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#1 h Apr 18, 2025, 6:18 PM • 1 Y

Y by Svyatoslav

Find $\displaystyle\lim_{x\to\infty}\left|f(x)-\Gamma^{-1}(x)\right|$ where $\Gamma^{-1}(x)$ is the inverse gamma function, and $f^{-1}$ is the inverse of $f(x)=x^x.$

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#2 h Apr 20, 2025, 1:45 PM

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A nice problem! I'm sure there must be an elegant solution.

Meanwhile, we can do the following.
Denoting $f=e^{x_1\ln x_1}$ and $f=\Gamma (x_2)$ , we define the inverse functions $x_1(f)$ and $x_2(f)$ . What is important for us is that $x_1, \,x_2$ are monotonically increasing functions of $f$
Supposing that the limit exists, using the L'Hopital rule and taking the derivatives of inverse functions,
$L=\lim_{f\to\infty}\big(x_1(f)-x_2(f)\big)=\lim_{f\to\infty}\ln\left(\frac{e^{x_1(f)}}{e^{x_2(f)}}\right)=\lim_{f\to\infty}\ln\left(\frac{e^{x_1(f)}\frac{dx_1}{df}}{e^{x_2(f)}\frac{dx_2}{df}}\right)=\lim_{f\to\infty}\ln\left(\frac{e^{x_1(f)}\frac{df}{dx_2}}{e^{x_2(f)}\frac{df}{dx_1}}\right)$ $=\lim_{f\to\infty}\ln\left(\frac{e^{x_1(f)}\Gamma(x_2)\psi(x_2)}{e^{x_2(f)}e^{x_1\ln x_1}(\ln x_1+1)}\right)$ Using the fact that $x_1, x_2$ and $f$ are growing, and that $f=e^{x_1\ln x_1}=\Gamma(x_2)$ , we can get $x_2$ as a function of $x_1$ , and take the limit $x_1\to\infty$
Therefore,
$L=\lim_{f\to\infty}\big(x_1(f)-x_2(f)\big)=\lim_{x_1\to\infty}\ln\left(\frac{e^{x_1-x_2(x_1)}\psi\big(x_2(x_1)\big)}{\ln (x_1)+1}\right)$ Using the Stirling formula, we can get, for $x_1, x_2\gg1,\,\, x_2=x_1+\frac{x_1}{\ln(x_1)}+o\Big(\frac{x_1}{\ln(x_1)}\Big)$
Hence,
$L=\lim_{x_1\to\infty}\ln\left(\frac{e^{-\frac{x_1}{\ln(x_1)}+o\big(\frac{x_1}{\ln(x_1)}\big)}\psi\left(x_1+\frac{x_1}{\ln(x_1)}+o\Big(\frac{x_1}{\ln(x_1)}\Big)\right)}{\ln (x_1)+1}\right)=\lim_{x_1\to\infty}\ln\left(e^{-\frac{x_1}{\ln(x_1)}}\right)=-\infty$ Numeric check (WA) confirms the result

This post has been edited 2 times. Last edited by Svyatoslav, Apr 20, 2025, 3:36 PM
Reason: small corrections

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