Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Digamma function
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Integral representations== If the real part of {{mvar|z}} is positive then the digamma function has the following [[integral]] representation due to Gauss:<ref name="Whittaker and Watson, 12.3">Whittaker and Watson, 12.3.</ref> :<math>\psi(z) = \int_0^\infty \left(\frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}}\right)\,dt.</math> Combining this expression with an integral identity for the [[Euler–Mascheroni constant]] <math>\gamma</math> gives: :<math>\psi(z + 1) = -\gamma + \int_0^1 \left(\frac{1-t^z}{1-t}\right)\,dt.</math> The integral is Euler's [[harmonic number]] <math>H_z</math>, so the previous formula may also be written :<math>\psi(z + 1) = \psi(1) + H_z.</math> A consequence is the following generalization of the recurrence relation: :<math>\psi(w + 1) - \psi(z + 1) = H_w - H_z.</math> An integral representation due to Dirichlet is:<ref name="Whittaker and Watson, 12.3"/> :<math>\psi(z) = \int_0^\infty \left(e^{-t} - \frac{1}{(1 + t)^z}\right)\,\frac{dt}{t}.</math> Gauss's integral representation can be manipulated to give the start of the asymptotic expansion of <math>\psi</math>.<ref>Whittaker and Watson, 12.31.</ref> :<math>\psi(z) = \log z - \frac{1}{2z} - \int_0^\infty \left(\frac{1}{2} - \frac{1}{t} + \frac{1}{e^t - 1}\right)e^{-tz}\,dt.</math> This formula is also a consequence of Binet's first integral for the gamma function. The integral may be recognized as a [[Laplace transform]]. Binet's second integral for the gamma function gives a different formula for <math>\psi</math> which also gives the first few terms of the asymptotic expansion:<ref>Whittaker and Watson, 12.32, example.</ref> :<math>\psi(z) = \log z - \frac{1}{2z} - 2\int_0^\infty \frac{t\,dt}{(t^2 + z^2)(e^{2\pi t} - 1)}.</math> From the definition of <math>\psi</math> and the integral representation of the gamma function, one obtains :<math>\psi(z) = \frac{1}{\Gamma(z)} \int_0^\infty t^{z-1} \ln (t) e^{-t}\,dt,</math> with <math>\Re z > 0</math>.<ref>{{cite web | url=https://dlmf.nist.gov/5.9 | title=NIST. Digital Library of Mathematical Functions (DLMF), 5.9.}}</ref>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)