Limit of Buchstab function

Question

Let $\omega$ be the Buchstab function defined by $\omega(u)=\frac{1}{u}$ for $1\leqslant u\leqslant 2$ and $$ u\omega(u)=1+\int_1^{u-1}\omega(t)dt $$ How can one prove that $\lim\limits_{u\rightarrow +\infty}\omega(u)=e^{-\gamma}$ ? I can't find any paper where it is properly proved.

Answer 1

See section 4 (middle of page 9 to page 10) of the link here. The author creates an auxiliary function to generate an "adjoint equation" to the differential equation used to define the Buchstab function: $\frac{\partial}{\partial u}(uw(u))= w(u-1)$. This link was admittedly hard to find.

I also tried for a while to find an easier proof but did not succeed.

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Here's a proof for the last point: $\lim_{x \to +\infty} \left[\int_0^x \frac{e^{-t}-1}{t}dt+\ln x\right] = -\gamma$.

The term in brackets is $\int_0^1 \frac{e^{-t}-1}{t}dt+\int_1^x \frac{e^{-t}-1}{t}+\frac{1}{t}dt = \int_0^1 \frac{e^{-t}-1}{t}dt+\int_1^x \frac{e^{-t}}{t}dt$. Letting $x \to \infty$ gives $\int_0^1 \frac{e^{-t}-1}{t}dt+\int_1^\infty \frac{e^{-t}}{t}dt$. Now, integration by parts with $u = e^{-t}-1, dv = \frac{1}{t}dt$ for the first integral and integration by parts with $u = e^{-t}, dv = \frac{1}{t}dt$ for the second integral gives $\int_0^1 e^{-t}\ln t dt + \int_1^\infty e^{-t}\ln t dt = \int_0^\infty e^{-t}\ln t dt$, which is $-\gamma$.

Answer 2

Let $\omega:\left[1, \infty\right) \rightarrow \mathbb{R}$ be the Buchstab function. By definition we have that $$ \begin{align} \omega\!\left(x\right) &= \dfrac{1}{x},&& x \in \left[1, 2\right] \label{init cond} \tag{0} \\ \dfrac{\mathrm{d}}{\mathrm{d}x}\Big(x\omega\!\left(x\right)\!\Big) &= \omega\!\left(x - 1\right),&& x \geqslant 2 \label{diff eq} \tag{1} \end{align} $$ , where $\dot{\omega}\!\left(2\right)$ understood in sense of $\dot{\omega}\!\left(2+0\right)$.

We extend domain of $\omega$ to $\mathbb{R}$ (without any change in notations), letting $\omega\!\left(x\right) = 0$ for $x < 1$. With this extension we have that \eqref{diff eq} holds also for $x \in \mathbb{R}$ assuming right-sided derivatives at points of non differentiability.

Claim 1:

$$\omega\!\left(x\right) \geqslant 0 \text{ for } x \geqslant 1.$$

Integrating \eqref{diff eq} from $2$ to $x$ leads us to \begin{equation} x\omega\!\left(x\right) = 1 + \int\limits_{1}^{x-1} \!\omega\!\left(t\right)\,\mathrm{d}t \label{int eq} \tag{2} \end{equation} From equation \eqref{int eq} we see, that if $\omega \geqslant 0$ on the interval $\left[1, v\right]$, $v \geqslant 2$ then $\omega \geqslant 0$ also on the $\left[1, v + 1\right]$. It remains only to note that $\omega\!\left(t\right) \equiv x^{-1} \geqslant 0$ on $\left[1, 2\right]$.

