x-6-e-4x-2-dx-

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

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

Answered by Olaf last updated on 10/Oct/20

I = \int (- \frac{1}{8} x^{5}) (- 8 {x e}^{- 4 x^{2}}) d x

I = - \frac{x^{5}}{8} e^{- 4 x^{2}} + \frac{5}{8} \int x^{4} e^{- 4 x^{2}} d x

\dots and three other times by successive parts :

I = - \frac{x}{8} (x^{4} + \frac{5 x^{2}}{8} + \frac{15}{64}) e^{- 4 x^{2}} + \frac{15 \sqrt{π} \erf (2 x)}{2048}

Commented by bemath last updated on 10/Oct/20

\erf = error function ?

Commented by Olaf last updated on 10/Oct/20

yes sir .

Gauss error function .

\erf (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t

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

Put {4 x}^{2} = t \Rightarrow 8 xdx = dt \Rightarrow dx = \frac{dt}{4 \sqrt{t}}

\int x^{6} e^{- {4 x}^{2}} dx = \int {(\frac{t}{4})}^{3} e^{- t} \times \frac{dt}{4 \sqrt{t}}

= \frac{1}{2^{8}} \int t^{5 / 2} e^{- t} dt = - \frac{1}{2^{8}} \int t^{5 / 2} {de}^{- t}

= - \frac{1}{2^{8}} t^{5 / 2} e^{- t} - \frac{1}{2^{8}} \int \frac{5}{2} t^{3 / 2} {de}^{- t}

= - \frac{1}{2^{8}} t^{5 / 2} e^{- t} - \frac{1}{2^{8}} \times \frac{5}{2} t^{3 / 2} e^{- t} - \frac{5}{2^{9}} \times \frac{3}{2} \sqrt{t} e^{- t}

+ \frac{15}{2^{10}} \int e^{- t} \frac{1}{2 \sqrt{t}} dt =

Since e^{- t} \frac{1}{2 \sqrt{t}} dt = e^{- {4 x}^{2}} \times \frac{1}{2 \sqrt{t}} \times 4 \sqrt{t} dx = {2 e}^{- {4 x}^{2}} dx

\Rightarrow \int e^{- t} \sqrt{t} dt = \int e^{{(- 2 x)}^{2}} d (2 x)

= \frac{π}{2} \times \frac{2}{π} \int e^{{(- 2 x)}^{2}} d (2 x) = \frac{\sqrt{π}}{2} \erf (2 x)

t^{5 / 2} = {(2 x)}^{5} = {32 x}^{5}, t^{3 / 2} = {(2 x)}^{3} = {8 x}^{4}

We get :

F = \frac{1}{2^{8}} e^{- {4 x}^{2}} (- {32 x}^{5} - {20 x}^{3} - \frac{15}{2} x)

+ \frac{15}{2^{10}} \int \frac{1}{2 \sqrt{t}} e^{- {4 x}^{2}} dx

= \frac{- e^{- {4 x}^{2}}}{512} ({64 x}^{5} + {40 x}^{2} + 15 x) + \frac{15}{2048} \erf (2 x)