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

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

Answered by Olaf last updated on 10/Oct/20

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

Commented by bemath last updated on 10/Oct/20

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

Commented by Olaf last updated on 10/Oct/20

$$\mathrm{yes}\mathrm{sir}.$$$$\mathrm{Gauss}\mathrm{error}\mathrm{function}.$$$$\mathrm{erf}\left(x\right)=\frac{2}{\sqrt{\pi }}{\int }_0^x{e}^{-t^2}dt$$

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

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

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