calculate ∫_0 ^∞ ((ln(x))/((1+x)^3 ))dx


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Question Number 98305 by mathmax by abdo last updated on 12/Jun/20

$calculate \int_{0}^{\infty} \frac{\ln (x)}{{(1 + x)}^{3}} dx$
	Answered by maths mind last updated on 13/Jun/20
	$=∫_0 ^1 ((ln(t))/((1+t)^3 ))dt−∫_0 ^1 ((tln(t))/((1+t)^3 )) =2∫((ln(t))/((1+t)^3 ))dt−∫_0 ^1 ((ln(t))/((1+t)^2 ))=−(−2∫_0 ^1 ((ln(t)dt)/((1+t)^3 ))+∫_0 ^1 ((ln(t)dt)/((1+t)^2 )))=S ∫_y ^1 ((ln(t))/((a+t)^2 ))dt=[−((ln(t))/(a+t))]_y ^1 +∫_y ^1 (dt/(t(a+t))) =((ln(y))/(a+y))−(1/a){ln(y)+ln(1+a)−ln(a)} =((aln(y)−aln(y)−yln(y))/(a(a+y)))−((ln(1+a))/a)+((ln(a))/a) lim_(y→0) ∫_y ^1 ((ln(t))/((t+a)^2 ))dt=(1/a)ln((a/(a+1)))=f(a) f′(a)=−((ln((a/(a+1))))/a^2 )+(1/a)((1/a)−(1/(1+a))) S=−(f(1)+f′(1)) =−(ln((1/2))−ln((1/2))+(1/2))=−(1/2)$
	$= \int_{0}^{1} \frac{l n (t)}{{(1 + t)}^{3}} d t - \int_{0}^{1} \frac{t l n (t)}{{(1 + t)}^{3}}$ $= 2 \int \frac{l n (t)}{{(1 + t)}^{3}} d t - \int_{0}^{1} \frac{l n (t)}{{(1 + t)}^{2}} = - (- 2 \int_{0}^{1} \frac{l n (t) d t}{{(1 + t)}^{3}} + \int_{0}^{1} \frac{l n (t) d t}{{(1 + t)}^{2}}) = S$ $\int_{y}^{1} \frac{l n (t)}{{(a + t)}^{2}} d t = {[- \frac{l n (t)}{a + t}]}_{y}^{1} + \int_{y}^{1} \frac{d t}{t (a + t)}$ $= \frac{l n (y)}{a + y} - \frac{1}{a} {l n (y) + l n (1 + a) - l n (a)}$ $= \frac{a l n (y) - a l n (y) - y l n (y)}{a (a + y)} - \frac{l n (1 + a)}{a} + \frac{l n (a)}{a}$ $\lim_{y \to 0} \int_{y}^{1} \frac{l n (t)}{{(t + a)}^{2}} d t = \frac{1}{a} l n (\frac{a}{a + 1}) = f (a)$ $f^{'} (a) = - \frac{l n (\frac{a}{a + 1})}{a^{2}} + \frac{1}{a} (\frac{1}{a} - \frac{1}{1 + a})$ $S = - (f (1) + f^{'} (1))$ $= - (l n (\frac{1}{2}) - l n (\frac{1}{2}) + \frac{1}{2}) = - \frac{1}{2}$
	Commented by abdomathmax last updated on 13/Jun/20

	$thank you sir mind$
	Answered by abdomathmax last updated on 14/Jun/20

	$A = \int_{0}^{\infty} \frac{\ln (x)}{{(1 + x)}^{3}} dx \Rightarrow A = \int_{0}^{1} \frac{\ln (x)}{{(1 + x)}^{3}} dx$ $+ \int_{1}^{+ \infty} \frac{lnx}{{(1 + x)}^{3}} dx (\to x = \frac{1}{t})$ $= \int_{0}^{1} \frac{lnx}{{(1 + x)}^{3}} dx - \int_{0}^{1} \frac{- lnt}{{(1 + \frac{1}{t})}^{3}} (- \frac{dt}{t^{2}})$ $= \int_{0}^{1} \frac{lnx}{{(1 + x)}^{3}} dx - \int_{0}^{1} \frac{tlnt}{{(1 + t)}^{3}} dt$ $= \int_{0}^{1} \frac{(1 - x) lnx}{{(1 + x)}^{3}} dx = - \int_{0}^{1} (\frac{x + 1 - 2) lnx}{{(1 + x)}^{3}}) dx$ $= - \int_{0}^{1} \frac{lnx}{{(1 + x)}^{2}} dx + 2 \int_{0}^{1} \frac{lnx}{{(1 + x)}^{3}} dx$ $= \ln (2) + 2 \int_{0}^{1} \frac{lnx}{{(1 + x)}^{3}} dx by parts u^{^{'}} = {(1 + x)}^{- 3}$ $\int_{0}^{1} \frac{\ln (x)}{{(1 + x)}^{3}} dx = {[(\frac{1}{2} - \frac{1}{2 {(1 + x)}^{2}}) \ln (x)]}_{0}^{1}$ $- \int_{0}^{1} (\frac{1}{2} - \frac{1}{2 {(1 + x)}^{2}}) \frac{dx}{x}$ $= - \frac{1}{2} \int_{0}^{1} (\frac{x^{2} + 2 x}{x {(1 + x)}^{2}}) dx$ $= - \frac{1}{2} \int_{0}^{1} \frac{x + 2}{{(1 + x)}^{2}} dx$ $= - \frac{1}{2} \int_{0}^{1} \frac{x + 1 + 1}{{(1 + x)}^{2}} dx$ $= - \frac{1}{2} \int_{0}^{1} \frac{dx}{1 + x} - \frac{1}{2} \int_{0}^{1} \frac{dx}{{(1 + x)}^{2}}$ $= - \frac{1}{2} \ln (2) + \frac{1}{2} {[\frac{1}{x + 1}]}_{0}^{1} = - \frac{\ln (2)}{2} + \frac{1}{2} (- \frac{1}{2})$ $= - \frac{\ln (2)}{2} - \frac{1}{4} \Rightarrow$ $A = \ln (2) + 2 (- \frac{\ln 2}{2} - \frac{1}{4})$ $A = - \frac{1}{2}$
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