... nice calculus... prove that:: Ω=∫_0 ^( (π/2)) {tan^(−1) (ptan(x))−tan^(−1) (qtan(x))}(tan(x)+cot(x))dx =(π/2) log((p/q)) ( p , q >0 ) m.n.


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Question Number 122159 by mnjuly1970 last updated on 14/Nov/20
$... nice calculus... prove that:: Ω=∫_0 ^( (π/2)) {tan^(−1) (ptan(x))−tan^(−1) (qtan(x))}(tan(x)+cot(x))dx =(π/2) log((p/q)) ( p , q >0 ) m.n.$
$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $Ω = \int_{0}^{\frac{π}{2}} {{t a n}^{- 1} (p t a n (x)) - {t a n}^{- 1} (q t a n (x))} (t a n (x) + c o t (x)) d x$ $= \frac{π}{2} l o g (\frac{p}{q}) (p, q > 0)$ $m . n .$
	Answered by mathmax by abdo last updated on 14/Nov/20
	$Ω =∫_0 ^(π/2) (arctan(ptanx)dx−∫_0 ^(π/2) arctan(qtanx)dx =A_p −A_q A_p =∫_0 ^(π/2) arctan(ptanx)dx =_(tanx=t) ∫_0 ^∞ ((arctan(pt))/(1+t^2 ))dt =f(p) f^′ (p) =∫_0 ^∞ (t/((1+p^2 t^2 )(1+t^2 )))dt =_(pt=x) ∫_0 ^∞ (x/(p(1+x^2 )(1+(x^2 /p^2 ))))(dx/p) =∫_0 ^∞ ((xdx)/((x^2 +1)(x^2 +p^2 ))) let decompose F(x)=(x/((x^2 +1)(x^2 +p^2 ))) F(x)=((ax+b)/(x^2 +1)) +((cx +d)/(x^2 +p^2 )) F(−x)=−f(x) ⇒((−ax+b)/(x^2 +1)) +((−cx+d)/(x^2 +p^2 )) =((−ax−b)/(x^2 +1))+((−cx−d)/(x^2 +p^2 )) ⇒b=d=0 ⇒F(x) =((ax)/(x^2 +1))+((cx)/(x^2 +p^2 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)=((ax)/(x^2 +1))−((ax)/(x^2 +p^2 )) F(1)=(1/(2(p^2 +1))) =(a/2)−(a/(p^2 +1)) ⇒((1/2)−(1/(p^2 +1)))a =(1/(2(p^2 +1))) ⇒ ((p^2 −1)/(2(p^2 +1)))a =(1/(2(p^2 +1))) ⇒a =(1/(p^2 −1)) ⇒F(x)=(1/(p^2 −1)){(x/(x^2 +1))−(x/(x^2 +p^2 ))} ⇒ f^, (p) =(1/(p^2 −1))∫_0 ^∞ {(x/(x^2 +1))−(x/(x^2 +p^2 ))}dx =(1/(p^2 −1))×(1/2)[ln∣((x^2 +1)/(x^2 +p^2 ))∣]_0 ^∞ =(1/(2(p^2 −1)))(−2lnp) =−((lnp)/(p^2 −1)) =((lnp)/(1−p^2 )) ⇒ f(p) =∫_1 ^p ((lnt)/(1−t^2 ))dt +c c =f(1) =∫_0 ^∞ ((arctant)/(1+t^2 )) =[arctan^2 t]_0 ^∞ −∫_0 ^∞ ((arctant)/(1+t^2 ))dt ⇒ 2c =((π/2))^2 =(π^2 /4) ⇒c =(π^2 /8) ⇒f(p) =∫_1 ^p ((lnt)/(1−t^2 ))dt +(π^2 /8) if p<1 ⇒∫_1 ^p ((lnt)/(1−t^2 ))dt =−∫_p ^1 ((lnt)/(1−t^2 )) =−∫_p ^1 lntΣ_(n=0) ^∞ t^(2n) dt =−Σ_(n=0) ^∞ ∫_p ^1 t^(2n) lnt dt =−Σ_(n=0) ^∞ u_n u_n =[(t^(2n+1) /(2n+1))lnt]_p ^1 −∫_p ^1 (t^(2n+1) /(2n+1))(dt/t) =−(p^(2n+1) /(2n+1))−(1/(2n+1))∫_p ^1 t^(2n) dt =−(p^(2n+1) /(2n+1))−(1/((2n+1)^2 ))[t^(2n+1) ]_p ^1 =−(p^(2n+1) /(2n+1))−(1/((2n+1)^2 ))(1−p^(2n+1) ) ⇒ ∫_1 ^p ((lnt)/(1−t^2 ))dt =Σ_(n=0) ^∞ (p^(2n+1) /(2n+1)) +Σ_(n=0) ^∞ ((1−p^(2n+1) )/((2n+1)^2 ))....becontinued...$
	$Ω = \int_{0}^{\frac{π}{2}} (\arctan (ptanx) dx - \int_{0}^{\frac{π}{2}} \arctan (qtanx) dx$ $= A_{p} - A_{q}$ $A_{p} = \int_{0}^{\frac{π}{2}} \arctan (ptanx) dx =_{tanx = t} \int_{0}^{\infty} \frac{\arctan (pt)}{1 + t^{2}} dt = f (p)$ $f^{^{'}} (p) = \int_{0}^{\infty} \frac{t}{(1 + p^{2} t^{2}) (1 + t^{2})} dt =_{pt = x} \int_{0}^{\infty} \frac{x}{p (1 + x^{2}) (1 + \frac{x^{2}}{p^{2}})} \frac{dx}{p}$ $= \int_{0}^{\infty} \frac{xdx}{(x^{2} + 1) (x^{2} + p^{2})} let decompose F (x) = \frac{x}{(x^{2} + 1) (x^{2} + p^{2})}$ $F (x) = \frac{ax + b}{x^{2} + 1} + \frac{cx + d}{x^{2} + p^{2}}$ $F (- x) = - f (x) \Rightarrow \frac{- ax + b}{x^{2} + 1} + \frac{- cx + d}{x^{2} + p^{2}} = \frac{- ax - b}{x^{2} + 1} + \frac{- cx - d}{x^{2} + p^{2}}$ $\Rightarrow b = d = 0 \Rightarrow F (x) = \frac{ax}{x^{2} + 1} + \frac{cx}{x^{2} + p^{2}}$ $\lim_{x \to + \infty} xF (x) = 0 = a + c \Rightarrow c = - a \Rightarrow$ $F (x) = \frac{ax}{x^{2} + 1} - \frac{ax}{x^{2} + p^{2}}$ $F (1) = \frac{1}{2 (p^{2} + 1)} = \frac{a}{2} - \frac{a}{p^{2} + 1} \Rightarrow (\frac{1}{2} - \frac{1}{p^{2} + 1}) a = \frac{1}{2 (p^{2} + 1)} \Rightarrow$ $\frac{p^{2} - 1}{2 (p^{2} + 1)} a = \frac{1}{2 (p^{2} + 1)} \Rightarrow a = \frac{1}{p^{2} - 1} \Rightarrow F (x) = \frac{1}{p^{2} - 1} {\frac{x}{x^{2} + 1} - \frac{x}{x^{2} + p^{2}}} \Rightarrow$ $f^{,} (p) = \frac{1}{p^{2} - 1} \int_{0}^{\infty} {\frac{x}{x^{2} + 1} - \frac{x}{x^{2} + p^{2}}} dx$ $= \frac{1}{p^{2} - 1} \times \frac{1}{2} {[\ln ∣ \frac{x^{2} + 1}{x^{2} + p^{2}} ∣]}_{0}^{\infty} = \frac{1}{2 (p^{2} - 1)} (- 2 lnp) = - \frac{lnp}{p^{2} - 1} = \frac{lnp}{1 - p^{2}}$ $\Rightarrow f (p) = \int_{1}^{p} \frac{lnt}{1 - t^{2}} dt + c$ $c = f (1) = \int_{0}^{\infty} \frac{arctant}{1 + t^{2}} = {[\arctan^{2} t]}_{0}^{\infty} - \int_{0}^{\infty} \frac{arctant}{1 + t^{2}} dt \Rightarrow$ $2 c = {(\frac{π}{2})}^{2} = \frac{π^{2}}{4} \Rightarrow c = \frac{π^{2}}{8} \Rightarrow f (p) = \int_{1}^{p} \frac{lnt}{1 - t^{2}} dt + \frac{π^{2}}{8}$ $if p < 1 \Rightarrow \int_{1}^{p} \frac{lnt}{1 - t^{2}} dt = - \int_{p}^{1} \frac{lnt}{1 - t^{2}} = - \int_{p}^{1} lnt \sum_{n = 0}^{\infty} t^{2 n} dt$ $= - \sum_{n = 0}^{\infty} \int_{p}^{1} t^{2 n} lnt dt = - \sum_{n = 0}^{\infty} u_{n}$ $u_{n} = {[\frac{t^{2 n + 1}}{2 n + 1} lnt]}_{p}^{1} - \int_{p}^{1} \frac{t^{2 n + 1}}{2 n + 1} \frac{dt}{t} = - \frac{p^{2 n + 1}}{2 n + 1} - \frac{1}{2 n + 1} \int_{p}^{1} t^{2 n} dt$ $= - \frac{p^{2 n + 1}}{2 n + 1} - \frac{1}{{(2 n + 1)}^{2}} {[t^{2 n + 1}]}_{p}^{1} = - \frac{p^{2 n + 1}}{2 n + 1} - \frac{1}{{(2 n + 1)}^{2}} (1 - p^{2 n + 1}) \Rightarrow$ $\int_{1}^{p} \frac{lnt}{1 - t^{2}} dt = \sum_{n = 0}^{\infty} \frac{p^{2 n + 1}}{2 n + 1} + \sum_{n = 0}^{\infty} \frac{1 - p^{2 n + 1}}{{(2 n + 1)}^{2}} . . . . becontinued . . .$
	Answered by mindispower last updated on 14/Nov/20

	$n o t t r u e s i r$ $q = 1 \Rightarrow \int_{0}^{\frac{π}{2}} {t a n}^{-} (p t a n (x)) - x d x = \frac{π}{2} l n (p)$ $t a k p \to \infty \Rightarrow$ $\int_{0}^{\frac{π}{2}} {t a n}^{-} (p t a n (x)) - x d x = I \to + \infty$ $b u t {t a n}^{-} (z) < \frac{π}{2} \Rightarrow$ $I ⩽ \int_{0}^{\frac{π}{2}} (\frac{π}{2} - x) d x = \frac{π^{2}}{8}$
	Commented bymnjuly1970 last updated on 14/Nov/20

	$t h a n k y o u$ $c o r r e c t e d$
	Answered by mindispower last updated on 14/Nov/20
	$∫_0 ^(π/2) {tan^− (t.tan(x))}{tan(x)+cot(x)}dx=f(t) =∫_0 ^(π/2) {tan^− (t.tan(x)){((1+tan^2 (x))/(tan(x)))}dx,tan(x)=s ⇔ =∫_0 ^∞ tan^− (ts).