nice-calculus-prove-that-0-pi-2-tan-1-ptan-x-tan-1-qtan-x-tan-x-cot-x-dx-pi-2-log-p-q-p-q-gt-0-m-n-

Question Number 122159 by mnjuly1970 last updated on 14/Nov/20

\dots n i c e c a l c u l u s \dots

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 = \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}} \dots . becontinued \dots

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 by mnjuly1970 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

\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})