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Let $\psi(x)$ be the digamma function. Is the function which takes $\psi(x)-\log x$ for $x>0$ strictly increasing, and how could one show this if it is the case (link etc.)?

StackExchangeMMH
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 24 '20 at 08:16

1 Answers1

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Are you allowed to use $\psi(z) = -\gamma + \sum_{n=0}^\infty \left(\frac{1}{n + 1} - \frac{1}{n + z}\right)$ ? If so $$\psi'(x)-\frac1x = \sum_{n\ge 0} \frac1{(x+n)^2}-\int_{x+n}^{x+n+1} \frac1{t^2}dt$$

reuns
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  • Yes, and thanks a lot for the answer and the editting :) Im not quite sure how to obtain the right hand side of $$\psi'(x)-\frac1x = \sum_{n\ge 0} \frac1{(x+n)^2}-\int_{x+n}^{x+n+1} \frac1{t^2}dt$$ can you maybe elaborate on that? – StackExchangeMMH Oct 24 '20 at 10:05
  • What do you get when differientating $\psi(x)-\log x$ – reuns Oct 24 '20 at 10:48
  • Okay sry for not writting what I had tried. Okay, so i guess the derivative of $\psi(x)-\log(x)$ is equal to $\psi'(x)-\frac{1}{x}$ which is the left hand side. Since $\psi(x)=-\gamma+\sum_{n=0}^{\infty} \frac{1}{n+1}-\frac{1}{n+x}$ and the sum $\sum_{n=0}^{\infty}\frac{1}{(n+x)^2}$ is uniformly convergend for $x>0$ i also guess we also know that $\psi'(x)=\sum_{n=0}^{\infty} \frac{1}{(n+x)^2}$. But then I guess I end up with $-\int_{x+n}^{x+n+1} \frac{1}{t^2}dt$ which i get to be $\frac{1}{(x+n)(x+n+1)}$ should be equal to $1/x$? so i must have made a mistake somewhere. – StackExchangeMMH Oct 24 '20 at 12:35
  • $\sum_{n\ge 0}\int_{x+n}^{x+n+1} \frac1{t^2}dt = \int_x^\infty \frac1{t^2}dt$ @Mathisfun – reuns Oct 24 '20 at 12:40
  • Arhhh so the integral was inside the summation, thanks alot reuns :) – StackExchangeMMH Oct 24 '20 at 13:20