Question

prove that
$\gamma = 1 + 2 \div f$
where symbol have their usual meaning​

Answer 1

Answer:

1. The Gamma Function 1

1.1. Existence of Γ() 1

1.2. The Functional Equation of Γ() 3

1.3. The Factorial Function and Γ() 5

1.4. Special Values of Γ() 6

1.5. The Beta Function and the Gamma Function 14

2. Stirling’s Formula 17

2.1. Stirling’s Formula and Probabilities 18

2.2. Stirling’s Formula and Convergence of Series 20

2.3. From Stirling to the Central Limit Theorem 21

2.4. Integral Test and the Poor Man’s Stirling 24

2.5. Elementary Approaches towards Stirling’s Formula 25

2.6. Stationary Phase and Stirling 29

2.7. The Central Limit Theorem and Stirling 30

1. THE GAMMA FUNCTION

In this chapter we’ll explore some of the strange and wonderful properties of the Gamma function

Γ(), defined by

For > 0 (or actually ℜ() > 0), the Gamma function Γ() is

Γ() = ∫ ∞

0

−

−1

=

∫ ∞

0

−

.

There are countless integrals or functions we can define. Just looking at it, there is nothing to make

you think it will appear throughout probability and statistics, but it does. We’ll see where it occurs

and why, and discuss many of its most important properties.

1.1. Existence of Γ(). Looking at the definition of Γ(), it’s natural to ask: Why do we have re-

strictions on ? Whenever you are given an integrand, you must make sure it is well-behaved before

you can conclude the integral exists. Frequently there are two trouble points to check, near = 0

and near = ±∞ (okay, three points). For example, consider the function () =

−1/2 on the

interval [0, ∞). This function blows up at the origin, but only mildly. Its integral is 2

1/2

, and this is

integrable near the origin. This just means that

lim→0

∫ 1

−1/2

exists and is finite. Unfortunately, even though this function is tending to zero, it approaches zero so

slowly for large that it is not integrable on [0, ∞). The problem is that integrals such as

lim

→∞ ∫

1

−1/2

is infinite. Can the reverse problem happen, namely our function decays fast enough for large but

blows up too rapidly for small ? Sure – the following is a standard, albeit initially strange looking,

example. Consider

() = { 1

log2

if > 0

0 otherwise.