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′(x)+lnx 4) Find out the value lim_(x→∞) F(x) and lim_(x→1) F(x) 5)Can you prove that at least one of the both result is irrational???


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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
	$mistak at 3) F(x)=∫_0 ^1 e^(−x) ln(x−ln(t))dt F′(x)=∫_0 ^1 {−e^(−x) ln(x−ln(t))+(e^(−x) /(x−ln(t)))}dt by part F(x)=e^(−x) ∫_0 ^1 1.ln(x−ln(t))dt F(x)=e^(−x) {[tln(x−ln(t))]_0 ^1 −∫_0 ^1 t.((−1)/t).(dt/(x−ln(t)))} =e^(−x) ln(x)+e^(−x) ∫_0 ^1 (dt/(x−ln(t))) ⇒F(x)−F′(x)=∫_0 ^1 e^(−x) ln(x−ln(t))dt+e^(−x) ln(x) ⇔−F′(x)=e^(−x) ln(x) ⇒F′(x)=−e^(−x) ln(x)$
	$mistak at 3)$ $F (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt$ $F^{'} (x) = \int_{0}^{1} {- e^{- x} \ln (x - \ln (t)) + \frac{e^{- x}}{x - \ln (t)}} dt$ $by part F (x) = e^{- x} \int_{0}^{1} 1 . \ln (x - \ln (t)) dt$ $F (x) = e^{- x} {{[tln (x - \ln (t))]}_{0}^{1} - \int_{0}^{1} t . \frac{- 1}{t} . \frac{dt}{x - \ln (t)}}$ $= e^{- x} \ln (x) + e^{- x} \int_{0}^{1} \frac{dt}{x - \ln (t)}$ $\Rightarrow F (x) - F^{'} (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt + e^{- x} \ln (x)$ $\Leftrightarrow - F^{'} (x) = e^{- x} \ln (x)$ $\Rightarrow F^{'} (x) = - e^{- x} \ln (x)$
	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
	$F(x )=∫_0 ^1 e^(−x) ln(x−ln(t))dt uf x≥1⇒x−ln(t)≥1 ,∀t∈[0,1] 1≤ln(x−ln(t))=ln(x)+ln(1−((ln(t))/x))≤ln(x)−((ln(t))/x)≤ln(x) 0≤∫_0 ^1 e^(−x) ln(x−ln(t))dt≤e^(−x) ∫_0 ^1 (ln(x)−((ln(t))/x))dt=(ln(x)+(1/x))e^(−x) ⇒F(x) exist 2) ∀(x,t)∈[1,+∞[×[0,1] 0≤tln(x−ln(t))≤t ln(x)−((tln(t))/x) lim tln(x)−((tln(t))/x)→0 ⇒t ln(x−ln(t))→0 3) F is C_1 F is continous (x,t)→e^(−x) ln(x−ln(t)) is continous in each variable e^(−x) ln(x−ln(t))≤e^(−x) {ln(x)+ln(1−((ln(t))/x))}≤e^(−x) {ln(x)−((ln(t))/x)} ≤e^(−x) ln(x)+(e^(−x) /x){−ln(t)} e^(−x) ln(x)≤1,∀x≥1 (e^(−x) /x)≤1 ,∀x≥1⇒e^(−x) ln(x−ln(t))≤1−ln(t)=g(t) ⇒e^(−x) ln(x−ln(t))≤(1−ln(t))=g(t) witch is integrBl in ]0,1[ (t,x)→(∂/∂x)(e^(−x) ln(x−ln(t))) continus (t,x)→(∂/∂t)(e^(−x) ln(x−ln(t)))continus over [0,1[ (∂/∂x)(e^(−x) ln(x−ln(t)))=−e^(−x) ln(x−ln(t))+(e^(−x) /(x−ln(t))) x−ln(t)≥1⇒(e^(−x) /(x−ln(t)))≤e^(−x) ≤1 ln(x−ln(t))=ln(x)+ln(1−((ln(t))/x))≤ln(x)−((ln(t))/x) ⇒∣(∂/∂x)(e^(−x) ln(x−ln(t)))∣≤e^(−x) ln(x−ln(t))+1≤e^(−x) ln(x)+e^(−x) .((−ln(t))/x)+1 ≤3−ln(t) =g(t) ⇒∀x∈[1,+∞[ ∫∣(∂/∂x)(e^(−x) ln(x−ln(t))∣dt≤∫_0 ^1 g(t)dt=4 ⇒F(x) is C_1 g(x)=e^x F(x) g(x) is C_1 since F(x) and e^x are g(x)=∫_0 ^1 ln(x−ln(t))dt g(x)=∫_0 ^1 1.ln(x−ln(t))dt by part ⇒g(x)=[tln(x−ln(t))]_0 ^1 +∫(1/t).(t/(x−ln(t)))dt g(x)=ln(x)+∫(dt/(x−ln(t))) g′(x)=∫_0 ^1 (∂/∂x)(ln(x−ln(t))dt g′(x)=∫_0 ^1 (dt/(x−ln(t))) g(x)=ln(x)+g′(x)⇔g(x)=g′(x)+ln(x) g is continous lim_(x→∞) F(x)=lim_(x→∞) ∫_0 ^1 ln(x−ln(t))dt=lim_(x→∞) ∫_0 ^1 (dt/(x−ln(t))) x−ln(t)≥x⇒ ≤lim_(x→∞) ∫_0 ^1 (dt/x)=lim_(x→∞) (1/x)=0 since G(x) is C_1 ⇒lim_(x→1) G(x)=G(1)⇒ lim_(x→1) ∫_0 ^1 (dt/(x−ln(t)))=∫_0 ^1 (dt/(1−ln(t))) u=−ln(t)⇒G(1)=∫_0 ^(+∞) (e^(−u) /(1+u))$
	$F (x) = \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt$ $uf x ⩾ 1 \Rightarrow x - \ln (t) ⩾ 1, \forall t \in [0, 1]$ $1 ⩽ \ln (x - \ln (t)) = \ln (x) + \ln (1 - \frac{\ln (t)}{x}) ⩽ \ln (x) - \frac{\ln (t)}{x} ⩽ \ln (x)$ $0 ⩽ \int_{0}^{1} e^{- x} \ln (x - \ln (t)) dt ⩽ e^{- x} \int_{0}^{1} (\ln (x) - \frac{\ln (t)}{x}) dt = (\ln (x) + \frac{1}{x}) e^{- x}$ $\Rightarrow F (x) exist$ $2) \forall (x, t) \in [1, + \infty [\times [0, 1]$ $0 ⩽ tln (x - \ln (t)) ⩽ t \ln (x) - \frac{tln (t)}{x}$ $\lim tln (x) - \frac{tln (t)}{x} \to 0$ $\Rightarrow t \ln (x - \ln (t)) \to 0$ $3) F is C_{1}$ $F is continous$ $(x, t) \to e^{- x} \ln (x - \ln (t)) is continous in each variable$ $e^{- x} \ln (x - \ln (t)) ⩽ e^{- x} {\ln (x) + \ln (1 - \frac{\ln (t)}{x})} ⩽ e^{- x} {\ln (x) - \frac{\ln (t)}{x}}$ $⩽ e^{- x} \ln (x) + \frac{e^{- x}}{x} {- \ln (t)}$ $e^{- x} \ln (x) ⩽ 1, \forall x ⩾ 1$ $\frac{e^{- x}}{x} ⩽ 1, \forall x ⩾ 1 \Rightarrow e^{- x} \ln (x - \ln (t)) ⩽ 1 - \ln (t) = g (t)$ $\Rightarrow e^{- x} \ln (x - \ln (t)) ⩽ (1 - \ln (t)) = g (t) witch is integrBl in] 0, 1 [$ $(t, x) \to \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) continus$ $(t, x) \to \frac{\partial}{\partial t} (e^{- x} \ln (x - \ln (t))) continus over [0, 1 [$ $\frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) = - e^{- x} \ln (x - \ln (t)) + \frac{e^{- x}}{x - \ln (t)}$ $x - \ln (t) ⩾ 1 \Rightarrow \frac{e^{- x}}{x - \ln (t)} ⩽ e^{- x} ⩽ 1$ $\ln (x - \ln (t)) = \ln (x) + \ln (1 - \frac{\ln (t)}{x}) ⩽ \ln (x) - \frac{\ln (t)}{x}$ $\Rightarrow∣ \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t))) ∣⩽ e^{- x} \ln (x - \ln (t)) + 1 ⩽ e^{- x} \ln (x) + e^{- x} . \frac{- \ln (t)}{x} + 1$ $⩽ 3 - \ln (t) = g (t)$ $\Rightarrow \forall x \in [1, + \infty [\int ∣ \frac{\partial}{\partial x} (e^{- x} \ln (x - \ln (t)) ∣ dt ⩽ \int_{0}^{1} g (t) dt = 4$ $\Rightarrow F (x) is C_{1}$ $g (x) = e^{x} F (x)$ $g (x) is C_{1} since F (x) and e^{x} are$ $g (x) = \int_{0}^{1} \ln (x - \ln (t)) dt$ $g (x) = \int_{0}^{1} 1 . \ln (x - \ln (t)) dt by part$ $\Rightarrow g (x) = {[tln (x - \ln (t))]}_{0}^{1} + \int \frac{1}{t} . \frac{t}{x - \ln (t)} dt$ $g (x) = \ln (x) + \int \frac{dt}{x - \ln (t)}$ $g^{'} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\ln (x - \ln (t)) dt$ $g^{'} (x) = \int_{0}^{1} \frac{dt}{x - \ln (t)}$ $g (x) = \ln (x) + g^{'} (x) \Leftrightarrow g (x) = g^{'} (x) + \ln (x)$ $g is continous$ $\underset{x \to \infty}{\lim F} (x) = \lim_{x \to \infty} \int_{0}^{1} \ln (x - \ln (t)) dt = \lim_{x \to \infty} \int_{0}^{1} \frac{dt}{x - \ln (t)}$ $x - \ln (t) ⩾ x \Rightarrow$ $⩽ \lim_{x \to \infty} \int_{0}^{1} \frac{dt}{x} = \lim_{x \to \infty} \frac{1}{x} = 0$ $since G (x) is C_{1} \Rightarrow \underset{x \to 1}{\lim G} (x) = G (1) \Rightarrow$ $\lim_{x \to 1} \int_{0}^{1} \frac{dt}{x - \ln (t)} = \int_{0}^{1} \frac{dt}{1 - \ln (t)}$ $u = - \ln (t) \Rightarrow G (1) = \int_{0}^{+ \infty} \frac{e^{- u}}{1 + u}$
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