Claim 2:

$$\omega\!\left(x\right) \leqslant 1 \text{ for } x \geqslant 1.$$

By initial condition \eqref{init cond} we have that $\omega \leqslant 1$ on $\left[1, 2\right]$. Using that $\omega$ is nonnegative we bound \eqref{int eq}: \begin{align} x\omega\!\left(x\right) &= 1 + \int\limits_{1}^{x-1} \!\omega\!\left(t\right)\,\mathrm{d}t \\ &\leqslant 1 + \int\limits_{1}^{x} \omega\!\left(t\right)\mathrm{d}t = 1 + \int\limits_{1}^{x} \dfrac{t\omega\!\left(t\right)}{t}\,\mathrm{d}t \end{align} where $x \geqslant 2$. Applying Grönwall's lemma to function $x\omega\!\left(x\right)$ we have that for $x \geqslant 2$ \begin{align} x\omega\!\left(x\right) &\leqslant \exp\left(\int\limits_{1}^{x} t^{-1}\,\mathrm{d}t\right) \\ \omega\!\left(x\right) &\leqslant 1 \end{align}

Clearly inequality of Claim 2 holds true also for $x < 1$. Integrating it from $x-1$ to $x$, $x \in \mathbb{R}$ we have that \begin{equation} 1 - \int\limits_{x-1}^{x} \!\omega\!\left(t\right)\,\mathrm{d}t \geqslant 0 \label{geq 0} \tag{3} \end{equation}

Claim 3:

$$\exists \!\lim\limits_{x \,\rightarrow\, \infty} \!\omega\!\left(x\right).$$

Rewriting \eqref{int eq} as below \begin{equation} x\omega\!\left(x\right) - \int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t = 1 - \int\limits_{x-1}^{x} \!\omega\!\left(t\right)\,\mathrm{d}t \label{int eq2} \tag{2'} \end{equation} we have \begin{align} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{x}\int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t\right) &= \dfrac{1}{x^{2}}\left(x\omega\!\left(x\right) - \int\limits_{1}^{x}\omega\!\left(t\right)\,\mathrm{d}t \right) \\ &= \dfrac{1}{x^{2}}\left(1 - \int\limits_{x-1}^{x} \!\omega\!\left(t\right)\,\mathrm{d}t\right) \\ &\geqslant 0 \end{align} Thus function $\dfrac{1}{x}\displaystyle\int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t$ is nondecreasing for $x \geqslant 1$ and bounded \begin{equation} \dfrac{1}{x}\int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t \leqslant \dfrac{1}{x}\int\limits_{1}^{x} 1\,\mathrm{d}t \leqslant 1 \end{equation} Hence it has limit as $x \rightarrow \infty$. Dividing \eqref{int eq2} by $x$ and taking limit from both sides \begin{gather} \lim\limits_{x \,\rightarrow\, \infty} \left(\omega\!\left(x\right) - \dfrac{1}{x}\int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t\right) = 0 = \lim\limits_{x \,\rightarrow\, \infty} \!\omega\!\left(x\right) - \lim\limits_{x \,\rightarrow\, \infty} \!\dfrac{1}{x}\int\limits_{1}^{x} \omega\!\left(t\right)\,\mathrm{d}t \end{gather} and so $\lim\limits_{x \,\rightarrow\, \infty} \!\omega\!\left(x\right)$ exists.

Let \begin{equation} \omega_{\infty} = \lim\limits_{x \,\rightarrow\, \infty} \!\omega\!\left(x\right) \end{equation}

Claim 4:

$$\Big|\dot{\omega}\!\left(x\right)\!\Big| \leqslant \dfrac{4}{\Gamma\!\left(x+1\right)} \text{ for } x \geqslant 2.$$

