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Question-130317




Question Number 130317 by stelor last updated on 24/Jan/21
Answered by mathmax by abdo last updated on 24/Jan/21
I=∫_(−∞) ^(+∞)  x e^(−(x^2 /2)) dx  we do the changement (x/( (√2)))=t ⇒  I=∫_(−∞) ^(+∞)  (√2)t e^(−t^2 ) (√2)dt =2∫_(−∞) ^(+∞)  t e^(−t^2 ) dt =[−e^(−t^2 ) ]_(−∞) ^(+∞)  =0  also we can see that x→xe^(−((x2)/2))  is odd  J=∫_(−∞) ^(+∞)  x^2  e^(−(x^2 /2)) dx  =_((x/( (√2)))=t)    ∫_(−∞) ^(+∞)  2t^2  e^(−t^2 ) (√2)dt  =2(√2)∫_(−∞) ^(+∞)  t^2  e^(−t^2 ) dt   but  we get by parts  ∫_(−∞) ^(+∞)  t(te^(−t^2 ) )dt =[−(1/2)e^(−t^2 ) .t]_(−∞) ^(+∞) +∫_(−∞) ^(+∞)  (1/2)e^(−t^2 ) dt  =(1/2)∫_(−∞) ^(+∞)  e^(−t^2 ) dt =((√π)/2) ⇒J=2(√2)×((√π)/2)=(√(2π))
$$\mathrm{I}=\int_{−\infty} ^{+\infty} \:\mathrm{x}\:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}}}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}=\int_{−\infty} ^{+\infty} \:\sqrt{\mathrm{2}}\mathrm{t}\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \sqrt{\mathrm{2}}\mathrm{dt}\:=\mathrm{2}\int_{−\infty} ^{+\infty} \:\mathrm{t}\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:=\left[−\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \right]_{−\infty} ^{+\infty} \:=\mathrm{0} \\ $$$$\mathrm{also}\:\mathrm{we}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that}\:\mathrm{x}\rightarrow\mathrm{xe}^{−\frac{\mathrm{x2}}{\mathrm{2}}} \:\mathrm{is}\:\mathrm{odd} \\ $$$$\mathrm{J}=\int_{−\infty} ^{+\infty} \:\mathrm{x}^{\mathrm{2}} \:\mathrm{e}^{−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{dx}\:\:=_{\frac{\mathrm{x}}{\:\sqrt{\mathrm{2}}}=\mathrm{t}} \:\:\:\int_{−\infty} ^{+\infty} \:\mathrm{2t}^{\mathrm{2}} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \sqrt{\mathrm{2}}\mathrm{dt} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\mathrm{t}^{\mathrm{2}} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:\:\:\mathrm{but}\:\:\mathrm{we}\:\mathrm{get}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int_{−\infty} ^{+\infty} \:\mathrm{t}\left(\mathrm{te}^{−\mathrm{t}^{\mathrm{2}} } \right)\mathrm{dt}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } .\mathrm{t}\right]_{−\infty} ^{+\infty} +\int_{−\infty} ^{+\infty} \:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow\mathrm{J}=\mathrm{2}\sqrt{\mathrm{2}}×\frac{\sqrt{\pi}}{\mathrm{2}}=\sqrt{\mathrm{2}\pi} \\ $$
Commented by stelor last updated on 25/Jan/21
merci    cool.^
$$\mathrm{merci}\:\:\:\:\mathrm{cool}\overset{} {.} \\ $$

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