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Question Number 117148 by bemath last updated on 10/Oct/20

∫ x^6 e^(−4x^2 ) dx =?

$$\:\:\:\int\:\mathrm{x}^{\mathrm{6}} \:\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } \:\mathrm{dx}\:=? \\ $$

Answered by Olaf last updated on 10/Oct/20

I = ∫(−(1/8)x^5 )(−8xe^(−4x^2 ) )dx I = −(x^5 /8)e^(−4x^2 ) +(5/8)∫x^4 e^(−4x^2 ) dx ...and three other times by successive parts : I = −(x/8)(x^4 +((5x^2 )/8)+((15)/(64)))e^(−4x^2 ) +((15(√π)erf(2x))/(2048))

$$ \\ $$$$\mathrm{I}\:=\:\int\left(−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{5}} \right)\left(−\mathrm{8}{xe}^{−\mathrm{4}{x}^{\mathrm{2}} } \right){dx} \\ $$$$\mathrm{I}\:=\:−\frac{{x}^{\mathrm{5}} }{\mathrm{8}}{e}^{−\mathrm{4}{x}^{\mathrm{2}} } +\frac{\mathrm{5}}{\mathrm{8}}\int{x}^{\mathrm{4}} {e}^{−\mathrm{4}{x}^{\mathrm{2}} } {dx} \\ $$$$...\mathrm{and}\:\mathrm{three}\:\mathrm{other}\:\mathrm{times}\:\mathrm{by}\:\mathrm{successive}\:\mathrm{parts}\:: \\ $$$$\mathrm{I}\:=\:−\frac{{x}}{\mathrm{8}}\left({x}^{\mathrm{4}} +\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{15}}{\mathrm{64}}\right){e}^{−\mathrm{4}{x}^{\mathrm{2}} } +\frac{\mathrm{15}\sqrt{\pi}\mathrm{erf}\left(\mathrm{2}{x}\right)}{\mathrm{2048}} \\ $$

Commented by bemath last updated on 10/Oct/20

erf = error function?

$$\mathrm{erf}\:=\:\mathrm{error}\:\mathrm{function}? \\ $$

Commented by Olaf last updated on 10/Oct/20

yes sir. Gauss error function. erf(x) = (2/( (√π)))∫_0 ^x e^(−t^2 ) dt

$$\mathrm{yes}\:\mathrm{sir}. \\ $$$$\mathrm{Gauss}\:\mathrm{error}\:\mathrm{function}. \\ $$$$\mathrm{erf}\left({x}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$

Answered by 1549442205PVT last updated on 10/Oct/20

Put 4x^2 =t⇒8xdx=dt⇒dx=(dt/(4(√t))) ∫ x^6 e^(−4x^2 ) dx =∫((t/4))^3 e^(−t) ×(dt/(4(√t))) =(1/2^8 )∫t^(5/2) e^(−t) dt=−(1/2^8 )∫t^(5/2) de^(−t) =−(1/2^8 )t^(5/2) e^(−t) −(1/2^8 )∫(5/2)t^(3/2) de^(−t) =−(1/2^8 )t^(5/2) e^(−t) −(1/2^8 )×(5/2)t^(3/2) e^(−t) −(5/2^9 )×(3/2)(√t) e^(−t) +((15)/2^(10) )∫e^(−t) (1/(2(√t)))dt= Since e^(−t) (1/(2(√t))) dt=e^(−4x^2 ) ×(1/(2(√t)))×4(√t) dx=2e^(−4x^2 ) dx ⇒∫e^(−t) (√t) dt=∫e^((−2x)^2 ) d(2x) =(π/2)×(2/π)∫e^((−2x)^2 ) d(2x)=((√π)/2)erf(2x) t^(5/2) =(2x)^5 =32x^5 ,t^(3/2) =(2x)^3 =8x^4 We get: F=(1/2^8 )e^(−4x^2 ) (−32x^5 −20x^3 −((15)/2)x) +((15)/2^(10) )∫(1/(2(√t)))e^(−4x^2 ) dx =((−e^(−4x^2 ) )/(512))(64x^5 +40x^2 +15x)+((15)/(2048))erf(2x)

