let ϕ(a) =∫_1 ^(√3) arctan((a/x))dx 1) calculate ϕ(a) interms of a 2) calculate ϕ^′ (a) at form of integral. 3) determine ϕ^((n)) (a) at form of integral. 4) find the value of ∫


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Question Number 55274 by maxmathsup by imad last updated on 20/Feb/19

$l e t φ (a) = \int_{1}^{\sqrt{3}} a r c t a n (\frac{a}{x}) d x$ $1) c a l c u l a t e φ (a) i n t e r m s o f a$ $2) c a l c u l a t e φ^{^{'}} (a) a t f o r m o f i n t e g r a l .$ $3) d e t e r m i n e φ^{(n)} (a) a t f o r m o f i n t e g r a l .$ $4) f i n d t h e v a l u e o f \int_{1}^{\sqrt{3}} a r c t a n (\frac{2}{x}) d x .$
Commented by maxmathsup by imad last updated on 24/Feb/19
$1) changement (a/x) =t give x =(a/t) ⇒ϕ(a)=∫_a ^(a/(√3)) arctan(t)(((−a)/t^2 ))dt ⇒ ((ϕ(a))/a)= ∫_(a/(√3)) ^a ((arctan(t))/t^2 ) dt by parts we get ((ϕ(a))/a) =[−(1/t) arctan(t)]_(a/(√3)) ^a +∫_(a/(√3)) ^a (1/(t(1+t^2 )))dt =((√3)/a) arctan((a/(√3))) −((arctan(a))/a) +∫_(a/(√3)) ^a (dt/(t(1+t^2 )))dt but ∫_(a/(√3)) ^a (dt/(t(1+t^2 )))dt =∫_(a/(√3)) ^a ((1/t) −(t/(1+t^2 )))dt = [ln((t/(√(1+t^2 ))))]_(a/(√3)) ^a =ln((a/(√(1+a^2 )))) −ln((a/((√3)((√(1+(a^2 /3))))))=ln((a/(√(1+a^2 ))))−ln((a/(√(a^2 +3)))) =ln(a)−(1/2)ln(1+a^2 )−ln(a)+(1/2)ln(a^2 +3) =(1/2){ln(a^2 +3)−ln(a^2 +1)} ⇒ ((ϕ(a))/a) =((√3)/a) arctan((a/(√3)))−((arctan(a))/a) +ln((√((a^2 +3)/(a^2 +1)))) ⇒ ϕ(a) =(√3)arctan((a/(√3)))−arctan(a) +aln((√((a^2 +3)/(a^2 +1)))) .$
$1) c h a n g e m e n t \frac{a}{x} = t g i v e x = \frac{a}{t} \Rightarrow φ (a) = \int_{a}^{\frac{a}{\sqrt{3}}} a r c t a n (t) (\frac{- a}{t^{2}}) d t \Rightarrow$ $\frac{φ (a)}{a} = \int_{\frac{a}{\sqrt{3}}}^{a} \frac{a r c t a n (t)}{t^{2}} d t b y p a r t s w e g e t \frac{φ (a)}{a} = {[- \frac{1}{t} a r c t a n (t)]}_{\frac{a}{\sqrt{3}}}^{a} + \int_{\frac{a}{\sqrt{3}}}^{a} \frac{1}{t (1 + t^{2})} d t$ $= \frac{\sqrt{3}}{a} a r c t a n (\frac{a}{\sqrt{3}}) - \frac{a r c t a n (a)}{a} + \int_{\frac{a}{\sqrt{3}}}^{a} \frac{d t}{t (1 + t^{2})} d t b u t$ $\int_{\frac{a}{\sqrt{3}}}^{a} \frac{d t}{t (1 + t^{2})} d t = \int_{\frac{a}{\sqrt{3}}}^{a} (\frac{1}{t} - \frac{t}{1 + t^{2}}) d t = {[l n (\frac{t}{\sqrt{1 + t^{2}}})]}_{\frac{a}{\sqrt{3}}}^{a} = l n (\frac{a}{\sqrt{1 + a^{2}}})$ $- l n (\frac{a}{\sqrt{3} (\sqrt{1 + \frac{a^{2}}{3}}}) = l n (\frac{a}{\sqrt{1 + a^{2}}}) - l n (\frac{a}{\sqrt{a^{2} + 3}})$ $= l n (a) - \frac{1}{2} l n (1 + a^{2}) - l n (a) + \frac{1}{2} l n (a^{2} + 3) = \frac{1}{2} {l n (a^{2} + 3) - l n (a^{2} + 1)} \Rightarrow$ $\frac{φ (a)}{a} = \frac{\sqrt{3}}{a} a r c t a n (\frac{a}{\sqrt{3}}) - \frac{a r c t a n (a)}{a} + l n (\sqrt{\frac{a^{2} + 3}{a^{2} + 1}}) \Rightarrow$ $φ (a) = \sqrt{3} a r c t a n (\frac{a}{\sqrt{3}}) - a r c t a n (a) + a l n (\sqrt{\frac{a^{2} + 3}{a^{2} + 1}}) .$
Commented by maxmathsup by imad last updated on 24/Feb/19

