∫_0 ^∞ ((ln x)/((x^2 +a^2 )^5 )) dx


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Question Number 140388 by benjo_mathlover last updated on 07/May/21

$\int_{0}^{\infty} \frac{\ln x}{{(x^{2} + a^{2})}^{5}} dx$
	Answered by mathmax by abdo last updated on 07/May/21
	$let f(a)=∫_0 ^∞ ((logx)/((x^2 +a^2 )^4 )) ⇒f^′ (a)=−∫_0 ^∞ ((4(2a)(x^2 +a^2 )^3 )/((x^2 +a^2 )^8 ))logx dx =−8a ∫_0 ^∞ ((logx)/((x^2 +a^2 )^5 )) dx ⇒∫_0 ^∞ ((logx)/((x^2 +a^2 )^5 ))dx =−((f^′ (a))/(8a)) (suppose a>0) cha7gement x=at give f(a)=∫_0 ^∞ ((loga+logt)/(a^8 (t^2 +1)^4 )) adt =((loga)/a^7 )∫_0 ^∞ (dt/((t^2 +1)^4 )) +(1/a^7 )∫_0 ^∞ ((logt)/((t^2 +1)^4 ))dt we have ∫_0 ^∞ (dt/((t^2 +1)^4 )) =(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)^4 )) =(1/2)(2iπ)Res(w,i) with w(z)=(1/((z^2 +1)^4 ))=(1/((z−i)^4 (z+i)^4 )) Res(w,i)=lim_(z→i) (1/((4−1)!)){(z−i)^4 w(z)}^((3)) =(1/(3!))lim_(z→i) {(z+i)^(−4) }^((3)) =(1/(3!))lim_(z→i) {−4(z+i)^(−5) }^((2)) =(1/(3!))lim_(z→i) {20 (z+i)^(−6) }^((1)) =(1/(3!))lim_(z→i) (−120)(z+i)^(−7) =−20 (2i)^(−7 ) =((−20)/((2i)^7 )) =−((2^2 ×5)/(2^7 i^7 )) =−i(5/2^5 ) ⇒ ∫_0 ^∞ (dt/((t^2 +1)^4 )) =iπ(−i(5/2^5 )) =((5π)/(32)) and ∫_0 ^∞ ((logt)/((t^2 +1)^4 ))dt =_(t=(√x)) ∫_0 ^∞ ((log((√x)))/((x+1)^4 ))dt =(1/2)∫_0 ^∞ ((logx)/((x+1)^4 ))dx =(1/2)∫_0 ^1 ((logx)/((x+1)^4 ))dx +(1/2)∫_1 ^∞ ((logx)/((x+1)^4 ))dx(→x=(1/y)) =(1/2)∫_0 ^1 ((logx)/((x+1)^4 ))dx−(1/2)∫_0 ^1 ((−logy)/(((1/y)+1)^4 ))(−(dy/y^2 )) =(1/2)∫_0 ^1 ((logx)/((x+1)^4 ))dx−(1/2)∫_0 ^1 ((y^2 logy)/((y+1)^4 ))dy =(1/2)∫_0 ^1 (((1−x^2 )logx)/((x+1)^4 ))dx we ghave (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n ⇒ −(1/((1+x)^2 )) =Σ_(n=1) ^∞ n(−1)^n x^(n−1) ⇒ ((2(1+x))/((1+x)^4 )) =Σ_(n=2) ^∞ n(n−1)(−1)^n x^(n−2) =(2/((1+x)^3 )) ⇒ ((−2.3(1+x)^2 )/((1+x)^6 )) =Σ_(n=3) ^∞ n(n−1)(n−2)(−1)^n x^(n−3) =−(6/((1+x)^4 )) ⇒ ∫_0 ^1 (((1−x^2 )logx)/((x+1)^4 )) =−(1/6)Σ_(n=3) ^∞ n(n−1)(n−2)(−1)^n ∫_0 ^1 (1−x^2 )x^(n−3) dx =−(1/6) Σ_(n=3) ^∞ n(n−1)(n−2)(−1)^n ∫_0 ^1 (x^(n−3) −x^(n−1) )dx =−(1/6)Σ_(n=3) ^∞ n(n−1)(n−2)(−1)^n [(1/(n−2))x^(n−2) −(1/n)x^n ]_0 ^1 =−(1/6)Σ_(n=3) ^∞ n(n−1)(n−2)(−1)^n ((1/(n−2))−(1/n)) =−(1/6)Σ_(n=3) ^∞ n(n−1)(−1)^n +(1/6)Σ_(n=3) ^∞ (n−1)(n−2)(−1)^n ....be continued....$
	$let f (a) = \int_{0}^{\infty} \frac{logx}{{(x^{2} + a^{2})}^{4}} \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{4 (2 a) {(x^{2} + a^{2})}^{3}}{{(x^{2} + a^{2})}^{8}} logx dx$ $= - 8 a \int_{0}^{\infty} \frac{logx}{{(x^{2} + a^{2})}^{5}} dx \Rightarrow \int_{0}^{\infty} \frac{logx}{{(x^{2} + a^{2})}^{5}} dx = - \frac{f^{^{'}} (a)}{8 a} (suppose a > 0)$ $cha 7 gement x = at give f (a) = \int_{0}^{\infty} \frac{loga + logt}{a^{8} {(t^{2} + 1)}^{4}} adt$ $= \frac{loga}{a^{7}} \int_{0}^{\infty} \frac{dt}{{(t^{2} + 1)}^{4}} + \frac{1}{a^{7}} \int_{0}^{\infty} \frac{logt}{{(t^{2} + 1)}^{4}} dt we have$ $\int_{0}^{\infty} \frac{dt}{{(t^{2} + 1)}^{4}} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dt}{{(t^{2} + 1)}^{4}} = \frac{1}{2} (2 i π) Res (w, i) with$ $w (z) = \frac{1}{{(z^{2} + 1)}^{4}} = \frac{1}{{(z - i)}^{4} {(z + i)}^{4}}$ $Res (w, i) = \lim_{z \to i} \frac{1}{(4 - 1)!