prove that ln(z) = ∫_0 ^1 ((z−1)/(1+t(z−1)))dt .


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Question Number 54371 by maxmathsup by imad last updated on 02/Feb/19

$p r o v e t h a t l n (z) = \int_{0}^{1} \frac{z - 1}{1 + t (z - 1)} d t .$
Commented by maxmathsup by imad last updated on 02/Feb/19

$z i s a c o m p l e x n u m b e r .$
	Answered by Smail last updated on 02/Feb/19

	$l e t u = t (z - 1) \Rightarrow d u = (z - 1) d t$ $\int_{0}^{1} \frac{z - 1}{1 + t (z - 1)} d t = \int_{0}^{z - 1} \frac{d u}{1 + u} = {[l n ∣ 1 + u ∣]}_{0}^{z - 1}$ $= l n ∣ 1 + z - 1 ∣= l n ∣ z ∣$
	Commented by Abdo msup. last updated on 03/Feb/19

	$s i r w h a t d o y o u w o r k i n u s a . . .$
	Commented by Smail last updated on 03/Feb/19

	$I a m s t i l l s t u d y i n g a n d m y m a j o r i s$ $e l e c t r i c a l e n g i n e e r i n g$
	Commented by rahul 19 last updated on 03/Feb/19

	$w o w! w h i c h u n i v e r s i t y ?$
	Commented by maxmathsup by imad last updated on 03/Feb/19

	$g o o d l u c k s i r S m a i l . . . i h o p p e t o v i s i t u s a s o m e d a y s . . .$
	Commented by Smail last updated on 03/Feb/19

	$T h a n k y o u$
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