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Question Number 143709 by mnjuly1970 last updated on 17/Jun/21

Commented by mnjuly1970 last updated on 17/Jun/21

P rove ::⇑⇑⇑

Prove::⇑⇑⇑

Commented by TheHoneyCat last updated on 17/Jun/21

1°) Proof that it is well defined ∀x∈R_+ x≥1 ⇒ x^n ≥x so x≥1 ⇒0< e^(−x^n ) ≤e^(−x) also ∣cosλx∣≤1 so x≥1 ⇒ ∣e^(−x^n ) cosλx∣≤e^(−x) knowing that ∫_1 ^(+∞) e^(−x) dx=−e<∞ ∫_0 ^(+∞) e^(−x^n ) cos(λx)dx ∈R

1°)Proofthatitiswelldefined∀x∈R+x⩾1⇒xn⩾xsox⩾1⇒0<e−xn⩽e−xalso∣cosλx∣⩽1sox⩾1⇒∣e−xncosλx∣⩽e−xknowingthat∫1+∞e−xdx=−e<∞∫0+∞e−xncos(λx)dx∈R

Answered by TheHoneyCat last updated on 17/Jun/21

let (a,b)∈(R_+ )^2 and l be this limit with ∫_0 ^(+∞) e^(−x^n ) cosλx dx =Re(∫_0 ^(+∞) e^(−(x^n +λix) ) dx) =Re([((−1)/λ)e^(−x^n ) e^(−λix) ]_0 ^(+∞) −∫_0 ^(+∞) ((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) dx) =Re(((−1)/λ)−∫_0 ^(+∞) ((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) dx) =((−1)/λ)−Re(∫_0 ^(+∞) ((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) dx) with λ sugiciently high =∣((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) ∣ =((nx^(n−1) )/λ)e^(−x^n ) ≤e^(−P(x)) with P some polynomial such that P(x)→_(x→+∞) +∞ e^(−P(x)) can be integrated, and is independent from λ thus: lim_(λ→+∞) ((−1)/λ)−Re(∫_0 ^(+∞) ((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) dx) =lim_(λ→+∞) ((−1)/λ)−Re(∫_0 ^(+∞) lim_(λ→+∞) ((nx^(n−1) )/λ)e^(−x^n ) e^(−iλx) dx) =0−Re(∫_0 ^(+∞) 0dx) =0_■

let(a,b)∈(R+)2andlbethislimitwith∫0+∞e−xncosλxdx=Re(∫0+∞e−(xn+λix)dx)=Re([−1λe−xne−λix]0+∞−∫0+∞nxn−1λe−xne−iλxdx)=Re(−1λ−∫0+∞nxn−1λe−xne−iλxdx)=−1λ−Re(∫0+∞nxn−1λe−xne−iλxdx)withλsugicientlyhigh=∣nxn−1λe−xne−iλx∣=nxn−1λe−xn⩽e−P(x)withPsomepolynomialsuchthatP(x)→x→+∞+∞e−P(x)canbeintegrated,andisindependentfromλthus:limλ→+∞−1λ−Re(∫0+∞nxn−1λe−xne−iλxdx)=limλ→+∞−1λ−Re(∫0+∞limλ→+∞nxn−1λe−xne−iλxdx)=0−Re(∫0+∞0dx)=0◼

Commented by mnjuly1970 last updated on 17/Jun/21

thanks alot ....

thanksalot....

Commented by TheHoneyCat last updated on 17/Jun/21

your welcome :•)

yourwelcome:∙)

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