let-g-x-0-t-ln-t-dt-1-xt-3-with-x-gt-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-

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} .