advanced-mathematics-digamma-limit-if-k-gt-0-then-prove-that-lim-x-0-1-x-k-

Question Number 115193 by mnjuly1970 last updated on 24/Sep/20

\dots a d v a n c e d m a t h e m a t i c s \dots

:: d i g a m m a l i m i t ::

i f k > 0 t h e n

p r o v e t h a t

{l i m}_{x \to 0} \frac{1}{x} (ψ (\frac{k + x}{2 x}) - ψ (\frac{k}{2 x})) = \frac{1}{k} ✓

m . n . j u l y . 1970 \dots

Commented by Tawa11 last updated on 06/Sep/21

great

Answered by mathdave last updated on 24/Sep/20

s o l u t i o n

l e t I = \lim_{x \to 0} (\frac{1}{x} (ψ (\frac{m}{2 x} + \frac{1}{2}) - ψ (\frac{m}{2 x}))

w e k n o w n

\int_{0}^{1} \frac{t^{n - 1}}{1 + t} d t = \frac{1}{2} (ψ (\frac{k}{2} + \frac{1}{2}) - ψ (\frac{k}{2}))

l e t k = \frac{m}{x}

I = \frac{2}{x} \int_{0}^{1} \frac{t^{\frac{m}{x} - 1}}{1 + t} d t z = \frac{m}{x}

I = \frac{2}{m} \int_{0}^{1} \frac{{z t}^{z - 1}}{1 + t} d t (l e t \int d v = \int {z t}^{z - 1} d z, v = t^{z}

a n d u = \frac{1}{1 + t}, d u = - \frac{1}{{(1 + t)}^{2}}) u s i n g I B P

I = \frac{2}{m} {(\frac{t^{z}}{1 + t})}_{0}^{1} + \frac{2}{m} \int_{0}^{1} \frac{t^{z}}{{(1 + t)}^{2}} d t = \frac{1}{m} + \frac{2}{m} \lim_{z \to \infty} \int_{0}^{1} \frac{t^{z}}{{(1 + t)}^{2}} d t

l e t y = \frac{1}{t}, d y = - \frac{1}{t^{2}}

I = \frac{1}{m} + \frac{2}{m} \lim_{z \to \infty} \int_{\infty}^{1} \frac{y^{- z}}{{(1 + \frac{1}{y})}^{2}} \times - \frac{1}{y^{2}} d y

I = \frac{1}{m} + \frac{2}{m} \lim_{z \to \infty} \int_{0}^{\infty} \frac{y^{2}}{y^{z} {(1 + y)}^{2}} \times \frac{d y}{y^{2}} = (\frac{1}{m} + \frac{2}{m} \lim_{z \to \infty} \int_{1}^{\infty} \frac{d y}{y^{z} {(1 + y)}^{2}}) = \frac{1}{m}

∵ \lim_{x \to 0} \frac{1}{x} (ψ (\frac{m}{2 x} + \frac{1}{2}) - ψ (\frac{m}{2 x})) = \frac{1}{m} Q . E . D

b y m a t h d a v e (24 / 09 / 2020)

Commented by mnjuly1970 last updated on 24/Sep/20

g o o d w o r k