} {{(z - i)}^{4} w (z)}^{(3)}$ $= \frac{1}{3!} \lim_{z \to i} {{(z + i)}^{- 4}}^{(3)} = \frac{1}{3!} \lim_{z \to i} {- 4 {(z + i)}^{- 5}}^{(2)}$ $= \frac{1}{3!} \lim_{z \to i} {20 {(z + i)}^{- 6}}^{(1)} = \frac{1}{3!} \lim_{z \to i} (- 120) {(z + i)}^{- 7}$ $= - 20 {(2 i)}^{- 7} = \frac{- 20}{{(2 i)}^{7}} = - \frac{2^{2} \times 5}{2^{7} i^{7}} = - i \frac{5}{2^{5}} \Rightarrow$ $\int_{0}^{\infty} \frac{dt}{{(t^{2} + 1)}^{4}} = i π (- i \frac{5}{2^{5}}) = \frac{5 π}{32} and$ $\int_{0}^{\infty} \frac{logt}{{(t^{2} + 1)}^{4}} dt =_{t = \sqrt{x}} \int_{0}^{\infty} \frac{\log (\sqrt{x})}{{(x + 1)}^{4}} dt$ $= \frac{1}{2} \int_{0}^{\infty} \frac{logx}{{(x + 1)}^{4}} dx = \frac{1}{2} \int_{0}^{1} \frac{logx}{{(x + 1)}^{4}} dx + \frac{1}{2} \int_{1}^{\infty} \frac{logx}{{(x + 1)}^{4}} dx (\to x = \frac{1}{y})$ $= \frac{1}{2} \int_{0}^{1} \frac{logx}{{(x + 1)}^{4}} dx - \frac{1}{2} \int_{0}^{1} \frac{- logy}{{(\frac{1}{y} + 1)}^{4}} (- \frac{dy}{y^{2}})$ $= \frac{1}{2} \int_{0}^{1} \frac{logx}{{(x + 1)}^{4}} dx - \frac{1}{2} \int_{0}^{1} \frac{y^{2} logy}{{(y + 1)}^{4}} dy$ $= \frac{1}{2} \int_{0}^{1} \frac{(1 - x^{2}) logx}{{(x + 1)}^{4}} dx we ghave \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow$ $- \frac{1}{{(1 + x)}^{2}} = \sum_{n = 1}^{\infty} n {(- 1)}^{n} x^{n - 1} \Rightarrow$ $\frac{2 (1 + x)}{{(1 + x)}^{4}} = \sum_{n = 2}^{\infty} n (n - 1) {(- 1)}^{n} x^{n - 2} = \frac{2}{{(1 + x)}^{3}} \Rightarrow$ $\frac{- 2.3 {(1 + x)}^{2}}{{(1 + x)}^{6}} = \sum_{n = 3}^{\infty} n (n - 1) (n - 2) {(- 1)}^{n} x^{n - 3} = - \frac{6}{{(1 + x)}^{4}} \Rightarrow$ $\int_{0}^{1} \frac{(1 - x^{2}) logx}{{(x + 1)}^{4}} = - \frac{1}{6} \sum_{n = 3}^{\infty} n (n - 1) (n - 2) {(- 1)}^{n} \int_{0}^{1} (1 - x^{2}) x^{n - 3} dx$ $= - \frac{1}{6} \sum_{n = 3}^{\infty} n (n - 1) (n - 2) {(- 1)}^{n} \int_{0}^{1} (x^{n - 3} - x^{n - 1}) dx$ $= - \frac{1}{6} \sum_{n = 3}^{\infty} n (n - 1) (n - 2) {(- 1)}^{n} {[\frac{1}{n - 2} x^{n - 2} - \frac{1}{n} x^{n}]}_{0}^{1}$ $= - \frac{1}{6} \sum_{n = 3}^{\infty} n (n - 1) (n - 2) {(- 1)}^{n} (\frac{1}{n - 2} - \frac{1}{n})$ $= - \frac{1}{6} \sum_{n = 3}^{\infty} n (n - 1) {(- 1)}^{n} + \frac{1}{6} \sum_{n = 3}^{\infty} (n - 1) (n - 2) {(- 1)}^{n}$ $. . . . be continued . . . .$
	Answered by Dwaipayan Shikari last updated on 07/May/21

	$ν (ξ) = \int_{0}^{\infty} \frac{x^{ξ}}{{(x^{2} + a^{2})}^{5}} d x x = a u$ $\int_{0}^{\infty} \frac{l o g (x)}{{(x^{2} + a^{2})}^{5}} = ν^{'} (0)$ $ν (ξ) = a^{ξ - 9} \int_{0}^{\infty} \frac{u^{ξ}}{{(u^{2} + 1)}^{5}} d x = \frac{1}{2} a^{ξ - 9} \int_{0}^{\infty} \frac{t^{\frac{ξ}{2} - \frac{1}{2}}}{{(t + 1)}^{5}} d t$ $= \frac{a^{ξ - 9}}{2} \int_{0}^{\infty} \frac{t^{\frac{ξ + 1}{2} - 1}}{{(t + 1)}^{5 + \frac{ξ + 1}{2} - \frac{ξ + 1}{2}}} d t = \frac{a^{ξ - 9}}{2} . \frac{Γ (\frac{ξ + 1}{2}) Γ (\frac{9}{2} - \frac{ξ}{2})}{Γ (5)}$ $l o g (ν (ξ)) = l o g (\frac{a^{ξ - 9}}{48} Γ (\frac{ξ + 1}{2}) Γ (\frac{9}{2} - \frac{ξ}{2}))$ $\frac{ν^{'} (ξ)}{ν (ξ)} = l o g (a) a^{ξ - 9} + \frac{1}{2} (ψ (\frac{ξ + 1}{2}) - ψ (\frac{9}{2} - \frac{ξ}{2})$ $ν^{'} (0) = ν (0) (\frac{l o g (a)}{a^{9}} + \frac{1}{2} (ψ (\frac{1}{2}) - ψ (\frac{1}{2}) - \frac{2}{7} - \frac{2}{5} - \frac{2}{3} - 2))$ $ν^{'} (0) = \frac{1}{2 a^{9}} (\frac{\sqrt{π} . \frac{7}{2} . \frac{5}{2} . \frac{3}{2} . \frac{1}{2} \sqrt{π}}{4!}) = \frac{π}{768 a^{9}}$ $ν^{'} (0) = \frac{π}{768 a^{9}} (\frac{l o g (a)}{a^{9}} - \frac{12}{35} - \frac{4}{3}) = \frac{π}{768 a^{9}} (\frac{l o g (a)}{a^{9}} - \frac{176}{105})$
	Answered by qaz last updated on 07/May/21

	$A = \int_{0}^{\infty} \frac{l n x}{x^{2} + a^{2}} d x = \frac{1}{a} \int_{0}^{\infty} \frac{l n a + l n x}{x^{2} + 1} d x = \frac{π l n a}{2 a}$ $\frac{\partial A}{\partial a} = - 2 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{2}} d x = \frac{π}{2 a^{2}} (1 - l n a)$ $\Rightarrow B = \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{2}} d x = \frac{π}{4 a^{3}} (l n a - 1)$ $\frac{\partial B}{\partial a} = - 4 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{3}} d x = \frac{π}{4 a^{4}} (4 - 3 l n a)$ $\Rightarrow C = \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{3}} d x = \frac{π}{16 a^{5}} (3 l n a - 4)$ $\frac{\partial C}{\partial a} = - 6 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{4}} d x = \frac{π}{16 a^{6}} (23 - 15 l n a)$ $\Rightarrow D = \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{4}} d x = \frac{π}{96 a^{7}} (15 l n a - 23)$ $\frac{\partial D}{\partial a} = - 8 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{5}} d x = \frac{π}{96 a^{8}} (176 - 105 l n a)$ $\Rightarrow E = \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{5}} d x = \frac{π}{768 a^{9}} (105 l n a - 176)$ $\frac{\partial E}{\partial a} = - 10 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{6}} d x = \frac{π}{768 a^{10}} (1689 - 945 l n a)$ $\Rightarrow \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{6}} d x = \frac{π}{7680 a^{11}} (945 l n a - 1689)$ $. . . . . . .$
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