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Question Number 34222 by abdo imad last updated on 03/May/18
let give the sequence of integrals J_n =∫_0 ^∞ x^n e^(−(x^2 /2)) dx 1) prove that J_n =(n−1)J_(n−2) ∀n≥2 2) calculate J_(2p) and J_(2p+1) by using factoriels. 3) prove that ∀n≥1 J_n ^2 ≤J_(n−1) . J_(n+1) . 4)prove that ((2^(2p) (p!)^2 )/((2p)!)) (1/( (√(2p+1)))) ≤J_0 ≤ ((2^(2p) (p!)^2 )/((2p)!)) (1/( (√(2p)))) 5) find a equivalent of ((2^(2p) (p!)^2 )/((2p)!)) (p→+∞)
$${let}\:{give}\:{the}\:{sequence}\:{of}\:{integrals} \\ $$$${J}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{J}_{{n}} =\left({n}−\mathrm{1}\right){J}_{{n}−\mathrm{2}} \:\:\:\forall{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{J}_{\mathrm{2}{p}} \:{and}\:{J}_{\mathrm{2}{p}+\mathrm{1}} \:{by}\:{using}\:{factoriels}. \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:\:\forall{n}\geqslant\mathrm{1}\:\:\:{J}_{{n}} ^{\mathrm{2}} \:\:\leqslant{J}_{{n}−\mathrm{1}} \:.\:{J}_{{n}+\mathrm{1}} . \\ $$$$\left.\mathrm{4}\right){prove}\:{that}\:\:\frac{\mathrm{2}^{\mathrm{2}{p}} \left({p}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{p}\right)!}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{p}+\mathrm{1}}}\:\leqslant{J}_{\mathrm{0}} \:\leqslant\:\frac{\mathrm{2}^{\mathrm{2}{p}} \:\left({p}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{p}\right)!}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{p}}} \\ $$$$\left.\mathrm{5}\right)\:{find}\:{a}\:{equivalent}\:{of}\:\:\frac{\mathrm{2}^{\mathrm{2}{p}} \left({p}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{p}\right)!}\:\:\left({p}\rightarrow+\infty\right) \\ $$
Commented by candre last updated on 03/May/18
J_n =∫_0 ^∞ x^n e^(−(x^2 /2)) dx u=x^(n−1) ⇒du=(n−1)x^(n−2) dx dv=xe^(−(x^2 /2)) dx⇒v=−e^(−(x^2 /2)) =−[x^(n−1) e^(−(x^2 /2)) ]_0 ^∞ +(n−1)∫x^(n−2) e^(−(x^2 /2)) dx =(n−1)J_(n−2) ∫xe^(−(x^2 /2)) dx=−∫e^u du=−e^u =−e^(−(x^2 /2)) u=−(x^2 /2)⇒du=−xdx
$${J}_{{n}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}{x}^{{n}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$${u}={x}^{{n}−\mathrm{1}} \Rightarrow{du}=\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} {dx} \\ $$$${dv}={xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\Rightarrow{v}=−{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$=−\left[{x}^{{n}−\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right]_{\mathrm{0}} ^{\infty} +\left({n}−\mathrm{1}\right)\int{x}^{{n}−\mathrm{2}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$$=\left({n}−\mathrm{1}\right){J}_{{n}−\mathrm{2}} \\ $$$$\int{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}=−\int{e}^{{u}} {du}=−{e}^{{u}} =−{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${u}=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{du}=−{xdx} \\ $$
Commented by candre last updated on 03/May/18
J_0 =∫_0 ^∞ e^(−(x^2 /2)) dx J_0 ^2 =(∫_0 ^∞ e^(−(x^2 /2)) dx)^2 =∫_0 ^∞ e^(−x^2 /2) dx∫_0 ^∞ e^(−y^2 /2) dy =∫_0 ^∞ ∫_0 ^∞ e^(−((x^2 +y^2 )/2)) dxdy =∫_0 ^(π/2) ∫_0 ^∞ e^(−(r^2 /2)) rdrdθ =∫_0 ^(π/2) dθ∫_0 ^∞ e^(−r^2 /2) rdr u=−r^2 /2⇒du=−rdr =(π/2)∙∫_(−∞) ^0 e^u du =(π/2) J_0 =(√(π/2))
