show-that-1-gamma-function-lim-n-H-n-ln-n-

Question Number 2818 by prakash jain last updated on 27/Nov/15

show that

Γ^{'} (1) = - γ

Γ gamma function

γ = \lim_{n \to \infty} [H_{n} - \ln n]

Answered by 123456 last updated on 30/Nov/15

Γ (x) = \underset{0}{\int^{+ \infty}} t^{x - 1} e^{- t} d t (x > 0)

Γ^{'} (x) = \underset{0}{\int^{+ \infty}} t^{x - 1} \ln t e^{- t} d t

Γ^{'} (1) = \underset{0}{\int^{+ \infty}} \ln t e^{- t} d t (Q 2435)

e^{- t} = \lim_{n \to + \infty} {(1 - \frac{t}{n})}^{n}

Γ (x) = \lim_{n \to + \infty} \underset{0}{\int^{n}} t^{x - 1} {(1 - \frac{t}{n})}^{n} d t

u = t / n, d u = d t / n t = 0, u = 0 t = n, u = 1

Γ (x) = \lim_{n \to \infty} \underset{0}{\int^{1}} {(u n)}^{x - 1} {(1 - u)}^{n} (d u \cdot n)

= \lim_{n \to \infty} n^{x} \underset{0}{\int^{1}} u^{x - 1} {(1 - u)}^{n} d u

integral by parts

a = {(1 - u)}^{n}, d a = - n {(1 - u)}^{n - 1} d u

d b = u^{x - 1} d u, b = u^{x} / x

= \lim_{n \to \infty} n^{x} \frac{n}{x} \underset{0}{\int^{1}} u^{x} {(1 - u)}^{n - 1} d u

⋮

= \lim_{n \to \infty} \frac{n^{x} n \dots 1}{x (x + 1) \dots (x + n - 1)} \underset{0}{\int^{1}} u^{x + n - 1} d u

= \lim_{n \to \infty} \frac{n^{x} n!}{x (x + 1) \dots (x + n)} (Q 2845)

= \lim_{n \to \infty} \frac{n^{x} n!}{\underset{l = 0}{\prod^{n}} x + l}

Γ (z + 1) = z Γ (z)

\ln Γ (z + 1) = \ln z + \ln Γ (z)

= \underset{n \to + \infty}{\lim l n} (n! n^{z}) - \underset{l = 1}{\sum^{n}} \ln (z + k)

ψ (z + 1) = \lim_{n \to \infty} \ln n - \underset{l = 1}{\sum^{n}} \frac{1}{z + l}

\frac{1}{z + l} = \frac{l}{l (z + l)}

= \frac{1}{l} \cdot \frac{l}{z + l}

= \frac{1}{l} \cdot \frac{z + l - z}{z + l}

= \frac{1}{l} (1 - \frac{z}{z + l})

= \frac{1}{l} - \frac{z}{l (z + l)}

so

ψ (z + 1) = \lim_{n \to \infty} \ln n - \underset{l = 1}{\sum^{n}} \frac{1}{l} + \underset{l = 1}{\sum^{n}} \frac{z}{l (z + l)}

= - \lim_{n \to \infty} (\underset{l = 1}{\sum^{n}} \frac{1}{l} - \ln n) + \underset{l = 1}{\sum^{+ \infty}} \frac{z}{l (z + l)}

= - γ + \underset{l = 1}{\sum^{+ \infty}} \frac{z}{l (z + l)}

take z = 0

ψ (1) = - γ

ψ (x) = \frac{d}{d x} \ln Γ (x)

ψ (x) = \frac{Γ^{'} (x)}{Γ (x)}

Γ^{'} (x) = ψ (x) Γ (x)

Γ^{'} (1) = ψ (1) Γ (1) = ψ (1) = - γ

Facebook Tweet Pin