calculate A_p =∫_0 ^1 ((x^p −1)/(ln(x)))dx with p>0.


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Question Number 47850 by maxmathsup by imad last updated on 15/Nov/18

$c a l c u l a t e A_{p} = \int_{0}^{1} \frac{x^{p} - 1}{l n (x)} d x w i t h p > 0 .$
Commented bytanmay.chaudhury50@gmail.com last updated on 16/Nov/18

$i h a v e t r i e d f o l l o w i n g w a y$ $A_{p} = \int_{0}^{1} \frac{x^{p} - 1}{l n x} d x$ $\frac{{d A}_{p}}{d p} = \int_{0}^{1} \frac{\partial}{\partial p} (\frac{x^{p} - 1}{l n x}) d x$ $= \int_{0}^{1} \frac{x^{p} l n x}{l n x} d x$ $= \int_{0}^{1} x^{p} d x$ $=∣ \frac{x^{p + 1}}{p + 1} ∣_{0}^{1} = \frac{1}{p + 1}$ ${d A}_{p} = \frac{d p}{p + 1}$ $\int {d A}_{p} = \int \frac{d p}{p + 1}$ $A_{p} = l n (p + 1) + c$ $n o w i n q u e s t i o n p > 0$ $s o h e r e i s t o p p e d f u r t h e r . . .$ $b u t i f p = 0 a s s u m e d$ $t h e n \int_{0}^{1} \frac{x^{p} - 1}{l n x} d x$ $\int_{0}^{1} \frac{x^{0} - 1}{l n x} d x = 0 = A_{p = 0}$ $t h e n$ $A_{p} = l n (p + 1) + c$ $A_{p = 0} = l n (0 + 1) + c$ $0 = 0 + c$ $c = 0$ $A_{p} = l n (p + 1)$ $s o A_{p} = \int_{0}^{1} \frac{x^{p} - 1}{l n x} d x = l n (p + 1)$ $p l s c h e c k s i r . . .$
Commented bymaxmathsup by imad last updated on 17/Nov/18

$c h a n g e m e n t l n (x) = - t g i v e A_{p} = - \int_{0}^{\infty} \frac{e^{- p t} - 1}{- t} (- e^{- t}) d t$ $= - \int_{0}^{\infty} \frac{e^{- (p + 1) t} - e^{- t}}{t} d t = \int_{0}^{\infty} \frac{e^{- t} - e^{- (p + 1) t}}{t} d t l e t$ $f (p) = \int_{0}^{\infty} \frac{e^{- t} - e^{- (p + 1) t}}{t} d t \Rightarrow f^{^{'}} (p) = \int_{0}^{\infty} \frac{\partial}{\partial p} (\frac{e^{- t} - e^{- t} e^{- p t}}{t}) d t$ $= \int_{0}^{\infty} \frac{- e^{- t} (- t) e^{- p t}}{t} d t = \int_{0}^{\infty} e^{- (p + 1) t} d t = {[- \frac{1}{p + 1} e^{- (p + 1) t}]}_{t = 0}^{\infty}$ $= \frac{1}{p + 1} \Rightarrow f (p) = l n (p + 1) + λ b u t λ = {l i m}_{x \to 0} (f (p) - l n (p + 1)) = 0 \Rightarrow$ $A_{p} = f (p) = l n (p + 1)$ $★ \int_{0}^{1} \frac{x^{p} - 1}{l n (x)} d x = l n (p + 1) ★ w i t h p > 0 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Nov/18

	$w e k n o w$ $\int_{0}^{1} x^{k} d x =∣ \frac{x^{k + 1}}{k + 1} ∣_{0}^{1} = \frac{1}{k + 1}$ $n o w i n t r e g a t e b o t h s i d e w . r . t k i n t h e i n t e r v a l$ $s a y q t o p$ $\int_{q}^{p} d k [\int_{0}^{1} x^{k} d x] = \int_{q}^{p} \frac{d k}{1 + k}$ $t a k i n g h e l p f r o m a d v a n c e d l e v e l i n t r e g a t i o n$ $\int_{0}^{1} d x \int_{q}^{p} x^{k} d k =∣ l n (1 + k) ∣_{q}^{p}$ $\int_{0}^{1} d x [∣ \frac{x^{k}}{l n x} ∣_{q}^{p}] = l n (\frac{1 + p}{1 + q})$ $\int_{0}^{1} \frac{x^{p} - x^{q}}{l n x} d x = l n (\frac{1 + p}{1 + q}) = l n (1 + p) - l n (1 + q)$ $n o w p u t q = 0 b o t h s i d e$ $\int_{0}^{1} \frac{x^{p} - 1}{l n x} d x = l n (1 + p) a n s$
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