$${J}_{\mathrm{0}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$${J}_{\mathrm{0}} ^{\mathrm{2}} =\left(\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\right)^{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} /\mathrm{2}} {dx}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{y}^{\mathrm{2}} /\mathrm{2}} {dy} \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}} {dxdy} \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}} {rdrd}\theta \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{d}\theta\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{r}^{\mathrm{2}} /\mathrm{2}} {rdr} \\ $$$${u}=−{r}^{\mathrm{2}} /\mathrm{2}\Rightarrow{du}=−{rdr} \\ $$$$=\frac{\pi}{\mathrm{2}}\centerdot\underset{−\infty} {\overset{\mathrm{0}} {\int}}{e}^{{u}} {du} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$$${J}_{\mathrm{0}} =\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$
Commented by candre last updated on 03/May/18
J_(2p) : J_2 =1∙J_0 J_4 =3∙J_2 =3∙1∙J_0 J_6 =5∙J_4 =5∙3∙1∙J_0 J_(2p) =(2p−1)!!J_0 J_(2p+1) : J_3 =2∙J_1 J_5 =4∙J_3 =4∙2∙J_1 J_(2p+1) =(2p)!!J_1 =2^p p!J_1
$${J}_{\mathrm{2}{p}} : \\ $$$${J}_{\mathrm{2}} =\mathrm{1}\centerdot{J}_{\mathrm{0}} \\ $$$${J}_{\mathrm{4}} =\mathrm{3}\centerdot{J}_{\mathrm{2}} =\mathrm{3}\centerdot\mathrm{1}\centerdot{J}_{\mathrm{0}} \\ $$$${J}_{\mathrm{6}} =\mathrm{5}\centerdot{J}_{\mathrm{4}} =\mathrm{5}\centerdot\mathrm{3}\centerdot\mathrm{1}\centerdot{J}_{\mathrm{0}} \\ $$$${J}_{\mathrm{2}{p}} =\left(\mathrm{2}{p}−\mathrm{1}\right)!!{J}_{\mathrm{0}} \\ $$$${J}_{\mathrm{2}{p}+\mathrm{1}} : \\ $$$${J}_{\mathrm{3}} =\mathrm{2}\centerdot{J}_{\mathrm{1}} \\ $$$${J}_{\mathrm{5}} =\mathrm{4}\centerdot{J}_{\mathrm{3}} =\mathrm{4}\centerdot\mathrm{2}\centerdot{J}_{\mathrm{1}} \\ $$$${J}_{\mathrm{2}{p}+\mathrm{1}} =\left(\mathrm{2}{p}\right)!!{J}_{\mathrm{1}} =\mathrm{2}^{{p}} {p}!{J}_{\mathrm{1}} \\ $$
Commented by abdo mathsup 649 cc last updated on 05/May/18
1) let prove that J_(n+1) = nJ_(n−1) J_(n+1) = ∫_0 ^∞ x^(n+1) e^(−(x^2 /2)) dx =−∫_0 ^∞ x^n (−xe^(−(x^2 /2)) dx) =−([ x^n e^(−(x^2 /2)) ]_0 ^(+∞) −∫_0 ^∞ n x^(n−1) e^(−(x^2 /2)) dx) = n ∫_0 ^∞ x^(n−1) e^(−(x^2 /2)) dx = n J_(n−1) ⇒ J_n =(n−1)J_(n−2) ∀n≥2 2) J_(2p) = (2p−1)J_(2p−2) ⇒ Π_(p=1) ^n J_(2p) =Π_(p=1) ^n J_(2p−2) Π_(p=1) ^n (2p−1)⇒ J_2 .J_4 .....J_(2n) =1.3.5....(2n−1) J_0 .J_2 ....J_(2n−2) ⇒ J_(2n) =1.3.5....(2n−1) J_0 =((1.2.3.4......(2n−1)(2n))/(2^n n!)) J_0 =(((2n)!)/(2^n (n!))) J_0 J_0 = ∫_0 ^∞ e^(−(x^2 /2)) dx =_((√2) t=x) ∫_0 ^∞ e^(−t^2 ) (√2)dt = (√2) ((√π)/2) =(√(π/2)) ⇒ J_(2n) = (((2n)!)/(2^n n!)) (√(π/2)) J_(2p+1) = 2p J_(2p−1 ) ⇒ Π_(p=1) ^n J_(2p+1) = Π_(p=1) ^n (2p) Π_(p=1) ^n J_(2p−1) ⇒ J_3 .J_5 .....J_(2n+1) =2^n n! J_1 .J_3 ....J_(2n−1) ⇒ J_(2n+1) = 2^n n! J_1 J_1 = ∫_0 ^∞ x e^(−(x^2 /2)) dx =−∫_0 ^∞ (−x) e^(−(x^2 /2)) dx =−[ e^(−(x^2 /2)) ]_0 ^(+∞) = 1 ⇒ J_(2n+1) = 2^n (n!) .
