<s>[INST] Bounding an expression involving digamma function Let $\psi$ be the digamma function. I have a conjecture that
$$\frac ax > \log(x) - \psi(x)$$
holds for all $x > 0$ if (and only if) $a \ge 1$. I do not know how to prove it. Please help.
 [/INST] Let
$$f(x) = \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}.$$
Then for $x > 0$, we have
$$f'(x) = \frac{1}{x^2} - \frac{1}{4\sinh^2(x/2)} = \frac{1}{x^2}\left(1 - \left(\frac{x/2}{\sinh (x/2)} \right)^2 \right) > 0. $$
This shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\lim_{x\to\infty} f(x) = \frac{1}{2}$. Thus we find that $0 < f(x) < \frac{1}{2}$ for $0 < x < \infty$ and hence the Laplace transform $\mathcal{L}f(s) = \int_{0}^{\infty} f(x) \, e^{-sx} \; dx$ of $f(x)$ satisfies
$$ 0 < \mathcal{L}f(s) < \int_{0}^{\infty} \frac{1}{2} \, e^{-sx} \; dx = \frac{1}{2s}. $$
But we also have
$$ \begin{align*}
\mathcal{L}f(s)
&= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x} \right) e^{-sx} \; dx\\
&= \left[ \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \right]_{0}^{\infty} \\
&\qquad + s \int_{0}^{\infty} \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \; dx \\
&= s \int_{0}^{\infty} \left( \frac{x}{2} - \log (sx) + \log s +  \log \left(1 - e^{-x} \right) \right) e^{-sx} \; dx \\
&= \frac{1}{2s} + \gamma + \log s - s \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(n+s)x}}{n} \; dx \\
&= \frac{1}{2s} + \gamma + \log s - \sum_{n=1}^{\infty} \frac{s}{n(n+s)} \\
&= \frac{1}{2s} + \log s - \psi_0(1+s),
\end{align*}$$
where we have used the fact that $\int_{0}^{\infty} e^{-x} \log x \; dx = -\gamma$ and
$$ \psi_0(1+x) = -\gamma + \sum_{n=1}^{\infty} \frac{x}{n(n+x)}. $$
Finally, since $\psi_0(1+s) = \frac{1}{s} + \psi_0 (s)$, we have
$$ \frac{1}{2s} < \log s - \psi_0(s) < \frac{1}{s}. $$
Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$.
Now we prove that $a = 1$ is the sharp bound. We are to calculate the limit
$$\alpha = \lim_{x\to 0} x(\log x - \psi_0(x)).$$
But by our previous calculations,
$$\begin{align*}\alpha
&= \lim_{s\to 0} s(\log s - \psi_0(s)) \\
&= \lim_{s\to 0} s\left( \frac{1}{2s} + \mathcal{L}f(s) \right) \\
&= \frac{1}{2} + \lim_{s\to 0} s \int_{0}^{\infty} f(x) \, e^{-sx} \; dx \\
&= \frac{1}{2} + \lim_{s\to 0} \int_{0}^{\infty} f(u/s) \, e^{-u} \; du \qquad (u = sx) \\
&= \frac{1}{2} + \int_{0}^{\infty} \lim_{s\to 0} f(u/s) \, e^{-u} \; du \\
&= \frac{1}{2} + \int_{0}^{\infty} \frac{1}{2} \, e^{-u} \; du \\
&= 1.
\end{align*}$$
Thus if $\frac{a}{x} > \log x - \psi_0 (x)$ is true for all $x > 0$, then we must have $a \geq \alpha = 1$, as desired.
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