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Question Number 131529 by mnjuly1970 last updated on 05/Feb/21

\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 ::: ::

ϕ = \int_{0}^{\infty} \frac{s i n (x^{4}) l n (x)}{x} d x = - \frac{π γ}{32}

n o t e : \int_{0}^{\infty} \frac{s i n (x) l n (x)}{x} d x \overset{w h y ? ? ?}{=} \frac{- π γ}{2}

ϕ \overset{⟨ x^{4} = t ⟩}{=} \frac{1}{4} \int_{0}^{\infty} \frac{s i n (t) l n (t^{\frac{1}{4}})}{t^{\frac{1}{4}}} * \frac{1}{t^{\frac{3}{4}}} d t

= \frac{1}{16} \int_{0}^{\infty} \frac{s i n (t) l n (t)}{t} d t \overset{n o t e}{=} \frac{- π γ}{32}

(ϕ = - \frac{π γ}{32})

n o t i c e :: y o u w i l l p r o v e t h a t :: \int_{0}^{\infty} \frac{s i n (x) l n (x)}{x} d x = \frac{- π γ}{2}

H i n t :: \int_{0}^{\infty} \frac{s i n (x)}{x^{μ}} d x \overset{? ? ?}{=} \frac{π}{2 Γ (μ) s i n (\frac{μ π}{2})}

\dots m . n . j u l y . 1970 \dots

Answered by Dwaipayan Shikari last updated on 06/Feb/21

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

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

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

= \frac{1}{2 Γ (μ)} \int_{0}^{\infty} \frac{j^{\frac{μ}{2} - 1}}{(1 + u)} d j = \frac{1}{2 Γ (μ)} . \frac{π}{s i n (\frac{μ π}{2})}

I (μ) = \frac{π}{2 Γ (μ) s i n (\frac{μ π}{2})} \Rightarrow I^{'} (μ) = - \frac{π^{2}}{4 Γ (μ)} c o s e c (\frac{μ π}{2}) c o t (\frac{μ π}{2}) - \frac{π}{2 s i n (\frac{μ π}{2})} . \frac{Γ^{'} (μ)}{Γ^{2} (μ)}

I^{'} (μ) = \int_{0}^{\infty} \frac{s i n x}{x^{μ}} l o g (x) d x

I^{'} (1) = \int_{0}^{\infty} \frac{s i n x}{x} l o g (x) d x = - \frac{π}{2} . Γ^{'} (1) = - \frac{π γ}{2}

Commented by mnjuly1970 last updated on 05/Feb/21

t a y d b a l l a h m r p a y a n . .