calculate ∫_0 ^∞ ((lnx)/((1+x^2 )^2 ))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 82755 by mathmax by abdo last updated on 23/Feb/20

$c a l c u l a t e \int_{0}^{\infty} \frac{l n x}{{(1 + x^{2})}^{2}} d x$
Commented by Ajao yinka last updated on 24/Feb/20
0
Commented by mathmax by abdo last updated on 24/Feb/20
$let f(a) =∫_0 ^∞ ((lnx)/(a+x^2 ))dx with a>0 we can verify eazly that f is derivable on ]0,+∞[ and f^′ (a) =−∫_0 ^∞ ((lnx)/((a+x^2 )^2 ))dx ⇒ ∫_0 ^∞ ((lnx)/((a+x^2 )^2 ))dx =−f^′ (a) let explicit f changement x=(√a)t give f(a) =∫_0 ^∞ ((ln((√a)t))/(a(1+t^2 )))(√a)dt =(1/(√a))∫_0 ^∞ (((1/2)lna +lnt)/(t^2 +1))dt =(1/2)((lna)/(√a))×(π/2) +(1/(√a))∫_0 ^∞ ((lnt)/(1+t^2 ))dt but ∫_0 ^∞ ((ln(t))/(1+t^2 ))dt =0((put t=(1/u)) ⇒ f(a) =((πlna)/(4(√a))) ⇒f^′ (a) =(π/4){((((√a)/a)−lna ×(1/(2(√a))))/a)} =(π/(4a(√a)))−(π/4)×((lna)/(2a(√a))) ⇒∫_0 ^∞ ((lnx)/((a+x^2 )^2 ))dx =−(π/(4a(√a)))+((πlna)/(8a(√a))) a=1 ⇒∫_0 ^∞ ((lnx)/((1+x^2 )^2 ))dx =−(π/4)$
$l e t f (a) = \int_{0}^{\infty} \frac{l n x}{a + x^{2}} d x w i t h a > 0 w e c a n v e r i f y e a z l y t h a t f i s$ $d e r i v a b l e o n] 0, + \infty [a n d f^{^{'}} (a) = - \int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x \Rightarrow$ $\int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x = - f^{^{'}} (a) l e t e x p l i c i t f c h a n g e m e n t x = \sqrt{a} t g i v e$ $f (a) = \int_{0}^{\infty} \frac{l n (\sqrt{a} t)}{a (1 + t^{2})} \sqrt{a} d t = \frac{1}{\sqrt{a}} \int_{0}^{\infty} \frac{\frac{1}{2} l n a + l n t}{t^{2} + 1} d t$ $= \frac{1}{2} \frac{l n a}{\sqrt{a}} \times \frac{π}{2} + \frac{1}{\sqrt{a}} \int_{0}^{\infty} \frac{l n t}{1 + t^{2}} d t b u t \int_{0}^{\infty} \frac{l n (t)}{1 + t^{2}} d t = 0 ((p u t t = \frac{1}{u}) \Rightarrow$ $f (a) = \frac{π l n a}{4 \sqrt{a}} \Rightarrow f^{^{'}} (a) = \frac{π}{4} {\frac{\frac{\sqrt{a}}{a} - l n a \times \frac{1}{2 \sqrt{a}}}{a}}$ $= \frac{π}{4 a \sqrt{a}} - \frac{π}{4} \times \frac{l n a}{2 a \sqrt{a}} \Rightarrow \int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x = - \frac{π}{4 a \sqrt{a}} + \frac{π l n a}{8 a \sqrt{a}}$ $a = 1 \Rightarrow \int_{0}^{\infty} \frac{l n x}{{(1 + x^{2})}^{2}} d x = - \frac{π}{4}$
	Answered by mind is power last updated on 24/Feb/20

	$\int_{0}^{+ \infty} \frac{l n (t)}{{(1 + t^{2})}^{n}} d t = A (n)$ $l n (t) = \frac{\partial}{\partial x} t^{x} ∣_{x = 0}$ $A = \frac{\partial}{\partial x} \int_{0}^{+ \infty} \frac{t^{x}}{{(1 + t^{2})}^{n}} d t ∣_{x = 0}$ $= \frac{\partial}{\partial x} \int_{0}^{\frac{π}{2}} \frac{{t g}^{x} (t)}{{(1 + {t g}^{2} (t))}^{n}} (1 + {t g}^{2} (t)) d t ∣_{x = 0}$ $= \frac{\partial}{\partial x} \int_{0}^{\frac{π}{2}} \frac{{t g}^{x} (t)}{{(1 + {t g}^{2} (t))}^{n}} d t ∣_{x = 0}$ $= \frac{1}{2} \frac{\partial}{\partial x} . 2 \int_{0}^{\frac{π}{2}} {s i n}^{x} (t) {c o s}^{- x + 2 n} (t) d t ∣_{x = 0}$ $= \frac{1}{2} \frac{\partial}{\partial x} β (\frac{x}{2} + \frac{1}{2}; - \frac{x}{2} + n + \frac{1}{2}) ∣_{x = 0}$ $= \frac{1}{2} \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (n + \frac{1}{2} - \frac{x}{2})}{Γ (n + 1)} ∣_{x = 0} = \frac{1}{2} \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (n + \frac{1}{2} - \frac{x}{2})}{n!} ∣_{x = 0}$ $Γ (n + \frac{1}{2} - \frac{x}{2}) = Γ (\frac{1}{2} - \frac{x}{2}) \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2})$ $= \frac{1}{2} . \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (\frac{1}{2} - \frac{x}{2})}{n!} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2}) ∣_{x = 0}$ $= \frac{1}{2 (n!)} \frac{\partial}{\partial x} . \frac{π}{s i n (\frac{π}{2} (x + 1))} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2}) ∣_{x = 0}$ $f (x) = \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2})$ $l n (f (x)) = \underset{k = 0}{\sum^{n - 1}} l n (k + \frac{1}{2} - \frac{x}{2})$ $\Rightarrow \frac{f^{'} (0)}{f (0)} = - \underset{k = 0}{\sum^{n - 1}} \frac{1}{2 (k + \frac{1}{2})} = - \underset{k = 0}{\sum^{n - 1}} \frac{1}{2 k + 1}$ $= \underset{k = 1}{\sum^{n}} \frac{1}{2 k} - \underset{k = 1}{\sum^{2 n}} \frac{1}{k} = \frac{H_{n}}{2} - H_{2 n}$ $f (0) = \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2}) = \frac{(2 n - 1)!!}{2^{n}} = \frac{(2 n)!}{2^{2 n} . n!}$ $\Rightarrow f^{'} (0) = \frac{(2 n)!}{2^{2 n} n!} . (\frac{H_{n}}{2} - H_{2 n})$ $A (n) = \frac{(2 n)!}{2^{2 n + 1} {(n!)}^{2}} (\frac{H_{n}}{2} - H_{2 n}) π = \int_{0}^{+ \infty} \frac{l n (x)}{{(1 + x^{2})}^{n}} d x$ $n = 2 \Rightarrow \int_{0}^{+ \infty} \frac{l n (x)}{{(1 + x^{2})}^{2}} d x = A (2)$ $A (2) = \frac{4!}{2^{5} {(2!)}^{2}} (- 1 - \frac{1}{3}) π = \frac{24}{32.4} . \frac{- 4}{3} π = \frac{- 8}{32} . π = - \frac{π}{4}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum