Question Number 186321 by Mingma last updated on 03/Feb/23

Answered by normans last updated on 03/Feb/23

∫_1 ^e^((n+1)) log (x)=e^(n+1) (log(e^(n+1) )−1+1 for e^n ≥0∧(e^(n+1) ∉R∨Re(e^(n+1) )>0)

Commented byMJS_new last updated on 03/Feb/23

this is no answer to the question

Answered by MJS_new last updated on 03/Feb/23

assuming n∈N ∫_1 ^e^(n+1) ⌊ln x⌋dx= [t=ln x → dx=e^t dt] =∫_0 ^(n+1) e^t ⌊t⌋dt=0[e^t ]_0 ^1 +1[e^t ]_1 ^2 +...n[e^t ]_n ^(n+1) = =Σ_(k=0) ^n (k(e^(k+1) −e^k )) =(e−1)Σ_(k=0) ^n (ke^k ) = =(e−1)(((1+(ne−n−1)e^n )e)/((e−1)^2 ))= =ne^(n+1) −(((e^n −1)e)/(e−1))

Answered by Mathspace last updated on 04/Feb/23

changement lnx=t give x=e^t ⇒ I=∫_0 ^(n+1) [t]e^t dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) k e^t dt =Σ_(k=0) ^(n−1) k(e^(k+1) −e^k ) =(e−1)Σ_(k=1) ^(n−1) ke^k f(x)=Σ_(k=0) ^(n−1) x^k =((x^n −1)/(x−1)) for x≠1 f^′ (x)=Σ_(k=1) ^(n−1) kx^(k−1) ⇒ xf^′ (x)=Σ_(k=1) ^(n−1) k x^k =x.((nx^(n−1) (x−1)−(x^n −1))/((x−1)^2 )) =x.((nx^n −nx^(n−1) −x^n +1)/((x−1)^2 )) =x.(((n−1)x^n −nx^(n−1) +1)/((x−1)^2 )) =(((n−1)x^(n+1) −nx^n +x)/((x−1)^2 )) x=e ⇒Σ_(k=1) ^(n−1) ke^k =(((n−1)e^(n+1) −ne^n +e)/((e−1)^2 )) ⇒ ⇒I=(e−1).(((n−1)e^(n+1) −ne^n +e)/((e−1)^2 )) =(((n−1)e^(n+1) −ne^n +e)/(e−1))