... ◂advanced calculus▶... prove that ::: Ω=∫_0 ^( 1) {((cos(log(x))−1)/(log(x)))}dx=((log(2))/2) ...∗adopted from youtube∗... ∗ ∗ youtube solution is not considered ∗ ∗


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Question Number 125133 by mnjuly1970 last updated on 08/Dec/20
$... ◂advanced calculus▶... prove that ::: Ω=∫_0 ^( 1) {((cos(log(x))−1)/(log(x)))}dx=((log(2))/2) ...∗adopted from youtube∗... ∗ ∗ youtube solution is not considered ∗ ∗$
$. . . ◂ a d v a n c e d c a l c u l u s ▸ . . .$ $p r o v e t h a t :::$ $Ω = \int_{0}^{1} {\frac{c o s (l o g (x)) - 1}{l o g (x)}} d x = \frac{l o g (2)}{2}$ $. . . * a d o p t e d f r o m y o u t u b e * . . .$ $* * y o u t u b e s o l u t i o n i s n o t c o n s i d e r e d * *$
	Answered by mathmax by abdo last updated on 08/Dec/20
	$I=∫_0 ^1 ((cos(logx)−1)/(logx)) we do the chamgement logx=−t ⇒x=e^(−t) I=∫_∞ ^0 ((cost−1)/(−t))(−e^(−t) )dt =−∫_0 ^∞ ((cost−1)/t)e^(−t) dt let f(a) =∫_0 ^∞ ((cost−1)/t)e^(−at) dt with a>0 we have f^′ (a)=−∫_0 ^∞ (cost−1)e^(−at) dt =−∫_0 ^∞ e^(−at) cost +∫_0 ^∞ e^(−at) dt but ∫_0 ^∞ e^(−at) dt =[−(1/a)e^(−at) ]_0 ^∞ =(1/a) ∫_0 ^∞ e^(−at) cost dt =Re(∫_0 ^∞ e^(−at+it) dt) snd ∫_0 ^∞ e^((−a+i)t) dt =[(1/(−a+i))e^((−a+i)t) ]_0 ^∞ =((−1)/(a−i)){−1} =(1/(a−i))=((a+i)/(a^2 +1)) ⇒∫_0 ^∞ e^(−at) cost dt =(a/(a^2 +1)) ⇒f^′ (a)=−(a/(1+a^2 ))+(1/a) ⇒ f(a)=lna−(1/2)ln(1+a^2 )+C =ln((a/( (√(a^2 +1))))) +c ∣f(a)∣≤(m/a)→0 (a→+∞) ⇒c=0 ⇒f(a)=ln((a/( (√(1+a^2 ))))) I =−f(1)=−ln((1/( (√2))))=ln((√2))=((ln(2))/2)$
	$I = \int_{0}^{1} \frac{\cos (logx) - 1}{logx} we do the chamgement logx = - t \Rightarrow x = e^{- t}$ $I = \int_{\infty}^{0} \frac{cost - 1}{- t} (- e^{- t}) dt = - \int_{0}^{\infty} \frac{cost - 1}{t} e^{- t} dt$ $let f (a) = \int_{0}^{\infty} \frac{cost - 1}{t} e^{- at} dt with a > 0 we have$ $f^{^{'}} (a) = - \int_{0}^{\infty} (cost - 1) e^{- a t} dt = - \int_{0}^{\infty} e^{- a t} cost + \int_{0}^{\infty} e^{- a t} dt$ $but \int_{0}^{\infty} e^{- a t} dt = {[- \frac{1}{a} e^{- a t}]}_{0}^{\infty} = \frac{1}{a}$ $\int_{0}^{\infty} e^{- a t} cost dt = Re (\int_{0}^{\infty} e^{- a t + it} dt) snd$ $\int_{0}^{\infty} e^{(- a + i) t} dt = {[\frac{1}{- a + i} e^{(- a + i) t}]}_{0}^{\infty} = \frac{- 1}{a - i} {- 1} = \frac{1}{a - i} = \frac{a + i}{a^{2} + 1}$ $\Rightarrow \int_{0}^{\infty} e^{- at} cost dt = \frac{a}{a^{2} + 1} \Rightarrow f^{^{'}} (a) = - \frac{a}{1 + a^{2}} + \frac{1}{a} \Rightarrow$ $f (a) = lna - \frac{1}{2} \ln (1 + a^{2}) + C = \ln (\frac{a}{\sqrt{a^{2} + 1}}) + c$ $∣ f (a) ∣⩽ \frac{m}{a} \to 0 (a \to + \infty) \Rightarrow c = 0 \Rightarrow f (a) = \ln (\frac{a}{\sqrt{1 + a^{2}}})$ $I = - f (1) = - \ln (\frac{1}{\sqrt{2}}) = \ln (\sqrt{2}) = \frac{\ln (2)}{2}$
	Commented by mnjuly1970 last updated on 08/Dec/20

	$t h a n k y o u s o m u c h$ $m r m a x . g r a t e f u l . . . .$
	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

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