First of all, from \eqref{diff eq} we have that \begin{equation} \Big|\dot{\omega}\!\left(x\right)\!\Big| = \dfrac{1}{x}\Big|\omega\!\left(x\right) - \omega\!\left(x - 1\right)\!\Big| \leqslant \dfrac{2}{x} \end{equation} But from another side, for $x \geq 2$ using Mean value theorem \begin{align} \Big|\dot{\omega}\!\left(x\right)\!\Big| &= \dfrac{1}{x}\Big|\dot{\omega}\!\left(x_{1}\right)\!\Big| \\ &\leqslant \dfrac{1}{x}\sup\limits_{\left[x-1, x\right]} \Big|\dot{\omega}\!\left(x_{1}\right)\!\Big| \\ &\leqslant \dfrac{1}{x}\sup\limits_{\left[x-1, x\right]}\left(\dfrac{1}{x_{1}}\sup\limits_{\left[x_{1}-1, x_{1}\right]}\Big|\dot{\omega}\!\left(x_{2}\right)\!\Big|\right) \\ &\leqslant \dfrac{1}{x\left(x-1\right)}\sup\limits_{\left[x-1, x\right]}\left(\sup\limits_{\left[x_{1}-1, x_{1}\right]}\Big|\dot{\omega}\!\left(x_{2}\right)\!\Big|\right) \\ &\leqslant \dfrac{1}{x\left(x-1\right)}\sup\limits_{\left[x-2, x\right]}\Big|\dot{\omega}\!\left(x_{1}\right)\!\Big| \\ &\leqslant \dots \\ &\leqslant \dfrac{1}{x\left(x-1\right)\dots\left(\left\{x\right\} + 3\right)}\sup\limits_{\left[\left\{x\right\}+2, x\right]}\Big|\dot{\omega}\!\left(x_{1}\right)\!\Big| \\ &\leqslant \dfrac{2}{x\left(x-1\right)\dots\left(\left\{x\right\} + 2\right)} \\ &\leqslant \dfrac{4}{x\left(x-1\right)\dots \left(\left\{x\right\} + 2\right)\left(\left\{x\right\} + 1\right)\Gamma\!\left(\left\{x\right\} + 1\right)} \\ &\leqslant \dfrac{4}{\Gamma\!\left(x+1\right)} \end{align}

Claim 5:

$$ \mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) = \int\limits_{0}^{\infty} \omega\left(x\right)e^{-\lambda x}\,\mathrm{d}x = \left(\exp \circ E_{1}\right)\!\left(\lambda\right) - 1 $$ , where $\lambda > 0$ and $E_{1}$ is En-Function with $n=1$.

Note that Laplace transform $\mathcal{L}\!\left(\omega\right)\!\left(\lambda\right)$ converges to zero as $\lambda \rightarrow \infty$. It follows from double inequality $$ 0 \leqslant \int\limits_{0}^{\infty} \omega\left(x\right)e^{-\lambda x}\,\mathrm{d}x \leqslant \int\limits_{0}^{\infty} e^{-\lambda x}\,\mathrm{d}x = \dfrac{1}{\lambda} $$

Now integrate \eqref{diff eq} from $0$ to $\infty$ \begin{align} \int\limits_{0}^{\infty} \dfrac{\mathrm{d}}{\mathrm{d}x}\Big(x\omega\!\left(x\right)\!\Big)e^{-\lambda x}\,\mathrm{d}x &= \int\limits_{1}^{\infty} \dfrac{\mathrm{d}} {\mathrm{d}x}\Big(x\omega\!\left(x\right)\!\Big)e^{-\lambda x}\,\mathrm{d}x \\ &= x\omega\!\left(x\right)e^{-\lambda x}\Bigg|_{1}^{\infty} + \lambda\int\limits_{1}^{\infty} x\omega\!\left(x\right)e^{-\lambda x}\,\mathrm{d}x \\ &= -e^{-\lambda} + \lambda\int\limits_{0}^{\infty} x\omega\!\left(x\right)e^{-\lambda x}\,\mathrm{d}x \\ &= -e^{-\lambda} - \lambda\dot{\mathcal{L}}\!\left(\omega\right)\!\left(\lambda\right) \\ &= \int\limits_{0}^{\infty} \omega\!\left(x - 1\right)e^{-\lambda x}\,\mathrm{d}x \\ &= \int\limits_{1}^{\infty} \omega\!\left(x - 1\right)e^{-\lambda x}\,\mathrm{d}x \\ &= e^{-\lambda}\mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) \\ \end{align} So $\mathcal{L}$ satisfy next linear differential equation: \begin{equation} -\lambda\dot{\mathcal{L}}\!\left(\omega\right)\!\left(\lambda\right) = e^{-\lambda}\Big(\mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) + 1\Big) \end{equation} Rewriting it and integrating from $\lambda$ to $\infty$: \begin{gather} \int\limits_{\lambda}^{\infty} \dfrac{\dot{\mathcal{L}}\!\left(\omega\right)\!\left(t\right)}{\mathcal{L}\!\left(\omega\right)\!\left(t\right) + 1}\,\mathrm{d}t = -\int\limits_{\lambda}^{\infty} \dfrac{e^{-t}}{t}\,\mathrm{d}t \\[1 ex] \ln\!\left(\mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) + 1\right) = E_{1}\!\left(\lambda\right) \\[1 ex] \mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) = \exp\!\left(E_{1}\!\left(\lambda\right)\right) - 1 \end{gather}