$$\mathrm{Put}\:\mathrm{4x}^{\mathrm{2}} =\mathrm{t}\Rightarrow\mathrm{8xdx}=\mathrm{dt}\Rightarrow\mathrm{dx}=\frac{\mathrm{dt}}{\mathrm{4}\sqrt{\mathrm{t}}} \\ $$$$\:\:\:\int\:\mathrm{x}^{\mathrm{6}} \:\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } \:\mathrm{dx}\:=\int\left(\frac{\mathrm{t}}{\mathrm{4}}\right)^{\mathrm{3}} \mathrm{e}^{−\mathrm{t}} ×\frac{\mathrm{dt}}{\mathrm{4}\sqrt{\mathrm{t}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\int\mathrm{t}^{\mathrm{5}/\mathrm{2}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\int\mathrm{t}^{\mathrm{5}/\mathrm{2}} \mathrm{de}^{−\mathrm{t}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\mathrm{t}^{\mathrm{5}/\mathrm{2}} \mathrm{e}^{−\mathrm{t}} −\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\int\frac{\mathrm{5}}{\mathrm{2}}\mathrm{t}^{\mathrm{3}/\mathrm{2}} \mathrm{de}^{−\mathrm{t}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\mathrm{t}^{\mathrm{5}/\mathrm{2}} \mathrm{e}^{−\mathrm{t}} −\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }×\frac{\mathrm{5}}{\mathrm{2}}\mathrm{t}^{\mathrm{3}/\mathrm{2}} \mathrm{e}^{−\mathrm{t}} −\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{9}} }×\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \\ $$$$+\frac{\mathrm{15}}{\mathrm{2}^{\mathrm{10}} }\int\mathrm{e}^{−\mathrm{t}} \frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}}}\mathrm{dt}= \\ $$$$\mathrm{Since}\:\mathrm{e}^{−\mathrm{t}} \frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}}}\:\mathrm{dt}=\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } ×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}}}×\mathrm{4}\sqrt{\mathrm{t}}\:\mathrm{dx}=\mathrm{2e}^{−\mathrm{4x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\Rightarrow\int\mathrm{e}^{−\mathrm{t}} \sqrt{\mathrm{t}}\:\mathrm{dt}=\int\mathrm{e}^{\left(−\mathrm{2x}\right)^{\mathrm{2}} } \mathrm{d}\left(\mathrm{2x}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}×\frac{\mathrm{2}}{\pi}\int\mathrm{e}^{\left(−\mathrm{2x}\right)^{\mathrm{2}} } \mathrm{d}\left(\mathrm{2x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erf}\left(\mathrm{2x}\right) \\ $$$$\mathrm{t}^{\mathrm{5}/\mathrm{2}} =\left(\mathrm{2x}\right)^{\mathrm{5}} =\mathrm{32x}^{\mathrm{5}} ,\mathrm{t}^{\mathrm{3}/\mathrm{2}} =\left(\mathrm{2x}\right)^{\mathrm{3}} =\mathrm{8x}^{\mathrm{4}} \\ $$$$\mathrm{We}\:\mathrm{get}: \\ $$$$\mathrm{F}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{8}} }\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } \left(−\mathrm{32x}^{\mathrm{5}} −\mathrm{20x}^{\mathrm{3}} −\frac{\mathrm{15}}{\mathrm{2}}\mathrm{x}\right) \\ $$$$+\frac{\mathrm{15}}{\mathrm{2}^{\mathrm{10}} }\int\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{t}}}\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\frac{−\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}} } }{\mathrm{512}}\left(\mathrm{64x}^{\mathrm{5}} +\mathrm{40x}^{\mathrm{2}} +\mathrm{15x}\right)+\frac{\mathrm{15}}{\mathrm{2048}}\mathrm{erf}\left(\mathrm{2x}\right) \\ $$

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