let-f-x-0-dt-x-t-t-2-3-with-x-gt-1-4-1-calculate-f-x-2-calculate-also-g-x-0-dt-x-t-t-2-4-3-find-the-values-of-0-dt-1-t-t-2-3-and-0-

Question Number 65061 by mathmax by abdo last updated on 24/Jul/19

l e t f (x) = \int_{0}^{\infty} \frac{d t}{{(x - t + t^{2})}^{3}} w i t h x > \frac{1}{4}

1) c a l c u l a t e f (x)

2) c a l c u l a t e a l s o g (x) = \int_{0}^{\infty} \frac{d t}{{(x - t + t^{2})}^{4}}

3) f i n d t h e v a l u e s o f \int_{0}^{\infty} \frac{d t}{{(1 - t + t^{2})}^{3}} a n d \int_{0}^{\infty} \frac{d t}{{(2 - t + t^{2})}^{4}}

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

t^{2} - t + x = {(t - \frac{1}{2})}^{2} + \frac{4 x - 1}{4} = \frac{4 x - 1}{4} [{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1] c a u s e x > \frac{1}{4} \Rightarrow 4 x - 1 > 0

f (x) = {(\frac{4}{4 x - 1})}^{3} \int_{0}^{\infty} \frac{d t}{{[{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1]}^{3}}

l e t c h a n g e u = a r c t a n (\frac{2 t - 1}{\sqrt{4 x - 1}}) t h e n d u = \frac{\frac{2}{\sqrt{4 x - 1}}}{{(\frac{2 t - 1}{\sqrt{4 x - 1}})}^{2} + 1} d t

i f t = 0 t h e n u = a r c t a n (\frac{- 1}{\sqrt{4 x - 1}}) = \frac{- π}{2} + a r c t a n (\sqrt{4 x - 1}) = θ_{0}

w h e n t = \infty u = \frac{π}{2}

T h e n f (x) = {(\frac{4}{4 x - 1})}^{3} . \frac{2}{\sqrt{4 x - 1}} \int_{θ_{0}}^{\frac{π}{2}} \frac{d u}{{[{t a n}^{2} u + 1]}^{2}}

= {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . \int_{θ_{0}}^{\frac{π}{2}} {c o s}^{4} u d u

i t i s e a s y t o p r o v e t h a t {_{{c o s}^{4} u - {s i n}^{4} u = c o s 2 u}^{{c o s}^{4} u + {s i n}^{4} u = 1 - \frac{s i n 2 u}{2}} t h e n w e g o t {c o s}^{4} u = \frac{1}{2} - \frac{s i n 2 u}{4} + \frac{c o s 2 u}{2}

S o

f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . {[\frac{u}{2} + \frac{c o s 2 u}{8} + \frac{s i n 2 u}{4}]}_{θ_{0}}^{\frac{π}{2}}

k n o w i n g t h a t {_{s i n 2 u = \frac{2 t a n u}{1 + {t a n}^{2} u}}^{c o s 2 u = \frac{1 - {t a n}^{2} u}{1 + {t a n}^{2} u}} w e g o t

f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . [\frac{π}{4} - \frac{1}{8} - (\frac{a r c t a n \sqrt{4 x - 1}}{2} - \frac{π}{4} + \frac{1}{8} . \frac{1 - \frac{1}{4 x - 1}}{1 + \frac{1}{4 x - 1}} + \frac{1}{4} . \frac{2 . \frac{- 1}{\sqrt{4 x - 1}}}{1 + \frac{1}{4 x - 1}})]

f (x) = {(\frac{4}{4 x - 1})}^{\frac{7}{2}} . [\frac{π}{2} - \frac{1}{4} - \frac{a r c t a n \sqrt{4 x - 1}}{2} - \frac{2 x - 1 - 2 \sqrt{4 x - 1}}{16 x}]

2) \frac{d f}{d x} = \int_{0}^{\infty} \frac{d (\frac{1}{{[x - t + t^{2}]}^{3}})}{d x} d t = - 3 g (x)

s o g (x) = \frac{- 1}{3} \frac{d f}{d x} .

3) b y r e p l a c i n g x = 1 o n t h e f i n a l e x p r e s s i o n o f f (x) w e g o t

f (1) = {(\frac{4}{3})}^{\frac{7}{2}} [\frac{π}{2} - \frac{1}{4} - \frac{π}{3 \times 2} - \frac{1 - 2 \sqrt{3}}{16}]

f (2) = {(\frac{4}{7})}^{\frac{7}{2}} . [\frac{π}{2} - \frac{1}{4} - \frac{a r c t a n \sqrt{7}}{2} - \frac{3 - 2 \sqrt{7}}{16}]

Commented by mathmax by abdo last updated on 25/Jul/19

t h a n k y o u s i r .