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I have taken the limit of both sides of an equation for x going toward infinity. There is a digamma (psi(x)) function on the RHS, and the limit of the term is supposed to be (at least close to) 0. This has to be in order for the term to vanish from the RHS.

My professor said that indeed, the digamma function is supposed to be around 0 for large arguments and he wants me to justify that with the digamma function's integral form. However, when I research digamma online, it appears to slowly diverge or converge to a number larger than 0. Can someone clarify this?

Judy
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  • How się you research it? Looking at graphs can be misleading as they only show portion of domain and range. – Blazej Jul 12 '15 at 14:35
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    @Blazej: men can be even more misleading than graphs. – Jack D'Aurizio Jul 12 '15 at 14:46
  • That's true. According to MATLAB and Wolfram Alpha, the limit of Psi approaching infinity is infinity. The minimum appears to occur between 1 and 2, not at large arguments. – Judy Jul 12 '15 at 15:02

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We have: $$ \Gamma(t+1) = e^{-\gamma t}\prod_{n\geq 1}\left(1+\frac{t}{n}\right)^{-1}e^{\frac{t}{n}} \tag{1}$$ hence: $$ \log\Gamma(t+1) = -\gamma t+\sum_{n\geq 1}\left(\frac{t}{n}-\log\left(1+\frac{t}{n}\right)\right)\tag{2} $$ and: $$ \psi(t+1) = -\gamma + \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+t}\right) \tag{3} $$ so: $$ \psi'(t+1) = \sum_{n\geq 1}\frac{1}{(t+n)^2} \approx \frac{1}{t+1}\tag{4} $$ gives that $\psi(t+1)$ behaves like $\color{red}{\log(t+1)}$ for large real values of $t$.

Your professor was simply wrong, or he just said digamma in place of trigamma.

Jack D'Aurizio
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