calculste ∫_0 ^∞ ((ln(x))/((1+x)^4 )) dx


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Question Number 108749 by mathmax by abdo last updated on 19/Aug/20

$calculste \int_{0}^{\infty} \frac{\ln (x)}{{(1 + x)}^{4}} dx$
	Answered by mnjuly1970 last updated on 19/Aug/20

	$f (a) = \int_{0}^{\infty} \frac{x^{a}}{{(1 + x)}^{4}} \Rightarrow goal := f^{^{'}} (0)$ $f^{^{'}} (a) = \frac{d}{d a} [β (a + 1, 3 - a)] = \frac{1}{6} \frac{d}{d a} [Γ (a + 1) . Γ (3 - a)]$ $= \frac{1}{6} [Γ^{^{'}} (1) . Γ (3) - Γ^{^{'}} (3) Γ (1)]$ $= \frac{1}{6} [- 2 γ - 2 ψ (3)] = \frac{1}{6} [- 2 γ - 2 (\frac{3}{2} - γ)]$ $\frac{- 1}{2} . . . .$
	Commented by mathmax by abdo last updated on 19/Aug/20

	$thankx sir$
	Commented by mnjuly1970 last updated on 19/Aug/20

	$t h a n k y o u m a s t e r . . y o u r q u e s t i o n s$ $a r e v e r y n i c e . . .$
	Answered by mathmax by abdo last updated on 19/Aug/20
	$we use ∫_0 ^∞ q(x)ln(x) =−(1/2)Re(Σ_i Re(q(z)ln^2 z ,a_i ) we have q(z)ln^2 z =((ln^2 z)/((z+1)^4 ))=w(z) ⇒−1 is pole of (with ordr 4) Res(w,−1) =lim_(z→−1) (1/((4−1)!)){(z+1)^4 w(z)}^((3)) =(1/6)lim_(z→−1) {ln^2 (z)}^((3)) =(1/6)lim_(z→−1) {((2lnz)/z)}^((2)) =(1/3)lim_(z→−1) {((1−lnz)/z^2 )}^((1)) =(1/3)lim_(z→−1) {((−z−2z(1−lnz))/z^4 )} =(1/3)lim_(z→−1) {((−1−2(1−lnz))/z^3 )} =(1/3)lim_(z→−1) {((−3+2ln(z))/z^3 )} =(1/3).(3−2ln(−1)) =1−(2/3)ln(e^(iπ) ) =1−((2iπ)/3) ⇒Re(...)=1 ⇒ ∫_0 ^∞ ((lnx)/((1+x)^4 ))dx =−(1/2)$
	$we use \int_{0}^{\infty} q (x) \ln (x) = - \frac{1}{2} Re (\sum_{i} Re (q (z) \ln^{2} z, a_{i})$ $we have q (z) \ln^{2} z = \frac{\ln^{2} z}{{(z + 1)}^{4}} = w (z) \Rightarrow - 1 is pole of (with ordr 4)$ $Res (w, - 1) = \lim_{z \to - 1} \frac{1}{(4 - 1)!} {{(z + 1)}^{4} w (z)}^{(3)}$ $= \frac{1}{6} \lim_{z \to - 1} {\ln^{2} (z)}^{(3)} = \frac{1}{6} \lim_{z \to - 1} {\frac{2 lnz}{z}}^{(2)}$ $= \frac{1}{3} \lim_{z \to - 1} {\frac{1 - lnz}{z^{2}}}^{(1)} = \frac{1}{3} \lim_{z \to - 1} {\frac{- z - 2 z (1 - lnz)}{z^{4}}}$ $= \frac{1}{3} \lim_{z \to - 1} {\frac{- 1 - 2 (1 - lnz)}{z^{3}}} = \frac{1}{3} \lim_{z \to - 1} {\frac{- 3 + 2 \ln (z)}{z^{3}}}$ $= \frac{1}{3} . (3 - 2 \ln (- 1)) = 1 - \frac{2}{3} \ln (e^{i π}) = 1 - \frac{2 i π}{3} \Rightarrow Re (. . .) = 1 \Rightarrow$ $\int_{0}^{\infty} \frac{lnx}{{(1 + x)}^{4}} dx = - \frac{1}{2}$
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