$2) w e h a v e φ^{^{'}} (a) = \int_{1}^{\sqrt{3}} \frac{1}{x (1 + \frac{a^{2}}{x^{2}})} d x = \int_{1}^{\sqrt{3}} \frac{x}{x^{2} + a^{2}} d x .$
Commented by maxmathsup by imad last updated on 24/Feb/19

$4) w e h a v e φ (a) = \int_{1}^{\sqrt{3}} a r c t a n (\frac{a}{x}) d x \Rightarrow$ $\int_{1}^{\sqrt{3}} a r c t a n (\frac{2}{x}) d x = φ (2) = \sqrt{3} a r c t a n (\frac{2}{\sqrt{3}}) - a r c t a n (2) + 2 l n (\sqrt{\frac{7}{5}})$ $= \sqrt{3} a r c t a n (\frac{2}{\sqrt{3}}) - a r c t a n (2) + l n (7) - l n (5) .$
Commented by maxmathsup by imad last updated on 24/Feb/19
$3) we have ϕ^((1)) (a) =∫_1 ^(√3) (x/(x^2 +a^2 ))dx ⇒ϕ^((n)) (a) =∫_1 ^(√3) (d^(n−1) /da^(n−1) ) ((x/(x^2 +a^2 ))) dx let =∫_1 ^(√3) x {(1/(x^2 +a^2 ))}_(/a) ^((n−1)) dx =∫_1 ^(√3) (1/(2i)) {(1/(a−ix)) −(1/(a+ix))}_(/a) ^((n−1)) dx =(1/(2i)) ∫_1 ^(√3) { (((−1)^(n−1) (n−1)!)/((a−ix)^n )) −(((−1)^(n−1) (n−1)!)/((a+ix)^n ))}dx =(((−1)^(n−1) (n−1)!)/(2i)) ∫_1 ^(√3) (((a+ix)^n −(a−ix)^n )/((a^2 +x^2 )^n )) dx =(−1)^(n−1) (n−1)! ∫_1 ^(√3) ((Im((a+ix)^n ))/((a^2 +x^2 )^n )) dx .$
$3) w e h a v e φ^{(1)} (a) = \int_{1}^{\sqrt{3}} \frac{x}{x^{2} + a^{2}} d x \Rightarrow φ^{(n)} (a) = \int_{1}^{\sqrt{3}} \frac{d^{n - 1}}{{d a}^{n - 1}} (\frac{x}{x^{2} + a^{2}}) d x l e t$ $= \int_{1}^{\sqrt{3}} x {\frac{1}{x^{2} + a^{2}}}_{/ a}^{(n - 1)} d x = \int_{1}^{\sqrt{3}} \frac{1}{2 i} {\frac{1}{a - i x} - \frac{1}{a + i x}}_{/ a}^{(n - 1)} d x$ $= \frac{1}{2 i} \int_{1}^{\sqrt{3}} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(a - i x)}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(a + i x)}^{n}}} d x$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} \int_{1}^{\sqrt{3}} \frac{{(a + i x)}^{n} - {(a - i x)}^{n}}{{(a^{2} + x^{2})}^{n}} d x$ $= {(- 1)}^{n - 1} (n - 1)! \int_{1}^{\sqrt{3}} \frac{I m ({(a + i x)}^{n})}{{(a^{2} + x^{2})}^{n}} d x .$
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