... nice calculus... prove that::: ∫_0 ^( 1) {((1+(1−x)^(1/2) )/x) +(2/(ln(1−x)))}dx =^(???) 2(γ−1+log(2))


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 124064 by mnjuly1970 last updated on 30/Nov/20
$... nice calculus... prove that::: ∫_0 ^( 1) {((1+(1−x)^(1/2) )/x) +(2/(ln(1−x)))}dx =^(???) 2(γ−1+log(2))$
$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t :::$ $\int_{0}^{1} {\frac{1 + {(1 - x)}^{\frac{1}{2}}}{x} + \frac{2}{l n (1 - x)}} d x$ $\overset{? ? ?}{=} 2 (γ - 1 + l o g (2))$
	Answered by mindispower last updated on 01/Dec/20
	$let ln(1−x)=−t ⇔∫_0 ^∞ {((1+e^(−(t/2)) )/(1−e^(−t) ))−(2/t)}e^(−t) dt =∫_0 ^∞ −((2e^(−t) )/t)+((1+e^(−(t/2)) )/(1−e^(−t) ))e^(−t) dt =−2∫{(e^(−t) /t)−(1/2)((1+e^(−(t/2)) )/(1−e^(−t) ))e^(−t) }dt =−2∫(e^(−t) /t)−(e^(−3(t/(2 ))) /(1−e^(−t) ))+(1/2).((e^(−(t/2)) −1)/(1−e^(−t) )).e^(−t) dt =−2∫_0 ^∞ {(e^(−t) /t)−(e^(−3(t/2)) /(1−e^(−t) ))}dt+∫_0 ^∞ ((1−e^(−(t/2)) )/(1−e^(−t) ))e^(−t) dt Ψ(z)=∫_0 ^∞ ((e^(−t) /t)−(e^(−zt) /(1−e^(−t) )))dt we get −2Ψ((3/2))+∫_0 ^∞ ((e^(−t) dt)/(1+e^(−(t/2)) ))=−2Ψ((3/2))+A A=∫_0 ^∞ {e^(−(t/2)) −1+(1/(1+e^(−(t/2)) ))}dt A=lim_(x→∞) ∫_0 ^x {e^(−(t/2)) −1+(1/(1+e^(−(t/2)) ))}dt =lim_(x→∞) [−2e^(−(x/2)) +2−x+2ln(e^(x/2) +1)−2ln(2)] =lim_(x→∞) [2−x+2.(x/2)+2ln(1+e^(−(x/2)) )−2ln(2)] =2−2ln(2) Ψ((3/2))=Ψ((1/2))+2=−2ln(2)−γ+2 we get −2(−2ln(2)−γ)+2−2ln(2) =2ln(2)+2γ−2=2(−1+γ+ln(2)) ∫((1+(√(1−x)))/x)+(2/(ln(1−x)))dx=2(γ−1+ln(2))$
	$l e t l n (1 - x) = - t$ $\Leftrightarrow \int_{0}^{\infty} {\frac{1 + e^{- \frac{t}{2}}}{1 - e^{- t}} - \frac{2}{t}} e^{- t} d t$ $= \int_{0}^{\infty} - \frac{2 e^{- t}}{t} + \frac{1 + e^{- \frac{t}{2}}}{1 - e^{- t}} e^{- t} d t$ $= - 2 \int {\frac{e^{- t}}{t} - \frac{1}{2} \frac{1 + e^{- \frac{t}{2}}}{1 - e^{- t}} e^{- t}} d t$ $= - 2 \int \frac{e^{- t}}{t} - \frac{e^{- 3 \frac{t}{2}}}{1 - e^{- t}} + \frac{1}{2} . \frac{e^{- \frac{t}{2}} - 1}{1 - e^{- t}} . e^{- t} d t$ $= - 2 \int_{0}^{\infty} {\frac{e^{- t}}{t} - \frac{e^{- 3 \frac{t}{2}}}{1 - e^{- t}}} d t + \int_{0}^{\infty} \frac{1 - e^{- \frac{t}{2}}}{1 - e^{- t}} e^{- t} d t$ $Ψ (z) = \int_{0}^{\infty} (\frac{e^{- t}}{t} - \frac{e^{- z t}}{1 - e^{- t}}) d t$ $w e g e t$ $- 2 Ψ (\frac{3}{2}) + \int_{0}^{\infty} \frac{e^{- t} d t}{1 + e^{- \frac{t}{2}}} = - 2 Ψ (\frac{3}{2}) + A$ $A = \int_{0}^{\infty} {e^{- \frac{t}{2}} - 1 + \frac{1}{1 + e^{- \frac{t}{2}}}} d t$ $A = \lim_{x \to \infty} \int_{0}^{x} {e^{- \frac{t}{2}} - 1 + \frac{1}{1 + e^{- \frac{t}{2}}}} d t$ $= \lim_{x \to \infty} [- 2 e^{- \frac{x}{2}} + 2 - x + 2 l n (e^{\frac{x}{2}} + 1) - 2 l n (2)]$ $= \lim_{x \to \infty} [2 - x + 2 . \frac{x}{2} + 2 l n (1 + e^{- \frac{x}{2}}) - 2 l n (2)]$ $= 2 - 2 l n (2)$ $Ψ (\frac{3}{2}) = Ψ (\frac{1}{2}) + 2 = - 2 l n (2) - γ + 2$ $w e g e t - 2 (- 2 l n (2) - γ) + 2 - 2 l n (2)$ $= 2 l n (2) + 2 γ - 2 = 2 (- 1 + γ + l n (2))$ $\int \frac{1 + \sqrt{1 - x}}{x} + \frac{2}{l n (1 - x)} d x = 2 (γ - 1 + l n (2))$
	Commented by mnjuly1970 last updated on 01/Dec/20

