advanced-calculus-prove-that-0-1-cos-log-x-1-log-x-dx-log-2-2-adopted-from-youtube-youtube-solution-is-not-considered

Question Number 125133 by mnjuly1970 last updated on 08/Dec/20

\dots ◂ a d v a n c e d c a l c u l u s ▸ \dots

p r o v e t h a t :::

Ω = \int_{0}^{1} {\frac{c o s (l o g (x)) - 1}{l o g (x)}} d x = \frac{l o g (2)}{2}

\dots * a d o p t e d f r o m y o u t u b e * \dots

* * y o u t u b e s o l u t i o n i s n o t c o n s i d e r e d * *

Answered by mathmax by abdo last updated on 08/Dec/20

I = \int_{0}^{1} \frac{\cos (logx) - 1}{logx} we do the chamgement logx = - t \Rightarrow x = e^{- t}

I = \int_{\infty}^{0} \frac{cost - 1}{- t} (- e^{- t}) dt = - \int_{0}^{\infty} \frac{cost - 1}{t} e^{- t} dt

let f (a) = \int_{0}^{\infty} \frac{cost - 1}{t} e^{- at} dt with a > 0 we have

f^{^{'}} (a) = - \int_{0}^{\infty} (cost - 1) e^{- a t} dt = - \int_{0}^{\infty} e^{- a t} cost + \int_{0}^{\infty} e^{- a t} dt

but \int_{0}^{\infty} e^{- a t} dt = {[- \frac{1}{a} e^{- a t}]}_{0}^{\infty} = \frac{1}{a}

\int_{0}^{\infty} e^{- a t} cost dt = Re (\int_{0}^{\infty} e^{- a t + it} dt) snd

\int_{0}^{\infty} e^{(- a + i) t} dt = {[\frac{1}{- a + i} e^{(- a + i) t}]}_{0}^{\infty} = \frac{- 1}{a - i} {- 1} = \frac{1}{a - i} = \frac{a + i}{a^{2} + 1}

\Rightarrow \int_{0}^{\infty} e^{- at} cost dt = \frac{a}{a^{2} + 1} \Rightarrow f^{^{'}} (a) = - \frac{a}{1 + a^{2}} + \frac{1}{a} \Rightarrow

f (a) = lna - \frac{1}{2} \ln (1 + a^{2}) + C = \ln (\frac{a}{\sqrt{a^{2} + 1}}) + c

∣ f (a) ∣⩽ \frac{m}{a} \to 0 (a \to + \infty) \Rightarrow c = 0 \Rightarrow f (a) = \ln (\frac{a}{\sqrt{1 + a^{2}}})

I = - f (1) = - \ln (\frac{1}{\sqrt{2}}) = \ln (\sqrt{2}) = \frac{\ln (2)}{2}

Commented by mnjuly1970 last updated on 08/Dec/20

t h a n k y o u s o m u c h

m r m a x . g r a t e f u l \dots .

Commented by mathmax by abdo last updated on 08/Dec/20

you are welcome sir

Answered by Dwaipayan Shikari last updated on 08/Dec/20

\int_{0}^{1} \frac{c o s (l o g x) - 1}{l o g x} d x

= \int_{- \infty}^{0} e^{t} \frac{c o s (t) - 1}{t} d t l o g x = t \Rightarrow \frac{1}{x} = \frac{d t}{d x}

I (a) = 2 \int_{0}^{\infty} \frac{c o s t - 1}{t} e^{- a t} d t

I^{'} (a) = - \int_{0}^{\infty} {c o s t e}^{- a t} + \int_{0}^{\infty} e^{- a t} d t

= - \frac{1}{2} \int_{0}^{\infty} e^{- t (a - i)} + e^{- t (i + a)} d t + \frac{1}{a}

= - \frac{1}{2} (\frac{1}{a - i} + \frac{1}{i + a}) + \frac{1}{a} = - \frac{a}{a^{2} + 1} + \frac{1}{a}

I (a) = - \frac{1}{2} l o g (a^{2} + 1) + l o g (a) + C

a \to \infty C = 0

I (1) = \frac{1}{2} l o g (2)

Commented by mnjuly1970 last updated on 08/Dec/20

b r a v o m r p a y a n . t h a n k y o u

s o m u c h . .

Answered by mnjuly1970 last updated on 08/Dec/20

s o l u t i o n :

Ω = R e (\int_{0}^{1} \frac{x^{i} - 1}{l o g (x)} d x) = R e (l n (1 + i))

= R e (l n (\sqrt{2} e^{\frac{i π}{4}})) = R e [l n (\sqrt{2}) + i \frac{π}{4}]

∴ Ω = \frac{l n (2)}{2} ✓

Commented by mathmax by abdo last updated on 08/Dec/20

you are taking route 66 in america . .!

Commented by mnjuly1970 last updated on 08/Dec/20

s i n c e r e l y y o u r s

m . n . 1970

Commented by mnjuly1970 last updated on 09/Dec/20

s i r m a x

s i n c e : s o m e i n t e g r a l s a n d p r o b l e m s

w e r e s o l v e d m a n y t i m e t h e r e f o r e i t r i e d

t o s h o r t e n t h e s o l u t i o n .

e x a m p l e :

l n (a + 1) = \int_{0}^{1} (\frac{x^{a} - 1}{l n (x)}) d x, \dots .