prove-that-I-0-pi-2-arccosh-sin-x-cos-x-dx-pi-2-ln-2-

Question Number 141916 by mnjuly1970 last updated on 24/May/21

p r o v e t h a t ::

I := \int_{0}^{\frac{π}{2}} a r c c o s h (s i n (x) + c o s (x)) d x = \frac{π}{2} l n (2) . .

Answered by mindispower last updated on 25/May/21

b y p a r t w i t h e d (a r c c o h (x)) = \frac{d x}{\sqrt{1 - x^{2}}}

{I = x a r c c o s h (s i n x + c o s (x))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{x (c o s (x) - s i n (x))}{\sqrt{s i n (2 x)}} d x

- \int_{0}^{\frac{π}{2}} \frac{x c o s (x) - x s i n (x)}{\sqrt{s i n (2 x)}} d x

= - \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} {x s i n}^{- \frac{1}{2}} (x) {c o s}^{\frac{1}{2}} (x) + \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} {x s i n}^{\frac{1}{2}} (x) {c o s}^{- \frac{1}{2}} (x)

\frac{1}{\sqrt{2}} \int_{0}^{\infty} \frac{a r c t g (x) (\sqrt{x} - \frac{1}{\sqrt{x}})}{1 + x^{2}} d x = A

\int_{0}^{\infty} \frac{x^{s} a r c t a n (t x)}{1 + x^{2}} d x = f (t), s \in] - 1, 1 [

f^{'} (t) = \int_{0}^{\infty} \frac{x^{s + 1}}{(1 + x^{2}) (1 + t^{2} x^{2})} d x

\int_{- \infty}^{\infty} \frac{x^{s + 1}}{(1 + x^{2}) (1 + t^{2} x^{2})} d x = a = 2 i π (. \frac{i^{s + 1}}{2 i (1 - t^{2})} + \frac{i^{s + 1} t^{2}}{t^{s + 1} . (t^{2} - 1)} . \frac{1}{2 i t})

= \frac{π i^{s + 1}}{1 - t^{2}} (1 - t^{- s})

a = f (t) + \int_{0}^{\infty} \frac{{(- x)}^{s + 1}}{(1 + x^{2}) (1 + t^{2} x^{2})} d x

= (1 + {(- 1)}^{s + 1}) f (t)

= (1 + e^{i π (s + 1)}) f (t)

f (t) = \frac{π}{1 - t^{2}} (1 - t^{- s}) . \frac{e^{i \frac{π}{2} (s + 1)}}{1 + e^{i π (s + 1)}}

= \frac{π (1 - t^{- s})}{2 (1 - t^{2}) c o s (\frac{π}{2} (s + 1))} = f^{'} (t)

\int_{0}^{1} f^{'} (t) = \frac{x^{s} a r c t a n (x)}{1 + x^{2}}

= \frac{π}{2 c o s (\frac{π}{2} (s + 1))} \int_{0}^{1} \frac{1 - t^{- s}}{1 - t^{2}} {d t}_{t^{2} = x}

= - \frac{π}{4 s i n (\frac{π s}{2})} \int_{0}^{1} \frac{1 - x^{- \frac{s}{2}}}{1 - x} x^{- \frac{1}{2}}

= - \frac{π}{4 s i n (\frac{π s}{2})} \int_{0}^{1} \frac{x^{- \frac{1}{2}} - x^{- \frac{s + 1}{2}}}{1 - x} d x = g (s)

Ψ (s + 1) = - γ + \int_{0}^{1} \frac{1 - x^{s}}{1 - x} d x

g (s) = - \frac{π}{4 s i n (\frac{π s}{2})} . (Ψ (\frac{1 - s}{2}) - Ψ (\frac{1}{2}))

A = - \frac{π}{4 s i n (\frac{π}{4}) \sqrt{2}} (Ψ (\frac{1}{4}) - 2 Ψ (\frac{1}{2}) + Ψ (\frac{3}{4}))

Ψ (\frac{1}{2}) = - 2 l n (2) - γ

Ψ (\frac{1}{4}) = (- \frac{π}{2} - 3 l n (2) - γ)

Ψ (\frac{3}{4}) = Ψ (\frac{1}{4}) + π

= - \frac{π}{4} (- π - 6 l n (2) - 2 γ + π + 4 l n (2) + 2 γ)

= - \frac{π}{4} . - 2 l n (2) = \frac{π}{2} l n (2)

Commented by mnjuly1970 last updated on 25/May/21

v e r y n i c e . t h a n k s a l o t \dots

Commented by mindispower last updated on 25/May/21

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