x-x-dx-

Question Number 6118 by sanusihammed last updated on 15/Jun/16

\int x^{x} d x

Commented by FilupSmith last updated on 15/Jun/16

ANSWER :

x^{x} = e^{x \ln (x)}

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots

∴ e^{x \ln (x)} = 1 + {(x \ln (x))}^{1} + \frac{{(x \ln (x))}^{2}}{2!} + \dots

e^{x \ln x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n} \ln {(x)}^{n}}{n!}

∴ \int x^{x} d x = \int \underset{n = 0}{\sum^{\infty}} \frac{x^{n} \ln {(x)}^{n}}{n!} d x

= \underset{n = 0}{\sum^{\infty}} \int \frac{x^{n} \ln {(x)}^{n}}{n!} d x

= \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int x^{n} \ln {(x)}^{n} d x

let x = e^{- \frac{u}{n + 1}}

∴ d x = - \frac{1}{n + 1} e^{- \frac{u}{n + 1}} d u

= \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int e^{- \frac{u n}{n + 1}} \ln {(e^{- \frac{u}{n + 1}})}^{n} (- \frac{1}{n + 1} e^{- \frac{u}{n + 1}}) d u

= \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int e^{- \frac{u n}{n + 1}} (\frac{{(- u)}^{n}}{{(n + 1)}^{n}}) (- \frac{1}{n + 1} e^{- \frac{u}{n + 1}}) d u

= \underset{n = 0}{\sum^{\infty}} - \frac{1}{n!} \int e^{- (\frac{u n}{n + 1} + \frac{u}{n + 1})} (\frac{{(- 1)}^{n} {(u)}^{n}}{{(n + 1)}^{n}}) \frac{1}{n + 1} d u

\frac{u n}{n + 1} + \frac{u}{n + 1} = \frac{u n + u}{n + 1} = \frac{u (n + 1)}{(n + 1)} = u

= \underset{n = 0}{\sum^{\infty}} - \frac{{(- 1)}^{n}}{n!} \int e^{- u} \frac{{(u)}^{n}}{{(n + 1)}^{n + 1}} d u

∴ \int x^{x} d x = \underset{n = 0}{\sum^{\infty}} (- \frac{1}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int e^{- u} u^{n} d u)

f r o m Q 6042

\int e^{- u} u^{n} d u = C n! - e^{- u} n! (\underset{r = 0}{\sum^{n}} \frac{u^{n - r}}{(n - r)!}), C = constant

Commented by FilupSmith last updated on 15/Jun/16

a n i n t e r e s t i n g / s p e c i a l r e s u l t :

\int_{0}^{1} x^{x} d x = \underset{n = 0}{\sum^{\infty}} (- \frac{1}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int_{a}^{b} e^{- u} u^{n} d u)

x = \exp (- \frac{u}{n + 1})

\Rightarrow u = - (n + 1) \ln (x)

a = - (n + 1) \ln (0)

∴ a = - (- \infty) = \infty

b = - (n + 1) \ln (1)

∴ b = 0

∴ \int_{0}^{1} x^{x} d x = \underset{n = 0}{\sum^{\infty}} (- \frac{1}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int_{\infty}^{^{0}} e^{- u} u^{n} d u)

= \underset{n = 0}{\sum^{\infty}} (+ \frac{1}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)} \int_{0}^{^{\infty}} e^{- u} u^{n} d u)

\int_{0}^{^{\infty}} e^{- u} u^{n} d u = n!

= \underset{n = 0}{\sum^{\infty}} (\frac{n!}{n!} {(- 1)}^{n} {(n + 1)}^{- (n + 1)})

∴ \int_{0}^{1} x^{x} d x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {(n + 1)}^{- (n + 1)}

Commented by sanusihammed last updated on 15/Jun/16

T h a n k s s o m u c h .