nice-calculus-prove-that-0-sin-x-log-2-x-x-pi-24-12-2-pi-2-

Question Number 132877 by mnjuly1970 last updated on 17/Feb/21

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

p r o v e t h a t :

ϕ = \int_{0}^{\infty} \frac{s i n (x) . {l o g}^{2} (x)}{x}

= \frac{π}{24} (12 γ^{2} + π^{2}) \dots . .

Answered by Dwaipayan Shikari last updated on 17/Feb/21

ϕ (α) = \int_{0}^{\infty} \frac{s i n x}{x^{α}} d x

ϕ (α) = \frac{1}{Γ (α)} \int_{0}^{\infty} \int_{0}^{\infty} t^{α - 1} e^{- x t} s i n (x) d t d x

= \frac{1}{2 i Γ (α)} \int_{0}^{\infty} \int_{0}^{\infty} t^{α - 1} e^{- (t - i) x} - t^{α - 1} e^{- (t + i) x} d x

= \frac{1}{2 i Γ (α)} \int_{0}^{\infty} \frac{t^{α - 1}}{t - i} - \frac{t^{α - 1}}{t + i} d t

= \frac{1}{Γ (α)} \int_{0}^{\infty} \frac{t^{α - 1}}{t^{2} + 1} d t = \frac{1}{2 Γ (α)} \int_{0}^{\infty} \frac{u^{\frac{α}{2} - 1}}{{(u + 1)}^{1 - \frac{α}{2} + \frac{α}{2}}} d u

= \frac{1}{2 Γ (α)} . \frac{Γ (\frac{α}{2}) Γ (1 - \frac{α}{2})}{Γ (1)} = \frac{π}{2 Γ (α) s i n (\frac{π α}{2})}

ϕ^{″} (α) ∣_{α = 1} = \int_{0}^{\infty} \frac{s i n (x) {l o g}^{2} (x)}{x} d x = \frac{π}{24} (12 γ^{2} + π^{2})

Commented by Dwaipayan Shikari last updated on 17/Feb/21

ϕ^{'} (α) = - (\frac{π^{2}}{4 Γ (α)} . \frac{c o s (\frac{π}{2} α)}{{s i n}^{2} (\frac{π}{2} α)} + \frac{\frac{π^{2}}{2} Γ (α) . \frac{c o s (\frac{π}{2} α)}{{s i n}^{2} (\frac{π}{2} α)}}{2 Γ^{2} (α)} + \frac{π Γ^{'} (α) c o s e c (\frac{π}{2} 2 α)}{2 Γ^{2} (α)})

ϕ^{'} (1) = \frac{π γ}{2}

Commented by mnjuly1970 last updated on 17/Feb/21

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