(ds/s) f(p)−f(q)=∫_0 ^∞ (tan^− (ps)−tan^− (qs))(ds/s) =[(tan^− (ps)−tan^− (qs))ln(x)]_0 ^∞ −∫_0 ^∞ {p((ln(s))/(1+p^2 s2))−q((ln(s))/(1+q^2 s^2 ))}ds tan^− (ps)−tan^− (qs)=((π/2)−(1/(ps))+o((1/s))−((π/2)−(1/(qs))+o((1/s))) =(1/s)((1/q)−(1/p))+o((1/s))⇒lim_(x→∞) ln(x){tan^− (ps)−tan^− (qs)}=0 ⇔ f(p)−f(q)=−∫_0 ^∞ ((pln(s))/(1+p^2 s2))−((qln(s))/(1+q^2 s^2 ))ds=−(g(p)−g(q)) g(z)=∫_0 ^∞ ((zln(x))/(1+z^2 x^2 ))dx,z>0 withe Quation llet zx=t ⇔g(z)=∫_0 ^∞ ((ln(t)−ln(z))/(1+t^2 ))dt =∫_0 ^∞ ((ln(t))/(1+t^2 ))dt_(=0) −ln(z)∫_0 ^∞ (dt/(1+t^2 )) =−ln(z)[tan^(−1) (t)]_0 ^∞ =−(π/2)ln(z) ∫_0 ^∞ {tan^(−1) (ptan(x))−tan^− (qtan(x)}(tan(x)+cot(x))dx =−g(p)+g(q)=(π/2)ln(p)−(π/2)ln(q)=(π/2)ln((p/q))$
	$\int_{0}^{\frac{π}{2}} {{t a n}^{-} (t . t a n (x))} {t a n (x) + c o t (x)} d x = f (t)$ $= \int_{0}^{\frac{π}{2}} {{t a n}^{-} (t . t a n (x)) {\frac{1 + {t a n}^{2} (x)}{t a n (x)}} d x, t a n (x) = s$ $\Leftrightarrow$ $= \int_{0}^{\infty} {t a n}^{-} (t s) . \frac{d s}{s}$ $f (p) - f (q) = \int_{0}^{\infty} ({t a n}^{-} (p s) - {t a n}^{-} (q s)) \frac{d s}{s}$ $= {[({t a n}^{-} (p s) - {t a n}^{-} (q s)) l n (x)]}_{0}^{\infty} - \int_{0}^{\infty} {p \frac{l n (s)}{1 + p^{2} s 2} - q \frac{l n (s)}{1 + q^{2} s^{2}}} d s$ ${t a n}^{-} (p s) - {t a n}^{-} (q s) = (\frac{π}{2} - \frac{1}{p s} + o (\frac{1}{s}) - (\frac{π}{2} - \frac{1}{q s} + o (\frac{1}{s}))$ $= \frac{1}{s} (\frac{1}{q} - \frac{1}{p}) + o (\frac{1}{s}) \Rightarrow \lim_{x \to \infty} l n (x) {{t a n}^{-} (p s) - {t a n}^{-} (q s)} = 0$ $\Leftrightarrow$ $f (p) - f (q) = - \int_{0}^{\infty} \frac{p l n (s)}{1 + p^{2} s 2} - \frac{q l n (s)}{1 + q^{2} s^{2}} d s = - (g (p) - g (q))$ $g (z) = \int_{0}^{\infty} \frac{z l n (x)}{1 + z^{2} x^{2}} d x, z > 0 w i t h e Q u a t i o n$ $l l e t z x = t$ $\Leftrightarrow g (z) = \int_{0}^{\infty} \frac{l n (t) - l n (z)}{1 + t^{2}} d t$ $= \int_{0}^{\infty} \frac{l n (t)}{1 + t^{2}} {d t}_{= 0} - l n (z) \int_{0}^{\infty} \frac{d t}{1 + t^{2}}$ $= - l n (z) {[\tan^{- 1} (t)]}_{0}^{\infty} = - \frac{π}{2} l n (z)$ $\int_{0}^{\infty} {\tan^{- 1} (p t a n (x)) - {t a n}^{-} (q t a n (x)} (t a n (x) + c o t (x)) d x$ $= - g (p) + g (q) = \frac{π}{2} l n (p) - \frac{π}{2} l n (q) = \frac{π}{2} l n (\frac{p}{q})$
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