The digamma function is defined to be $\psi^{(0)}(x)=\frac{d}{dx}ln(\Gamma(x))$, from which we can derive: $$\psi^{(0)}(x)=\frac{d}{dx}ln(\Gamma(x))=\frac{\Gamma'(x)}{\Gamma(x)}$$ also, $$\frac{d}{dx}\Gamma(x)=\int_{0}^{\infty}\frac{d}{dx}t^{x-1}e^{-t}dt=\int_{0}^{\infty}t^{x-1}ln(t)e^{-t}dt$$ therefore, $$\psi^{(0)}(x)=\frac{\int_{0}^{\infty}t^{x-1}ln(t)e^{-t}dt}{\int_{0}^{\infty}t^{x-1}e^{-t}dt}$$ that's as far as I got, so my question is, is there a way to simply this expression so that the digamma function can be expressed as a single integral, as opposed to a quotient of integrals?
Asked
Active
Viewed 1,796 times
2 Answers
4
As a rule of thumb, it is best to apply a logarithmic derivative to a product.
For instance, if we start with the Weierstrass product for the $\Gamma$ function
$$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1} $$
we instantly get:
$$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{z+n}\right) = \sum_{n\geq 1}\left[\log(n+1)-\log(n)+\frac{1}{n+z}\right]\tag{2} $$
and by the inverse Laplace transform and Frullani's theorem we have
$$ \log(n+1)-\log(n) = \int_{0}^{+\infty}\frac{e^{-nt}-e^{-(n+1)t}}{t}\,dt,\qquad \frac{1}{n+z} = \int_{0}^{+\infty} e^{-(n+z)t}\,dt \tag{3}$$
hence by plugging these integral representations into $(2)$ and summing over $n$ we recover the integral representation shown above.
Jack D'Aurizio
- 353,855
2
There is a well-known intergral representation for the digamma function. $$ {\psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt} $$ There are other integral representations listed here.
user153012
- 12,240