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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.

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2 Answers 2

up vote 3 down vote accepted

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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This is a cool technique. (There's a typo after you switch summation and integral.) –  Tunococ Aug 16 '12 at 18:25
    
@Tunococ, Thanks. Also I fixed the typo. Even I cannot believe how I made such a mistake! :) –  sos440 Aug 17 '12 at 3:36

Recently I happened to find out that Qi et al (J. Math. Anal. Appl. 310 (2005) 303–308) proved the following inequality, among others, for digamma function $\psi(x)$ when $x>0$:

$$\frac{1}{2x} < \log x - \psi(x) < \frac{1}{2x} + \frac{1}{12x^2}$$

This inequality is sharper than the one given by @sos440 above when x>1/6.

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