let-f-x-pi-3-pi-2-d-1-xtan-with-x-real-1-find-a-explicit-form-for-f-x-2-determine-also-g-x-pi-3-pi-2-tan-1-xtan-2-d-3-let-U-n-x-f-n-x-give-U-n-x-

Question Number 58774 by maxmathsup by imad last updated on 29/Apr/19

l e t f (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + x t a n θ} w i t h x r e a l

1) f i n d a e x p l i c i t f o r m f o r f (x)

2) d e t e r m i n e a l s o g (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ}{{(1 + x t a n θ)}^{2}} d θ

3) l e t U_{n} (x) = f^{(n)} (x) g i v e U_{n} (x) a t f o r m o f i n t e g r a l .

4) c a l c u l a t e \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 t a n θ} a n d \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ d θ}{{(1 + 2 t a n θ)}^{2}}

Commented by maxmathsup by imad last updated on 01/May/19

1) w e h a v e f (x) = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + x \frac{2 t a n (\frac{θ}{2})}{1 - {t a n}^{2} (\frac{θ}{2})}} =_{t a n (\frac{θ}{2}) = t} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{1 + \frac{2 x t}{1 - t^{2}}} \frac{2 d t}{1 + t^{2}}

= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (1 - t^{2}) d t}{(1 - t^{2} + 2 x t) (1 + t^{2})} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (t^{2} - 1)}{(t^{2} - 2 x t - 1) (t^{2} + 1)} d t r o o t s o f

t^{2} - 2 x t - 1 \to Δ^{^{'}} = x^{2} + 1 \Rightarrow t_{1} = x + \sqrt{1 + x^{2}} a n d t_{2} = x - \sqrt{1 + x^{2}} l e t d e c o m p o s e

F (t) = \frac{2 (t^{2} - 1)}{(t^{2} - 2 x t - 1) (t^{2} + 1)} \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} = \frac{2 (t^{2} - 1)}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)}

a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{2 (t_{1}^{2} - 1)}{(t_{1} - t_{2}) (t_{1}^{2} + 1)} = \frac{2 (t_{1}^{2} - 1)}{2 \sqrt{1 + x^{2}} (t_{1}^{2} + 1)}

b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{2 (t_{2}^{2} - 1)}{- 2 \sqrt{1 + x^{2}} (t_{2}^{2} + 1)}

{l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - a - b \Rightarrow F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{(- a - b) t + d}{t^{2} + 1}

F (0) = 2 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d \Rightarrow d = \frac{a}{t_{1}} + \frac{b}{t_{2}} + 2 \Rightarrow

\int_{\frac{1}{\sqrt{3}}}^{1} F (t) d t = a \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{1}} + b \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{2}} - \frac{a + b}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 d t}{t^{2} + 1} + d \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t^{2} + 1}

= {[a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣]}_{\frac{1}{\sqrt{3}}}^{1} - \frac{a + b}{2} {[l n (t^{2} + 1)]}_{\frac{1}{\sqrt{3}}}^{1} + d {[a r c t a n (t)]}_{\frac{1}{\sqrt{3}}}^{1}

= a l n ∣ 1 - t_{1} ∣ + b l n ∣ 1 - t_{2} ∣ - a l n ∣ \frac{1}{\sqrt{3}} - t_{1} ∣ - b l n ∣ \frac{1}{\sqrt{3}} - t_{2} ∣ - \frac{a + b}{2} {l n (2) - l n (\frac{4}{3})}

+ d {\frac{π}{4} - \frac{π}{6}} = a l n ∣ \frac{1 - t_{1}}{\frac{1}{\sqrt{3}} - t_{1}} ∣ + b l n ∣ \frac{1 - t_{2}}{\frac{1}{\sqrt{3}} - t_{2}} ∣ - \frac{a + b}{2} (l n (3) - l n (2)) + \frac{π d}{12}

a = \frac{2 ({(x + \sqrt{1 + x^{2}})}^{2} - 1)}{2 \sqrt{1 + x^{2}} ({(x + \sqrt{1 + x^{2}})}^{2} + 1)} = \frac{{x^{2} + 2 x \sqrt{1 + x^{2}} + x^{2}}}{\sqrt{1 + x^{2}} (x^{2} + 2 x \sqrt{1 + x^{2}} + x^{2} + 2)}

= \frac{{x^{2} + x \sqrt{1 + x^{2}}}}{\sqrt{1 + x^{2}} {x^{2} + x \sqrt{1 + x^{2} + 1}}}

b = \frac{2 ({(x - \sqrt{1 + x^{2}})}^{2} - 1)}{- 2 \sqrt{1 + x^{2}} ({(x - \sqrt{1 + x^{2}})}^{2} + 1)} = - \frac{x^{2} - 2 x \sqrt{1 + x^{2}} + x^{2}}{\sqrt{1 + x^{2}} (x^{2} - 2 x \sqrt{1 + x^{2}} + x^{2} + 2)}

= - \frac{(x^{2} - x \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}} (x^{2} - x \sqrt{1 + x^{2} + 1} + 1)} \dots . b e c o n t i n u e d \dots .

