∫(x)^(1/x) dx=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 111558 by Study last updated on 04/Sep/20

$\int \sqrt[x]{x} d x = ?$
Commented by Her_Majesty last updated on 04/Sep/20

$w e c a n n o t s o l v e \int x^{x} d x, \int \sqrt[x]{x} d x$
Commented by bemath last updated on 04/Sep/20

$w a w . . . t h i s v e r y h a r d$
Commented by mathdave last updated on 04/Sep/20

$w e c a n s o l v e t h e m i t v e r y v e r y s i m p l e$
Commented by Her_Majesty last updated on 04/Sep/20

$s h o w u s$
Commented by mathdave last updated on 04/Sep/20

$l e t m e s t a r t b y s o l v i n g \int x^{x} d x b e f o r e$ $c o m i n g t o s o l v e \int \sqrt[x]{x} d x$ $s o l u t i o n t o \int x^{x} d x$ $l e t I = \int x^{x} d x = \int e^{x \ln x} d x$ $b u t t h e s e r i e s t e r m o f$ $e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} a n d t h a t o f$ $e^{x \ln x} = \underset{n = 0}{\sum^{\infty}} \frac{{(x \ln x)}^{n}}{n!}$ $I = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int {(x \ln x)}^{n} d x$ $l e t y = - \ln x, x = e^{- y} a n d d x = - e^{- y} d y$ $n o w w e h a v e$ $I = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int {(e^{- y})}^{n} {(- y)}^{n} (- e^{- y}) d y$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} \int e^{- n y} e^{- y} {(- 1)}^{n} {(y)}^{n} d y$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int y^{n} e^{- y (n + 1)} d y$ $l e t z = y (n + 1), y = \frac{z}{n + 1} a n d d y = \frac{d z}{n + 1}$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int {(\frac{z}{n + 1})}^{n} e^{- z} \times \frac{d z}{n + 1}$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n!} \int \frac{z^{n} e^{- z}}{{(1 + n)}^{n} (1 + n)} d z$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! {(1 + n)}^{n + 1}} \int z^{n} e^{- z} d z$ $f r o m g a m m a f u n c t i o n$ $Γ (l, x) = \int x^{l - 1} e^{- x} d$ $w h e n c o m p a r e t h i s t o t h e o r i g i n a l$ $f u n c t i o n w e h a v e$ $l - 1 = n t h e n l = (n + 1) a n d z = x s o$ $Γ (n + 1, z) = \int z^{n} e^{- z} d x$ $b u t z = y (n + 1) a n d y = - \ln x f i n a l l y$ $z = - (k + 1) \ln x$ $i f p l u g g i n t h e s e w e h a v e$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! {(n + 1)}^{n + 1}} \times - Γ [n + 1, (n + 1) \ln x]$ $∵ \int x^{x} d x = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n! {(n + 1)}^{n + 1}} Γ [n + 1, (n + 1) \ln x]$ $b y m a t h d a v e (04 / 09 / 2020)$
Commented by Her_Majesty last updated on 04/Sep/20

$t h a n k y o u!$ ${i t}^{'} s g r e a t b u t n o t v e r y v e r y s i m p l e t o e v e r y b o d y . . .$
Commented by mathdave last updated on 04/Sep/20

$u a r e w e l c o m e u r m a j e s t y$
Commented by bemath last updated on 04/Sep/20

$w a w . . a m a z i n g . . . t h i s i s s u p e r v e r y s i m p l e$
Commented by Tawa11 last updated on 06/Sep/21

$great sir$
	Answered by mathdave last updated on 04/Sep/20

	$f o l l o w t h i s p s t t e r n t o g e t r e s u l t f o r$ $\int \sqrt[x]{x} d x$ $b y s a y i n g l e t$ $I = \int \sqrt[x]{x} d x = \int x^{\frac{1}{x}} d x = \int e^{\frac{1}{x} \ln x} d x$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum