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Question Number 202406 by mou0113 last updated on 26/Dec/23

	Answered by witcher3 last updated on 26/Dec/23

	$f (s) = \int_{0}^{\infty} \frac{t^{s}}{{(1 + t^{2})}^{2}} dt \Rightarrow f^{'} (0) = \int_{0}^{\infty} \frac{\ln (x)}{{(1 + x^{2})}^{2}} dx$ $f (s) = \int_{0}^{\infty} \frac{t^{\frac{s}{2}} . t^{- \frac{1}{2}}}{{(1 + t)}^{2}} \frac{dt}{2}$ $= \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{s - 1}{2}}}{{(1 + t)}^{2}} dt = \frac{1}{2} β (\frac{s + 1}{2}, 2 - \frac{s + 1}{2})$ $= \frac{1}{2} . Γ (\frac{s + 1}{2}) Γ (2 - \frac{s + 1}{2}) = \frac{1}{2} (1 - \frac{s + 1}{2}) . \frac{π}{\sin (\frac{π}{2} (s + 1)}$ $= \frac{π}{4} . \frac{1 - s}{\cos (\frac{π}{2} s)} = f (s)$ $f^{'} (0) = \frac{π}{4} (\frac{- 1}{1}) = - \frac{π}{4}$
	Commented by mou0113 last updated on 26/Dec/23

	$t h a n k y o u$
	Answered by Mathspace last updated on 26/Dec/23
	$I=∫_0 ^∞ ((lnx)/((1+x^2 )^2 ))dx (x=(√t)) I=(1/2)∫_0 ^∞ ((lnt)/((1+t)^2 ))(dt/(2(√t))) ⇒4I=∫_0 ^∞ ((t^(−(1/2)) lnt)/((1+t)^2 ))dt B(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) ))dt (∂B/∂x)(x,y)=∫_0 ^∞ ((t^(x−1) lnt)/((1+t)^(x+y) ))dt we take x−1=−(1/2) and x+y=2 ⇒ x=(1/2) and y=(3/2) so4I=(∂B/∂x)((1/2),(3/2)) we have proved that (∂B/∂x)(x,y)=B(x,y){Ψ(x)−Ψ(x+y)} ⇒4I=B((1/2),(3/2)){Ψ((1/2))−Ψ(2)} B((1/2),(3/2))=((Γ((1/2)).Γ((3/2)))/(Γ(2))) =(((√π).(1/2)(√π))/(1!))=(π/2) Ψ(s)=−γ+ ∫_0 ^1 ((1−x^(s−1) )/(1−x))dx ⇒Ψ(2)=−γ+1 Ψ((1/2))=−γ+∫_0 ^1 ((1−x^(−(1/2)) )/(1−x))dx and ∫_0 ^1 ((1−(1/( (√x))))/(1−x))dx =∫_0 ^1 (((√x)−1)/( (√x)(1−(√x))(1+(√x))))dx =−∫_0 ^1 (dx/( (√x)(1+(√x))))dx ((√x)=t) =−∫_0 ^1 ((2tdt)/(t(1+t)))=−2∫_0 ^1 (dt/(1+t)) =−2ln2 ⇒ Ψ((1/2))=−γ−2ln2 ⇒ 4I=(π/2){−γ−2ln2+γ−1} ⇒ 4I=−πln2−(π/2) ⇒ I=−((πln2)/4)−(π/8)$
	$I = \int_{0}^{\infty} \frac{l n x}{{(1 + x^{2})}^{2}} d x (x = \sqrt{t})$ $I = \frac{1}{2} \int_{0}^{\infty} \frac{l n t}{{(1 + t)}^{2}} \frac{d t}{2 \sqrt{t}}$ $\Rightarrow 4 I = \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} l n t}{{(1 + t)}^{2}} d t$ $B (x, y) = \int_{0}^{\infty} \frac{t^{x - 1}}{{(1 + t)}^{x + y}} d t$ $\frac{\partial B}{\partial x} (x, y) = \int_{0}^{\infty} \frac{t^{x - 1} l n t}{{(1 + t)}^{x + y}} d t$ $w e t a k e x - 1 = - \frac{1}{2} a n d x + y = 2 \Rightarrow$ $x = \frac{1}{2} a n d y = \frac{3}{2}$ $s o 4 I = \frac{\partial B}{\partial x} (\frac{1}{2}, \frac{3}{2})$ $w e h a v e p r o v e d t h a t$ $\frac{\partial B}{\partial x} (x, y) = B (x, y) {Ψ (x) - Ψ (x + y)}$ $\Rightarrow 4 I = B (\frac{1}{2}, \frac{3}{2}) {Ψ (\frac{1}{2}) - Ψ (2)}$ $B (\frac{1}{2}, \frac{3}{2}) = \frac{Γ (\frac{1}{2}) . Γ (\frac{3}{2})}{Γ (2)}$ $= \frac{\sqrt{π} . \frac{1}{2} \sqrt{π}}{1!} = \frac{π}{2}$ $Ψ (s) = - γ +$ $\int_{0}^{1} \frac{1 - x^{s - 1}}{1 - x} d x$ $\Rightarrow Ψ (2) = - γ + 1$ $Ψ (\frac{1}{2}) = - γ + \int_{0}^{1} \frac{1 - x^{- \frac{1}{2}}}{1 - x} d x$ $a n d$ $\int_{0}^{1} \frac{1 - \frac{1}{\sqrt{x}}}{1 - x} d x = \int_{0}^{1} \frac{\sqrt{x} - 1}{\sqrt{x} (1 - \sqrt{x}) (1 + \sqrt{x})} d x$ $= - \int_{0}^{1} \frac{d x}{\sqrt{x} (1 + \sqrt{x})} d x (\sqrt{x} = t)$ $= - \int_{0}^{1} \frac{2 t d t}{t (1 + t)} = - 2 \int_{0}^{1} \frac{d t}{1 + t}$ $= - 2 l n 2 \Rightarrow$ $Ψ (\frac{1}{2}) = - γ - 2 l n 2 \Rightarrow$ $4 I = \frac{π}{2} {- γ - 2 l n 2 + γ - 1} \Rightarrow$ $4 I = - π l n 2 - \frac{π}{2} \Rightarrow$ $I = - \frac{π l n 2}{4} - \frac{π}{8}$
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