let g(x) =∫_0 ^∞ ((t ln(t)dt)/((1+xt)^3 )) with x>0 1) give a explicit form of g(x) 2) calculate ∫_0 ^∞ ((t ln(t))/((1+t)^3 ))dt 3) calculate ∫_0 ^∞ ((tln(t))/((1+2t)^3 )) dt 4) calculate A(θ) =∫_0 ^∞ ((t ln(t))/((1+t sinθ)^3 ))dt with 0<θ<(π/2)


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Question Number 44515 by maxmathsup by imad last updated on 30/Sep/18

$l e t g (x) = \int_{0}^{\infty} \frac{t l n (t) d t}{{(1 + x t)}^{3}} w i t h x > 0$ $1) g i v e a e x p l i c i t f o r m o f g (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{t l n (t)}{{(1 + t)}^{3}} d t$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{t l n (t)}{{(1 + 2 t)}^{3}} d t$ $4) c a l c u l a t e A (θ) = \int_{0}^{\infty} \frac{t l n (t)}{{(1 + t s i n θ)}^{3}} d t w i t h 0 < θ < \frac{π}{2}$
	Answered by maxmathsup by imad last updated on 02/Oct/18

	$1) l e t f (x) = \int_{0}^{\infty} \frac{l n (t)}{{(1 + t x)}^{2}} d t w e h a v e p r o v e d t h a t f (x) = - \frac{l n (x)}{x}$ $\Rightarrow f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 t (1 + t x) l n (t)}{{(1 + t x)}^{4}} = - \int_{0}^{\infty} \frac{2 t l n (t)}{{(1 + t x)}^{3}} d t \Rightarrow$ $g (x) = - \frac{1}{2} f^{^{'}} (x) b u t f^{^{'}} (x) = - \frac{1 - l n (x)}{x^{2}} \Rightarrow g (x) = \frac{1 - l n (x)}{2 x^{2}}$ $2) \int_{0}^{\infty} \frac{t l n (t)}{{(1 + t)}^{3}} = g (1) = 0$ $3) \int_{0}^{\infty} \frac{t l n (t)}{{(1 + 2 t)}^{3}} d t = g (2) = \frac{1 - l n (2)}{8}$ $4) \int_{0}^{\infty} \frac{t l n (t)}{{(1 + t s i n θ)}^{3}} d t = g (s i n θ) = \frac{1 - l n (s i n θ)}{2 {s i n}^{2} θ} .$
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