Commented by maxmathsup by imad last updated on 01/May/19

2) w e h a v e f^{^{'}} (x) = - \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ d θ}{{(1 + x t a n θ)}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x) r e s t t o c a l c u l a t e

f^{^{'}} (x)

3) w e h a v e f^{(1)} (x) = (- 1) \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{t a n θ}{{(1 + x t a n θ)}^{2}} d θ \Rightarrow f^{(2)} (x) = {(- 1)}^{2} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{2 (1 + x t a n θ)}{{(1 + x t a n θ)}^{4}} {t a n}^{2} θ d θ

= 2 . {(- 1)}^{2} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{2} θ}{{(1 + x t a n θ)}^{3}} d θ l e t s u p p o s e f^{(n)} (x) = {(- 1)}^{n} n! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{n} θ}{{(1 + x t a n θ)}^{n + 1}} d θ

\Rightarrow f^{(n + 1)} (x) = {(- 1)}^{n + 1} n! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{(n + 1) t a n θ {(1 + x t a n θ)}^{n} {t a n}^{n} θ}{{(1 + x t a n θ)}^{2 n + 2}} d θ

= {(- 1)}^{n + 1} (n + 1)! \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{t a n}^{n + 1} θ}{{(1 + x t a n θ)}^{n + 2}} d θ s o t h e r e s u l t i s p r o v e d .

Commented by maxmathsup by imad last updated on 01/May/19

4) l e t A = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 t a n θ} \Rightarrow A = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{d θ}{1 + 2 \frac{2 t a n (\frac{θ}{2})}{1 - {t a n}^{2} (\frac{θ}{2})}}

A =_{t a n (\frac{θ}{2}) = t} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{1 + \frac{4 t}{1 - t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (1 - t^{2})}{(1 - t^{2} + 4 t) (1 + t^{2})} d t

= \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 (t^{2} - 1)}{(t^{2} - 4 t - 1) (t^{2} + 1)} d t l e t d e c o m p o s e F (t) = \frac{2 (t^{2} - 1)}{(t^{2} - 4 t - 1) (t^{2} + 1)}

r o o t f t^{2} - 4 t - 1 \to Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5} \Rightarrow

F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1} = \frac{2 (t^{2} - 1)}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)}

a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{2 (t_{1}^{2} - 1)}{2 \sqrt{5} (t_{1}^{2} + 1)} = \frac{{(2 + \sqrt{5})}^{2} - 1}{\sqrt{5} ({(2 + \sqrt{5})}^{2} + 1)}

= \frac{4 + 4 \sqrt{5} + 5 - 1}{\sqrt{5} (4 + 4 \sqrt{5} + 5 + 1)} = \frac{8 + 4 \sqrt{5}}{\sqrt{5} (10 + 4 \sqrt{5})} = \frac{4 + 2 \sqrt{5}}{\sqrt{5} (5 + 2 \sqrt{5})}

b = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{2 (t_{2}^{2} - 1)}{- 2 \sqrt{5} (t_{2}^{2} + 1)} = - \frac{{(2 - \sqrt{5})}^{2} - 1}{\sqrt{5} ({(2 - \sqrt{5})}^{2} + 1)}

= - \frac{4 - 4 \sqrt{5} + 5 - 1}{\sqrt{5} (4 - 4 \sqrt{5} + 5 + 1)} = - \frac{8 - 4 \sqrt{5}}{\sqrt{5} (10 - 4 \sqrt{5})} = - \frac{4 - 2 \sqrt{5}}{\sqrt{5} (5 - 2 \sqrt{5})} = \frac{- 4 + 2 \sqrt{5}}{\sqrt{5} (5 - 2 \sqrt{5})}

{l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - (a + b)

F (0) = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d = 2 \Rightarrow d = 2 + \frac{a}{t_{1}} + \frac{b}{t_{2}} \Rightarrow

\int_{\frac{1}{\sqrt{3}}}^{1} F (t) d t = a \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{1}} + b \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{t - t_{2}} + \frac{c}{2} \int_{\frac{1}{\sqrt{3}}}^{1} \frac{2 t}{t^{2} + 1} d t + d \int_{\frac{1}{\sqrt{3}}}^{1} \frac{d t}{1 + t^{2}}

= {[a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + \frac{c}{2} l n (t^{2} + 1) + d a r c t a n (t)]}_{\frac{1}{\sqrt{3}}}^{1}

= a l n ∣ 1 - t_{1} ∣ + b l n ∣ 1 - t_{2} ∣ + \frac{c}{2} l n (2) + \frac{d π}{4} - a l n ∣ \frac{1}{\sqrt{3}} - t_{1} ∣ - b l n ∣ \frac{1}{\sqrt{3}} - t_{2} ∣

- \frac{c}{2} l n (\frac{4}{3}) - \frac{d π}{6} \dots .