Can-you-solve-the-indefinite-integral-e-u-u-n-du-

Question Number 6042 by FilupSmith last updated on 10/Jun/16

Can you solve the indefinite integral :

\int e^{- u} u^{n} d u

Commented by Yozzii last updated on 11/Jun/16

D e f i n e I (n) = \int e^{- u} u^{n} d u (n ⩾ 0) .

∴ I n t e g r a t i n g b y p a r t s g i v e s

I (n) = - e^{- u} u^{n} - \int - {n e}^{- u} u^{n - 1} d u

I (n) = - e^{- u} u^{n} + n \int e^{- u} u^{n - 1} d u

I (n) = - e^{- u} u^{n} + n I (n - 1)

I (n) = - e^{- u} u^{n} - {n e}^{- u} u^{n - 1} + n (n - 1) I (n - 2)

I (n) = - e^{- u} u^{n} - {n e}^{- u} u^{n - 1} - n (n - 1) e^{- u} u^{n - 2} + n (n - 1) (n - 2) I (n - 3) .

I (n) = - e^{- u} (\frac{n!}{(n - 0)!} u^{n - 0} + \frac{n!}{(n - 1)!} u^{n - 1} + \frac{n!}{(n - 2)!} u^{n - 2} + \frac{n!}{(n - 3)!} u^{n - 3} + \dots + \frac{n!}{2!} u^{2}) + \frac{n!}{1!} I (1)

I (1) = \int e^{- u} u d u = - e^{- u} u + \int e^{- u} d u = - e^{- u} u - e^{- u} + C .

I (n) = - e^{- u} (\frac{n!}{(n - 0)!} u^{n - 0} + \frac{n!}{(n - 1)!} u^{n - 1} + \frac{n!}{(n - 3)!} u^{n - 3} + \dots + \frac{n!}{2!} u^{2} + \frac{n!}{1!} u^{1} + \frac{n!}{0!} u^{0}) + D

I (n) = D - e^{- u} n! (\underset{r = 0}{\sum^{n}} \frac{u^{n - r}}{(n - r)!})

Commented by FilupSmith last updated on 11/Jun/16

is a D constant ?

Commented by Yozzii last updated on 11/Jun/16

D = n! C w h e r e n = c o n s t a n t

a n d C = c o n s t a n t . Y o u g e t t h a t

a f t e r s u b s t i t u t i n g t h e e x p r e s s i o n f o r I (1) .

Commented by FilupSmith last updated on 11/Jun/16

I see, thank you!