Question Number 101493 by bemath last updated on 03/Jul/20

∫(√(x.((x.((x.((x.((x.((x...))^(1/7) ))^(1/6) ))^(1/5) ))^(1/4) ))^(1/3) )) dx =

Answered by floor(10Β²Eta[1]) last updated on 03/Jul/20

let′s see a small case first to understand better whats going on (√(x.((x(x)^(1/4) ))^(1/3) ))=(x(x(x)^(1/4) )^(1/3) )^(1/2) =x^(1/2) .x^((1/2)×(1/3)) .x^((1/2)×(1/3)×(1/4)) =x^((1/2)+(1/(2.3))+(1/(2.3.4))) =x^((1/(2!))+(1/(3!))+(1/(4!))) so back to the initial problem, we have: =∫x^((1/(2!))+(1/(3!))+(1/(4!))+...) dx now we want to get a close form for this expression on the exponent: (1/(2!))+(1/(3!))+(1/(4!))+...=Σ_(n=2) ^∞ (1/(n!)) from the taylor series expansion to e^x we have that e^x =Σ_(n=0) ^∞ (x^n /(n!)) when x=1: e=Σ_(n=0) ^∞ (1/(n!))=(1/(0!))+(1/(1!))+Σ_(n=2) ^∞ (1/(n!))=2+Σ_(n=2) ^∞ (1/(n!)) ⇒Σ_(n=2) ^∞ (1/(n!))=e−2 ⇒∫x^((1/(2!))+(1/(3!))+...) dx=∫x^(e−2) dx=(x^(e−1) /(e−1))+C

Commented by1549442205 last updated on 03/Jul/20

A great work !

Commented bybemath last updated on 03/Jul/20

coll 😎😎😎