calculate-A-p-0-1-x-p-1-ln-x-dx-with-p-gt-0-

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 by tanmay.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 \dots

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 \dots

Commented by maxmathsup 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