	$g r a t e f u l s i r m i n d s p o w e r . . .$
	Commented by mindispower last updated on 02/Dec/20

	$w i t h e p l e a s u r s i r$
	Answered by mnjuly1970 last updated on 01/Dec/20
	$solution we know that .. i: ∫_0 ^( 1) ((1/(ln(x)))+(1/(1−x)))dx=_([IMAGE⇓⇓]) ^(why?) γ ✓ ii:∫_0 ^( 1) ((t^n −1)/(ln(t)))dt=^(why?) ln(n+1) ✓ Φ=∫_0 ^( 1) (((1+(√(1−x)))/x)+(2/(ln(1−x))))dx =^(1−x=t^2 ) 2∫_0 ^( 1) {((1+t)/(1−t^2 ))+(2/(2ln(t)))}tdt =2∫_0 ^( 1) ((1/(1−t))+(1/(ln(t))))tdt =2∫_0 ^( 1) (((t−1+1)/(1−t))+((t+1−1)/(ln(t))))dt =2∫_0 ^( 1) ((1/(1−t))+(1/(ln(t))))dt−2+∫_0 ^( 1) ((t^1 −1)/(ln(t)))dt =2γ−2+ln(2)=2(γ−1+ln(2))✓$
	$s o l u t i o n$ $w e k n o w t h a t . .$ $i : \int_{0}^{1} (\frac{1}{l n (x)} + \frac{1}{1 - x}) d x \underset{[I M A G E ⇓⇓]}{\overset{w h y ?}{=}} γ ✓$ $i i : \int_{0}^{1} \frac{t^{n} - 1}{l n (t)} d t \overset{w h y ?}{=} l n (n + 1) ✓$ $Φ = \int_{0}^{1} (\frac{1 + \sqrt{1 - x}}{x} + \frac{2}{l n (1 - x)}) d x$ $\overset{1 - x = t^{2}}{=} 2 \int_{0}^{1} {\frac{1 + t}{1 - t^{2}} + \frac{2}{2 l n (t)}} t d t$ $= 2 \int_{0}^{1} (\frac{1}{1 - t} + \frac{1}{l n (t)}) t d t$ $= 2 \int_{0}^{1} (\frac{t - 1 + 1}{1 - t} + \frac{t + 1 - 1}{l n (t)}) d t$ $= 2 \int_{0}^{1} (\frac{1}{1 - t} + \frac{1}{l n (t)}) d t - 2 + \int_{0}^{1} \frac{t^{1} - 1}{l n (t)} d t$ $= 2 γ - 2 + l n (2) = 2 (γ - 1 + l n (2)) ✓$
	Commented by mnjuly1970 last updated on 01/Dec/20

	Commented by mnjuly1970 last updated on 01/Dec/20

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum