Prove-that-0-1-x-a-1-log-x-dx-log-a-1-

Question Number 45117 by rahul 19 last updated on 09/Oct/18

P r o v e t h a t \int_{0}^{1} \frac{x^{a} - 1}{\log x} d x = \log (a + 1) .

Commented by maxmathsup by imad last updated on 09/Oct/18

f o r a > - 1 l e t f (a) = \int_{0}^{1} \frac{x^{a} - 1}{l n (x)} d x c h a n g e m e n t l n (x) = - t g i v e

f (a) = - \int_{0}^{+ \infty} \frac{e^{- a t} - 1}{- t} (- 1) e^{- t} d t = - \int_{0}^{\infty} \frac{e^{- (a + 1) t} - e^{- t}}{t} d t

= \int_{0}^{\infty} \frac{e^{- t} - e^{- (a + 1) t}}{t} d t \Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} e^{- (a + 1) t} d t = {[- \frac{1}{a + 1} e^{- (a + 1) t}]}_{t = 0}^{+ \infty}

= \frac{1}{a + 1} \Rightarrow f (a) = l n ∣ a + 1 ∣ + c = l n (a + 1) + c b u t

f (0) = 0 = l n (1) + c \Rightarrow c = 0 \Rightarrow f (a) = l n (a + 1) .

Commented by rahul 19 last updated on 09/Oct/18

thanks prof Abdo

Commented by turbo msup by abdo last updated on 09/Oct/18

y o u a r e w e l c o m e s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

I (a) = \int_{0}^{1} \frac{x^{a} - 1}{l n x} d x

\frac{d I}{d a} = \int_{0}^{1} \frac{1}{l n x} \frac{\partial}{\partial a} (x^{a} - 1) d x

= \int_{0}^{1} \frac{1}{l n x} \times x^{a} \times l n x d x

= \int_{0}^{1} x^{a} d x

=∣ \frac{x^{a + 1}}{a + 1} ∣_{0}^{1}

= \frac{1}{a + 1}

\int d I = \int \frac{d a}{a + 1} I = l n (a + 1)

Commented by rahul 19 last updated on 09/Oct/18

Sir , how will this come to my mind first we have to differentiate!! ��

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

i s h a l l p o s t s o m e e x a m p l e \dots t h i s t y p e i n t r e g a l

a r e a d v a n c e d i n t r e g a l \dots f r o m b o o k i s h a l l p o s t . .

Commented by rahul 19 last updated on 12/Oct/18

S i r, p l s s p o s t 2 - 3 e x a m p l e s \dots . .