Question Number 217855 by yamane last updated on 22/Mar/25
$${i}\:{need}\:{help}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left({e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right){dx}\:\:,\left({n}\in\mathbb{N}\right) \\ $$
Answered by mr W last updated on 23/Mar/25
![if n=0: I_0 =∫_0 ^1 e^(−(x^2 /2)) dx=(√(π/2))erf((1/( (√2)))) if n=1: I_1 =∫_0 ^1 xe^(−(x^2 /2)) dx=−∫_0 ^1 e^(−(x^2 /2)) d(−(x^2 /2)) =[e^(−(x^2 /2)) ]_1 ^0 =1−(1/( (√e))) I_n =∫_0 ^1 x^n e^(−(x^2 /2)) dx =(1/(n+1))∫_0 ^1 e^(−(x^2 /2)) dx^(n+1) =(1/(n+1)){[x^(n+1) e^(−(x^2 /2)) ]_0 ^1 +∫_0 ^1 x^(n+2) e^(−(x^2 /2)) dx} =(1/(n+1))((1/( (√e)))+I_(n+2) ) ⇒I_(n+2) =(n+1)I_n −(1/( (√e))) I_n =(n−1)I_(n−2) −(1/( (√e))) =(n−1)(n−3)I_(n−4) −(n−3)(1/( (√e))) =.... =(n−1)(n−3)...3×1I_0 −(1/( (√e)))[1+3!!+5!!+...+(n−3)!!] I_n =(√(π/2))erf((1/( (√2))))(n−1)!!−(1/( (√e)))Σ_(k=1) ^(n−3) k!! for n=even I_n =(1−(1/( (√e))))(n−1)!!−(1/( (√e)))Σ_(k=2) ^(n−3) k!! for n=odd](https://www.tinkutara.com/question/Q217865.png)
$${if}\:{n}=\mathrm{0}: \\ $$$${I}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}=\sqrt{\frac{\pi}{\mathrm{2}}}{erf}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${if}\:{n}=\mathrm{1}: \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {d}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\left[{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right]_{\mathrm{1}} ^{\mathrm{0}} =\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$$$ \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$$\:\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}^{{n}+\mathrm{1}} \\ $$$$\:\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\left[{x}^{{n}+\mathrm{1}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{2}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\right\} \\ $$$$\:\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\:\sqrt{{e}}}+{I}_{{n}+\mathrm{2}} \right) \\ $$$$\Rightarrow{I}_{{n}+\mathrm{2}} =\left({n}+\mathrm{1}\right){I}_{{n}} −\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$$${I}_{{n}} =\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} −\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$$$\:\:\:\:=\left({n}−\mathrm{1}\right)\left({n}−\mathrm{3}\right){I}_{{n}−\mathrm{4}} −\left({n}−\mathrm{3}\right)\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$$$\:\:\:\:=…. \\ $$$$\:\:\:\:=\left({n}−\mathrm{1}\right)\left({n}−\mathrm{3}\right)…\mathrm{3}×\mathrm{1}{I}_{\mathrm{0}} −\frac{\mathrm{1}}{\:\sqrt{{e}}}\left[\mathrm{1}+\mathrm{3}!!+\mathrm{5}!!+…+\left({n}−\mathrm{3}\right)!!\right] \\ $$$${I}_{{n}} =\sqrt{\frac{\pi}{\mathrm{2}}}{erf}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\left({n}−\mathrm{1}\right)!!−\frac{\mathrm{1}}{\:\sqrt{{e}}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{3}} {\sum}}{k}!!\:\:{for}\:{n}={even} \\ $$$${I}_{{n}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{e}}}\right)\left({n}−\mathrm{1}\right)!!−\frac{\mathrm{1}}{\:\sqrt{{e}}}\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{3}} {\sum}}{k}!!\:\:{for}\:{n}={odd} \\ $$
Answered by Frix last updated on 22/Mar/25
![∫_0 ^1 x^n e^(−(x^2 /2)) dx =^([t=(x^2 /2)]) 2^((n/2)−(1/2)) ∫_0 ^(1/2) t^((n/2)−(1/2)) e^(−t) dt= Let n=2z−1 =2^(z−1) ∫_0 ^(1/2) t^(z−1) e^(−t) dt= This is the lower incomplete Gamma function =2^(z−1) γ(z, (1/2))=(1/((2e)^z )) Σ_(k=0) ^∞ (1/(2^k Π_(j=0) ^k (z+j))); z=((n+1)/2) ...not nice...](https://www.tinkutara.com/question/Q217868.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{{n}} \mathrm{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx}\:\:\overset{\left[{t}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]} {=}\:\:\mathrm{2}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \:\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{t}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{e}^{−{t}} {dt}= \\ $$$$\mathrm{Let}\:{n}=\mathrm{2}{z}−\mathrm{1} \\ $$$$=\mathrm{2}^{{z}−\mathrm{1}} \underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}{t}^{{z}−\mathrm{1}} \mathrm{e}^{−{t}} {dt}= \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:{lower}\:{incomplete}\:{Gamma}\:{function} \\ $$$$=\mathrm{2}^{{z}−\mathrm{1}} \gamma\left({z},\:\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\left(\mathrm{2e}\right)^{{z}} }\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} \underset{{j}=\mathrm{0}} {\overset{{k}} {\prod}}\left({z}+{j}\right)};\:{z}=\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$…\mathrm{not}\:\mathrm{nice}… \\ $$