Articles of gamma function

Behavior of Gamma Distribution over time

I need to prove that the age behavior of the Gamma Distribution with probability density function $$ f(x)=\frac{\lambda^{\alpha}x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)}$$ For $x\geq 0$, $\lambda, \alpha >0$; I.e., the conditional pprobability $P(X>x+t|X>t)$, increases in $t$ whenever $\alpha >1$ and decreases in $t$ whenever $0<\alpha <1$. Thus far, I have the following: $$P(X>x+t|X>t)=\frac{P(X>x+t)\cap P(X>t)}{P(X>t)} = \frac{P(X>x+t)}{P(X>t)} =\frac{\int_{x+t}^{\infty}\frac{\lambda^{\alpha}u^{\alpha-1}e^{-\lambda u}}{\Gamma(\alpha)}du}{\int_{t}^{\infty}\frac{\lambda^{\alpha}u^{\alpha-1}e^{-\lambda […]

Is there any close form solution possible for the following integral?

I need to solve the following integral $$\int_{a}^{\infty}\Gamma\left(b,\frac{d}{x}\right)x^{k-1}e^{-cx}dx$$ where $a>0$, $b>0$ and $c>0$. I checked in wolfram alpha but it does not provide any answer. Further, I checked the book of Gradeshteyn (Eq. 6.456) but it can be seen that there are slight differences in my problem Eq. 6.456. Thanks in advance. My Attempt: In […]

How to prove convexity?

Let us consider the function $$I(p):= \frac {\Gamma(2-p)\Gamma(3p)}{(p\Gamma(p))^2} $$ on the interval $(0,1),$ where $\Gamma(x)$ denotes the gamma function. How to prove its convexity there?

differentiate log Gamma function

I am working with the log of negative binomial distribution $NegBin(r,p)$. I need to differentiate the following with respect to $r$ such that at the end, I am NOT left with $r$ in factorial form. $$\frac{d \;log( \Gamma(x + r) )}{dr} = \;\; ?$$ I read about digamma function, but I have $x+r$ within the […]

How do compare two ratios of gamma functions?

How would I show that for $x \ge 2$: $$\frac{\Gamma(x-1)}{[\Gamma(\frac{x}{2})]^2} \le \frac{\Gamma(x)}{[\Gamma(\frac{x+1}{2})]^2}$$ Any hints or suggestions are greatly appreciated.

Quotient of Gamma functions

I am trying to find a clever way to compute the quotient of two gamma functions whose inputs differ by some integer. In other words, for some real value $x$ and an integer $n < x$, I want to find a way to compute $$ \frac{\Gamma(x)}{\Gamma(x-n)} $$ For $n=1$, the quotient it is simply $(x-1)$ […]

How to prove Raabe's Formula

This question already has an answer here: Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$ 4 answers

Extending partial sums of the Taylor series of $e^x$ to a smooth function on $\mathbb{R}^2$?

Is there a smooth function $f:\mathbb{R}^2 \to \mathbb{R}$ such that $f(x,n)$, where $n\in\mathbb{N}$, is the truncated Taylor series of $e^x$, namely $1+ x + \frac{x^2}{2} + \dotsb + \frac{x^n}{n!}$, in a fashion similar to the gamma function?

Evaluating $\psi^{(1/2)}(x)$ of the extended polygamma function

The polygamma function is generally given by $$\psi^{(n)}(x)=\frac{d^{n+1}}{dx^{n+1}}\ln(\Gamma(x)),~n\in\mathbb N_{\ge0}$$ where $\Gamma$ is the gamma function. This can be extended to negative integers by letting $$\psi^{(n-1)}(x)=\int_a^x\psi^{(n)}(x)$$ Unfortunately, there is no fixed $a$ such that the above holds true for any $n$, but it does capture the general idea. Using fractional calculus, one can extend the polygamma […]

The Duplication Formula for the Gamma Function by logarithmic derivatives.

I was reading Ahlfors’ “Complex Analysis” (second edition) and in Chapter 5, section 2.4, where he studies the Gamma Function, he proves Legendre’s Duplication Formula: $$\sqrt{\pi}\Gamma(2z)=2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$His deduction begins with the fact that$$\frac{d}{dz}\left(\frac{\Gamma'(z)}{\Gamma(z)}\right)=\sum_{n=0}^{\infty}\frac{1}{(z+n)^2}$$(where I don’t have any problem deducing) and considers the sum $$ \frac{d}{dz}\left(\frac{\Gamma’\left(z\right)}{\Gamma\left(z\right)}\right)+\frac{d}{dz}\left(\frac{\Gamma’\left(z+\frac{1}{2}\right)}{\Gamma\left(z+\frac{1}{2}\right)}\right) = \sum_{n=0}^{\infty}\frac{1}{\left(z+n\right)^{2}}+\sum_{n=0}^{\infty}\frac{1}{\left(z+n+\frac{1}{2}\right)^{2}} = 4\left[\sum_{n=0}^{\infty}\frac{1}{\left(2z+2n\right)^{2}}+\sum_{n=0}^{\infty}\frac{1}{\left(2z+2n+1\right)^{2}}\right] = 4\sum_{m=0}^{\infty}\frac{1}{\left(2z+m\right)^{2}}. $$I fully understand these simple […]