Let-us-consider-the-function-F-x-0-1-e-x-ln-x-lnt-dt-1-Prove-that-for-all-x-1-F-x-exist-2-Prove-that-lim-t-0-tln-x-lnt-0-3-Prove-that-F-C-1-1-1-and-for-all-x-1-F-x-F-

Question Number 75220 by ~blr237~ last updated on 08/Dec/19

Let us consider the function

F (x) = \int_{0}^{1} e^{- x} \ln (x - lnt) dt

1) Prove that for all x ⩾ 1, F (x) exist

2) Prove that \lim_{t \to 0} tln (x - lnt) = 0

3) Prove that F \in C^{1} ([1 : \infty [, [1 : \infty [) and for

all x ⩾ 1 F (x) = F^{'} (x) + lnx

4) Find out the value

\lim_{x \to \infty} F (x) and \underset{x \to 1}{\lim F} (x)

5) Can you prove that at least one of the both result is irrational ? ? ?

Answered by mind is power last updated on 08/Dec/19

mistak at 3)

F (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt

F^{'} (x) = \int_{0}^{1} {- e^{- x} \ln (x - \ln (t)) + \frac{e^{- x}}{x - \ln (t)}} dt

by part F (x) = e^{- x} \int_{0}^{1} 1 . \ln (x - \ln (t)) dt

F (x) = e^{- x} {{[tln (x - \ln (t))]}_{0}^{1} - \int_{0}^{1} t . \frac{- 1}{t} . \frac{dt}{x - \ln (t)}}

= e^{- x} \ln (x) + e^{- x} \int_{0}^{1} \frac{dt}{x - \ln (t)}

\Rightarrow F (x) - F^{'} (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt + e^{- x} \ln (x)

\Leftrightarrow - F^{'} (x) = e^{- x} \ln (x)

\Rightarrow F^{'} (x) = - e^{- x} \ln (x)

Commented by ~blr237~ last updated on 08/Dec/19

Sorry {it}^{'} s because, i started my work with G (x) = e^{x} F (x)

and it was G (x) = G^{'} (x) + lnx .

Commented by mind is power last updated on 08/Dec/19

F (x) = \int_{0}^{1} \ln (x - \ln (t)) dt ?

Commented by ~blr237~ last updated on 09/Dec/19

yes

Answered by mind is power last updated on 09/Dec/19

F (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt

uf x ⩾ 1 \Rightarrow x - \ln (t) ⩾ 1, \forall t \in [0, 1]

1 ⩽ \ln (x - \ln (t)) = \ln (x) + \ln (1 - \frac{\ln (t)}{x}) ⩽ \ln (x) - \frac{\ln (t)}{x} ⩽ \ln (x)

0 ⩽ \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt ⩽ e^{- x} \int_{0}^{1} (\ln (x) - \frac{\ln (t)}{x}) dt = (\ln (x) + \frac{1}{x}) e^{- x}

\Rightarrow F (x) exist

2) \forall (x, t) \in [1, + \infty [\times [0, 1]

0 ⩽ tln (x - \ln (t)) ⩽ t \ln (x) - \frac{tln (t)}{x}

\lim tln (x) - \frac{tln (t)}{x} \to 0

\Rightarrow t \ln (x - \ln (t)) \to 0

3) F is C_{1}

F is continous

(x, t) \to e^{- x} \ln (x - \ln (t)) is continous in each variable

e^{- x} \ln (x - \ln (t)) ⩽ e^{- x} {\ln (x) + \ln (1 - \frac{\ln (t)}{x})} ⩽ e^{- x} {\ln (x) - \frac{\ln (t)}{x}}

⩽ e^{- x} \ln (x) + \frac{e^{- x}}{x} {- \ln (t)}

e^{- x} \ln (x) ⩽ 1, \forall x ⩾ 1

\frac{e^{- x}}{x} ⩽ 1, \forall x ⩾ 1 \Rightarrow e^{- x} \ln (x - \ln (t)) ⩽ 1 - \ln (t) = g (t)

\Rightarrow e^{- x} \ln (x - \ln (t)) ⩽ (1 - \ln (t)) = g (t) witch is integrBl in] 0, 1 [

(t, x) \to \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) continus

(t, x) \to \frac{\partial}{\partial t} (e^{- x} \ln (x - \ln (t))) continus over [0, 1 [

\frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) = - e^{- x} \ln (x - \ln (t)) + \frac{e^{- x}}{x - \ln (t)}

x - \ln (t) ⩾ 1 \Rightarrow \frac{e^{- x}}{x - \ln (t)} ⩽ e^{- x} ⩽ 1

\ln (x - \ln (t)) = \ln (x) + \ln (1 - \frac{\ln (t)}{x}) ⩽ \ln (x) - \frac{\ln (t)}{x}

\Rightarrow∣ \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) ∣⩽ e^{- x} \ln (x - \ln (t)) + 1 ⩽ e^{- x} \ln (x) + e^{- x} . \frac{- \ln (t)}{x} + 1

⩽ 3 - \ln (t) = g (t)

\Rightarrow \forall x \in [1, + \infty [\int ∣ \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t)) ∣ dt ⩽ \int_{0}^{1} g (t) dt = 4

\Rightarrow F (x) is C_{1}

g (x) = e^{x} F (x)

g (x) is C_{1} since F (x) and e^{x} are

g (x) = \int_{0}^{1} \ln (x - \ln (t)) dt

g (x) = \int_{0}^{1} 1 . \ln (x - \ln (t)) dt by part

\Rightarrow g (x) = {[tln (x - \ln (t))]}_{0}^{1} + \int \frac{1}{t} . \frac{t}{x - \ln (t)} dt

g (x) = \ln (x) + \int \frac{dt}{x - \ln (t)}

g^{'} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\ln (x - \ln (t)) dt

g^{'} (x) = \int_{0}^{1} \frac{dt}{x - \ln (t)}

g (x) = \ln (x) + g^{'} (x) \Leftrightarrow g (x) = g^{'} (x) + \ln (x)

g is continous

\underset{x \to \infty}{\lim F} (x) = \lim_{x \to \infty} \int_{0}^{1} \ln (x - \ln (t)) dt = \lim_{x \to \infty} \int_{0}^{1} \frac{dt}{x - \ln (t)}

x - \ln (t) ⩾ x \Rightarrow

⩽ \lim_{x \to \infty} \int_{0}^{1} \frac{dt}{x} = \lim_{x \to \infty} \frac{1}{x} = 0

since G (x) is C_{1} \Rightarrow \underset{x \to 1}{\lim G} (x) = G (1) \Rightarrow

\lim_{x \to 1} \int_{0}^{1} \frac{dt}{x - \ln (t)} = \int_{0}^{1} \frac{dt}{1 - \ln (t)}

u = - \ln (t) \Rightarrow G (1) = \int_{0}^{+ \infty} \frac{e^{- u}}{1 + u}