$$\left.\mathrm{1}\right)\:{let}\:{prove}\:{that} \\ $$$${J}_{{n}+\mathrm{1}} =\:{nJ}_{{n}−\mathrm{1}} \\ $$$${J}_{{n}+\mathrm{1}} =\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}+\mathrm{1}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:{x}^{{n}} \left(−{xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\right)\:=−\left(\left[\:{x}^{{n}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right]_{\mathrm{0}} ^{+\infty} \:−\int_{\mathrm{0}} ^{\infty} {n}\:{x}^{{n}−\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\right) \\ $$$$=\:{n}\:\int_{\mathrm{0}} ^{\infty} \:\:{x}^{{n}−\mathrm{1}} \:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:{dx}\:=\:{n}\:{J}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${J}_{{n}} =\left({n}−\mathrm{1}\right){J}_{{n}−\mathrm{2}} \:\:\:\forall{n}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{2}\right)\:{J}_{\mathrm{2}{p}} \:=\:\left(\mathrm{2}{p}−\mathrm{1}\right){J}_{\mathrm{2}{p}−\mathrm{2}} \:\Rightarrow \\ $$$$\prod_{{p}=\mathrm{1}} ^{{n}} \:{J}_{\mathrm{2}{p}} \:=\prod_{{p}=\mathrm{1}} ^{{n}} \:{J}_{\mathrm{2}{p}−\mathrm{2}} \:\prod_{{p}=\mathrm{1}} ^{{n}} \:\left(\mathrm{2}{p}−\mathrm{1}\right)\Rightarrow \\ $$$${J}_{\mathrm{2}} .{J}_{\mathrm{4}} …..{J}_{\mathrm{2}{n}} =\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{1}\right)\:{J}_{\mathrm{0}} .{J}_{\mathrm{2}} ….{J}_{\mathrm{2}{n}−\mathrm{2}} \:\Rightarrow \\ $$$${J}_{\mathrm{2}{n}} \:=\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{1}\right)\:{J}_{\mathrm{0}} \\ $$$$=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}……\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)}{\mathrm{2}^{{n}} {n}!}\:{J}_{\mathrm{0}} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} \:\left({n}!\right)}\:{J}_{\mathrm{0}} \\ $$$${J}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\:=_{\sqrt{\mathrm{2}}\:{t}={x}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{t}^{\mathrm{2}} } \:\sqrt{\mathrm{2}}{dt}\:=\:\sqrt{\mathrm{2}}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$$\Rightarrow\:{J}_{\mathrm{2}{n}} \:\:=\:\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{{n}} \:{n}!}\:\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$${J}_{\mathrm{2}{p}+\mathrm{1}} \:=\:\mathrm{2}{p}\:{J}_{\mathrm{2}{p}−\mathrm{1}\:} \:\Rightarrow\:\prod_{{p}=\mathrm{1}} ^{{n}} \:{J}_{\mathrm{2}{p}+\mathrm{1}} \:=\:\prod_{{p}=\mathrm{1}} ^{{n}} \left(\mathrm{2}{p}\right)\:\prod_{{p}=\mathrm{1}} ^{{n}} \:{J}_{\mathrm{2}{p}−\mathrm{1}} \\ $$$$\Rightarrow\:{J}_{\mathrm{3}} .{J}_{\mathrm{5}} …..{J}_{\mathrm{2}{n}+\mathrm{1}} =\mathrm{2}^{{n}} {n}!\:\:{J}_{\mathrm{1}} .{J}_{\mathrm{3}} ….{J}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\Rightarrow\:{J}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\mathrm{2}^{{n}} {n}!\:\:{J}_{\mathrm{1}} \\ $$$${J}_{\mathrm{1}} =\:\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\:=−\int_{\mathrm{0}} ^{\infty} \left(−{x}\right)\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$$=−\left[\:\:{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right]_{\mathrm{0}} ^{+\infty} =\:\mathrm{1}\:\Rightarrow\:\:{J}_{\mathrm{2}{n}+\mathrm{1}} =\:\mathrm{2}^{{n}} \:\left({n}!\right)\:. \\ $$

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