advanced-calculus-evaluate-0-pi-4-ln-ln-cot-x-dx-

Question Number 125668 by mnjuly1970 last updated on 12/Dec/20

\dots a d v a n c e d c a l c u l u s \dots

e v a l u a t e :::

Φ = \int_{0}^{\frac{π}{4}} l n (l n (c o t (x))) d x = ?

Answered by mindispower last updated on 13/Dec/20

Φ = \int_{1}^{\infty} \frac{l n (l n (x))}{1 + x^{2}} d x = \int_{0}^{\infty} \frac{l n (t)}{1 + e^{2 t}} e^{t} d t

= \int_{0}^{\infty} \frac{l n (t) e^{- t}}{1 + e^{- 2 t}} d t

= \sum_{k ⩾ 0} \int_{0}^{\infty} {(- 1)}^{k} l n (t) e^{- (1 + 2 k) t}

l n (t) = \partial_{s} t^{s} ∣ s = 0

g (s) = \sum_{k ⩾ 0} \int_{0}^{\infty} {(- 1)}^{k} t^{s} e^{- (1 + 2 k) t} d t

= \sum_{k ⩾ 0} \int_{0}^{\infty} {(- 1)}^{k} \frac{y^{s} e^{- y} d y}{{(1 + 2 k)}^{s + 1}}

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(1 + 2 k)}^{s + 1}} \int_{0}^{\infty} y^{s} e^{- y} d y

= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(1 + 2 k)}^{s + 1}} Γ (s + 1)

= Γ (s + 1) \sum_{k ⩾ 0} (\frac{1}{{(1 + 4 k)}^{s + 1}} - \frac{1}{{(3 + 4 k)}^{s + 1}})

= \frac{Γ (s + 1)}{4^{s + 1}} (\sum_{k ⩾ 0} \frac{1}{{(k + \frac{1}{4})}^{s + 1}} - \sum_{k ⩾ 0} \frac{1}{{(k + \frac{3}{4})}^{s + 1}})

= \frac{Γ (s + 1)}{4^{s + 1}} (ζ (s + 1, \frac{1}{4}) - ζ (s + 1, \frac{3}{4})

ζ (s, q) = \sum_{n ⩾ 0} \frac{1}{{(n + q)}^{s}}, Z e t s h r w i t z f u n c t i o n

w e w a n t g^{'} (0)

\frac{Γ^{'} (s + 1) 4^{s + 1} - 4^{s + 1} l n (4) Γ (s + 1)}{4^{2 s + 2}} ∣_{s = 0} . \lim_{s \to 0} (ζ (s + 1, \frac{1}{4}) - ζ (s + 1, \frac{3}{4}))

+ \lim_{s \to 0} \partial_{s} (ζ (s + 1, \frac{1}{4}) - ζ (s + 1, \frac{3}{4})) . \frac{Γ (s + 1)}{4^{s + 1}}

ζ (s + 1, q) = \frac{1}{s} + Σ \frac{{(- 1)}^{n}}{n!} γ_{n} (q) s^{n}, l a u r e n t s e r i e

e x p a n s i o n γ_{n} (q) S t i e l t j e s c o n s t a n t e

ζ (s + 1, \frac{1}{4}) - ζ (s + 1, \frac{3}{4}) = h (s) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{n!} (γ_{n} (\frac{1}{4}) - γ_{n} (\frac{3}{4})) s^{n}

\lim_{s \to 0} h (s) = γ_{0} (\frac{1}{4}) - γ_{0} (\frac{3}{4})

\partial_{s} h (s) = \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{(n - 1)!} (γ_{n} (\frac{1}{4}) - γ_{n} (\frac{3}{4}) s^{n - 1}

\lim_{s \to 0} h^{'} (s) = γ_{1} (\frac{3}{4}) - γ_{1} (\frac{1}{4})

γ_{1} (\frac{3}{4}) - γ_{1} (1 - \frac{3}{4}) = 2 π \underset{l = 1}{\sum^{3}} s i n (\frac{6 π l}{4}) l n Γ (\frac{l}{4}) - π (γ + l n (8 π)) c o t (\frac{3 π}{4})

= - 2 π Γ (\frac{1}{4}) + 2 π Γ (\frac{3}{4}) + π (γ + l n (8 π))

= 2 π (Γ (\frac{3}{4}) - Γ (\frac{1}{4})) + π (γ + l n (8 π))

γ_{0} (\frac{1}{4}) - γ_{0} (\frac{3}{4}) = Ψ (\frac{3}{4}) - Ψ (\frac{1}{4}) = - π c o t (\frac{3 π}{4})

= π

g^{'} (0) = \frac{4 Γ^{'} (1) - 4 l n (4) Γ (1)}{4^{2}} π + (2 π (Γ (\frac{3}{4}) - Γ (\frac{1}{4})) + π (γ + l n (8 π)) . \frac{1}{4}

= - \frac{γ}{4} π - \frac{l n (2)}{2} π + \frac{π}{2} (Γ (\frac{3}{4}) - Γ (\frac{1}{4})) + \frac{γ π}{4} + \frac{π}{4} l n (8 π)

= \frac{π}{4} l n (2 π) + \frac{π}{2} (Γ (\frac{3}{4}) - Γ (\frac{1}{4}))

Commented by mindispower last updated on 13/Dec/20

w e c a n

u s e Γ (z) Γ (z + \frac{1}{2}) = 2^{1 - 2 z} \sqrt{π} Γ (2 z)

Γ (\frac{1}{4}) . Γ (\frac{3}{4}) = 2^{\frac{1}{2}} \sqrt{π} . Γ (\frac{1}{2}) = π \sqrt{2}

\Rightarrow Γ (\frac{3}{4}) = π \sqrt{2} . \frac{1}{Γ (\frac{1}{4})} \dots t o e x p r e s s u s i n g

j u s t Γ (\frac{1}{4})

Commented by mnjuly1970 last updated on 13/Dec/20

t h a n k s a l o t

m r p o w e r \dots \dots

Commented by mindispower last updated on 13/Dec/20

w i t h e p l e a s u r