advanced-calculus-prove-that-0-sin-4-x-ln-x-x-2-dx-pi-4-1-euler-mascheroni-constant-m-n-july-1970-

Question Number 121674 by mnjuly1970 last updated on 10/Nov/20

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

p r o v e t h a t :

Ω = \int_{0}^{\infty} \frac{{s i n}^{4} (x) l n (x)}{x^{2}} d x = \frac{π}{4} (1 - γ)

γ : e u l e r - m a s c h e r o n i c o n s t a n t

m . n . j u l y . 1970

Answered by mindispower last updated on 11/Nov/20

Ω = \int_{0}^{\infty} \frac{{s i n}^{4} (x)}{x^{2}} l n (x) d x

{s i n}^{4} (x) = \frac{1}{8} (c o s (4 x) - 4 c o s (2 x) + 3)

8 Ω = \int_{0}^{\infty} (c o s (4 x) - 4 c o s (2 x) + 3) . \frac{l n (x) d x}{x^{2}}

b y p a r t u = c o s (4 x) - 4 c o s (2 x) + 3

d v = \frac{l n (x)}{x^{2}}, v = - \frac{l n (x)}{x} - \frac{1}{x}

8 Ω = {[- \frac{(l n (x) + 1)}{x} (c o s (4 x) - 4 c o s (2 x) + 3)]}_{0}^{\infty}

+ \int_{0}^{\infty} \frac{(l n (x) + 1)}{x} (- 4 s i n (4 x) + 8 s i n (2 x)) d x

8 Ω = - 4 \int_{0}^{\infty} \frac{l n (x) + 1}{x} (s i n (4 x) - 2 s i n (2 x)) d x

\Leftrightarrow - 2 Ω = {\int_{0}^{\infty} \frac{l n (x)}{x} s i n (4 x) - 2 \int \frac{l n (x)}{x} s i n (2 x)}_{= W}

+ {\int_{0}^{\infty} \frac{s i n (4 x)}{x} - 2 \int_{0}^{\infty} \frac{s i n (2 x)}{x} d x}_{= V}

a > 0, \int_{0}^{\infty} \frac{s i n (a x)}{x} d x = \int_{0}^{\infty} \frac{s i n (a x)}{a x} d (a x) = \frac{π}{2}

V = \frac{π}{2} - 2 . \frac{π}{2} = - \frac{π}{2}

\int_{0}^{\infty} \frac{s i n (a x)}{x} l n (x) = f (a) = \frac{\partial}{\partial t} \int_{0}^{\infty} s i n (a x) x^{t - 1} d x ∣_{t = 0}

f (t) = \int_{0}^{\infty} s i n (a x) x^{t - 1} d x = I m \int_{0}^{\infty} e^{i a x} x^{t - 1} d x

i a x = - z \Leftrightarrow I m \int_{0}^{i \infty} e^{- z} . {(\frac{i z}{a})}^{t} . \frac{a}{z} . \frac{d z}{a}

= I m a \int_{0}^{i \infty} e^{i \frac{π}{2} t} {(\frac{z}{a})}^{t - 1} e^{- z} d z

= I m - \frac{e^{i \frac{π}{2} t}}{a^{t}} \int_{0}^{\infty} z^{t - 1} e^{- z} d z

= - \frac{s i n (\frac{π t}{2}) Γ (t)}{a^{t}} = \frac{- s i n (\frac{π t}{2}) π}{a^{t} Γ (1 - t) s i n (\frac{π t}{2})} = - \frac{π}{2 c o s (\frac{π}{2} t) Γ (1 - t) a^{t}}

f^{'} (t) ∣_{t = 0} = \frac{π}{2} (- γ - l n (a))) = - \frac{π}{2} (γ + l n (a))

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

= \frac{π}{2} γ

- 2 Ω = W + V = \frac{π}{2} γ - \frac{π}{2} \Rightarrow Ω = - \frac{π γ}{4} + \frac{π}{4} = \frac{π}{4} (1 - γ)

Commented by mnjuly1970 last updated on 11/Nov/20

p e a c e b e u p o n y o u

m r p o w e r . t h a n k y o u .

Commented by mindispower last updated on 11/Nov/20

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

Answered by mnjuly1970 last updated on 11/Nov/20

Answered by mnjuly1970 last updated on 11/Nov/20