Approximation of $\ln(x+1)$ with $\Psi$ function

Question

I found the following approximation for the function $$f=\ln(x+1)$$ $$f\simeq\Psi\left(x+\dfrac{3}{2}\right)-2+\gamma+\ln(2)$$ where $\Psi(x)$ is the 'Digamma' function: $\Psi(x)=\dfrac{\dfrac{d}{dx}\Gamma(x)}{\Gamma(x)}$ and $\gamma$ is the Euler - Mascheroni constant. The following inequality holds: $$\left|\ln(x+1)-\Psi\left(x+\dfrac{3}{2}\right)-2+\gamma+\ln(2)\right|\lt 0.036$$ Is this a known result? Thanks

Answer 1

I'm not sure why there is a $(-2+\gamma+\ln 2)$, because $\left|\ln(x+1) - \Psi(x+\frac32)\right| < 0.0365$ itself is already satisfied.

The approximation comes from the series expansion of $\exp\left(\Psi(x+\frac12)\right)$, which states, for $x > 1$,

\begin{align} \Psi\left(x + \frac12\right) = \ln\left( x + \frac{1}{4!\cdot x} - \frac{37}{8\cdot 6!\cdot x^3} + o\left(\frac1{x^5}\right) \right) \end{align} and thus $\Psi(x+\frac32)$ is approximately $\ln(x+1)$. The largest difference happens at $x=0$ which is $\Psi(\frac32) = 2 - \ln 4 - \gamma \approx 0.03648997$.

Answer 2

I doubt your approximation. First I guess you missed a factor $2$ for $\ln 2$, the approximation should be $$f(x) =\ln(x+1)-\Psi(x+3/2)-2+\gamma+2\ln 2 \cdot$$ If you omit the $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ \ln 2 = -4 + 2\gamma + 3\ln 2 \approx -0.766$$ while with the factor $2$ you have $$f(0)= 0 -(2-\gamma-2\ln2) -2+\gamma+ 2\ln 2 = -4 + 2\gamma + 4\ln 2 \approx -0.0723$$ which is much smaller but still violates your given bound.

PS: A Taylor series at $x=0$ computed with Maple is $$\ln(x+1)-\Psi(x+\tfrac{3}{2}) = (-2+\gamma+2\ln2)+(5-\tfrac{1}{2}\pi^2)x+(-\tfrac{17}{2}+7\zeta(3))x^2 +O(x^3)$$