Claim 6:

$$\omega_{\infty} = e^{-\gamma}.$$

Let us look over next integral \begin{align} \mathfrak{I}\!\left(\lambda\right) = \int\limits_{1}^{\infty} \Big(\omega\left(x\right) - \omega_{\infty}\!\Big)e^{-\lambda x}\,\mathrm{d}x \end{align} , where $\lambda \geqslant 0$. It converges uniformly on whole parameter domain. Indeed, via Weierstrass test \begin{align} \int\limits_{1}^{\infty} \Big|\omega\left(x\right) - \omega_{\infty}\!\Big|e^{-\lambda x}\,\mathrm{d}x &\leqslant \int\limits_{1}^{\infty} \Big|\omega\left(x\right) - \omega_{\infty}\!\Big|\,\mathrm{d}x \\ &\leqslant \int\limits_{1}^{2} \Big|\omega\left(x\right) - \omega_{\infty}\!\Big|\,\mathrm{d}x + \int\limits_{2}^{\infty} \Big|\omega\left(x\right) - \omega_{\infty}\!\Big|\,\mathrm{d}x \\ &\leqslant 2 + \int\limits_{2}^{\infty}\!\left(\,\int\limits_{x}^{\infty} \Big|\dot{\omega}\left(t\right)\!\Big|\,\mathrm{d}t\right)\,\mathrm{d}x \\ &\leqslant 2 + 4\int\limits_{2}^{\infty}\!\left(\,\int\limits_{x}^{\infty} \dfrac{1}{\Gamma\!\left(1+t\right)}\,\mathrm{d}t\right)\,\mathrm{d}x \\ &< \infty \end{align} So $\mathfrak{I}\!\left(0\right)$ exists and finite. As integrand of $\mathfrak{I}$ is continuous as a function of two arguments, then $\mathfrak{I}$ continuous itself and \begin{align} \mathfrak{I}\!\left(0\right) &= \lim\limits_{\lambda \rightarrow 0^{+}} \mathfrak{I}\!\left(\lambda\right) \\ &= \lim\limits_{\lambda \rightarrow 0^{+}}\left(\mathcal{L}\!\left(\omega\right)\!\left(\lambda\right) - \omega_{\infty}\dfrac{e^{-\lambda}}{\lambda}\right) \\ &= \lim\limits_{\lambda \rightarrow 0^{+}} \dfrac{1}{\lambda}\Big(\lambda \left(\exp \circ E_{1}\right)\!\left(\lambda\right) - \omega_{\infty}e^{-\lambda}\Big) - 1 \end{align} So we must have $\lambda \left(\exp \circ E_{1}\right)\!\left(\lambda\right) \rightarrow \omega_{\infty}$ \begin{align} \omega_{\infty} &= \lim\limits_{\lambda \rightarrow 0^{+}} \lambda \left(\exp \circ E_{1}\right)\!\left(\lambda\right) \\ &= \lim\limits_{\lambda \rightarrow 0^{+}} \exp\!\left(\ln \lambda + \int\limits_{\lambda}^{\infty} \dfrac{e^{-t}}{t}\,\mathrm{d}t\right) \\ &= \lim\limits_{\lambda \rightarrow 0^{+}} \exp\!\left(\ln \lambda - e^{-\lambda}\ln \lambda + \int\limits_{\lambda}^{\infty} e^{-t}\ln t\,\mathrm{d}t\right) \\ &= \exp\!\left(\int\limits_{0}^{\infty} e^{-t}\ln t\,\mathrm{d}t\right) \\ &= e^{-\gamma} \end{align}