(1)$ and remain asymptotic, thus

$$ \begin{align} \int_0^\infty \frac{t - \lfloor t \rfloor - 1/2}{z + t} \,dt &\sim -\frac{1}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^n (2n)!}{(2\pi z)^{2n+1}} \sum_{k=1}^{\infty} \frac{1}{k^{2n+2}} \\ &= - \frac{1}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^n (2n)! \zeta(2n+2)}{(2\pi z)^{2n+1}} \\ &= -\sum_{n=0}^{\infty} \frac{B_{2n+2}}{(2n+1)(2n+2) z^{2n+1}} \\ &= - \frac{1}{12 z} + \frac{1}{360 z^3} - \frac{1}{1260 z^5} + \frac{1}{1680 z^7} - \cdots. \end{align} $$

This agrees with the known asymptotic for $\Gamma(z+1)$ in light of the result in Oliver's answer.

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[1] DLMF Eqn. 6.7.13

[2] N. Temme, Special Functions: An Introduction to the Classical Functions of Mathematical Physics, Wiley, 1996.