advanced-integral-i-0-1-1-ln-x-1-1-x-dx-ii-x-0-e-t-t-e-tx-1-e-t-dt-solution-2-ln-n-easy-0-1-x-n-

Question Number 122625 by mnjuly1970 last updated on 18/Nov/20

\dots a d v a n c e d i n t e g r a l \dots

i : \int_{0}^{1} (\frac{1}{l n (x)} + \frac{1}{1 - x}) d x = γ

i i : ψ (x) = \int_{0}^{\infty} (\frac{e^{- t}}{t} - \frac{e^{- t x}}{1 - e^{- t}}) d t

s o l u t i o n : {_{2 : l n (n) \overset{e a s y}{=} \int_{0}^{1} \frac{x^{n - 1} - 1}{l n (x)} d x (* *)}^{1 : H_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k} = \int_{0}^{1} \frac{1 - x^{n}}{1 - x} d x (*)}

(*) - (* *) : H_{n} - l n (n) = \int_{0}^{1} (\frac{1 - x^{n}}{1 - x} - \frac{x^{n - 1} - 1}{l n (x)}) d x

{l i m}_{n \to \infty} (x^{n}) \overset{0 < x < 1}{=} 0

{l i m}_{n \to \infty} (H_{n} - l n (n)) = \int_{0}^{1} (\frac{1}{1 n (x)} + \frac{1}{1 - x}) d x

γ = \int_{0}^{1} (\frac{1}{l n (x)} + \frac{1}{1 - x}) d x ✓

\dots \dots \dots \dots \dots \dots \dots \dots \dots . .

ψ (x) \overset{e a s y}{=} - γ + \int_{0}^{1} \frac{1 - t^{x - 1}}{1 - t} d t

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

= \int_{0}^{1} - \frac{1}{l n (t)} + \frac{1 - t^{x - 1} - 1}{1 - t} d t

= - \int_{0}^{1} \frac{1}{l n (t)} + \frac{t^{x - 1}}{1 - t} d t \overset{t = e^{- y}}{=}

= - \int_{\infty}^{0} (\frac{1}{- y} + \frac{e^{- y x + y}}{1 - e^{- y}}) (- e^{- y}) d y

= \int_{0}^{\infty} \frac{e^{- y}}{y} - \frac{e^{- y x}}{1 - e^{- y}} d y

∵ ψ (x) = \int_{0}^{\infty} (\frac{e^{- y}}{y} - \frac{e^{- y x}}{1 - e^{- y}}) d y ✓