0-sin-x-ln-x-x-dx-pi-2-

Question Number 99228 by M±th+et+s last updated on 19/Jun/20

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

Answered by maths mind last updated on 20/Jun/20

\frac{l n (x)}{x} = \frac{\partial}{\partial a} x^{a - 1} ∣_{a = 0}

\int_{0}^{+ \infty} s i n (x) x^{a - 1} d x = f (a)

W e w a n t \lim_{a \to 0} \frac{\partial f (a)}{\partial a}

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

i x = - y \Rightarrow d x = i d y

f (a) = I m \int_{0}^{- i \infty} e^{- y} i^{a} y^{a - 1} d y

= I m i^{a} \int_{0}^{+ \infty} e^{- y} y^{a - 1} d y = s i n (\frac{π}{2} a) Γ (a)

f (a) = \frac{s i n (\frac{π}{2} a) Γ (a) Γ (1 - a)}{Γ (1 - a)} \Leftrightarrow f (a) = \frac{π s i n (\frac{π}{2} a)}{s i n (π a) Γ (1 - a)} \dots E

= \frac{π}{2} . \frac{1}{c o s (\frac{π}{2} a) Γ (1 - a)}

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

f^{'} (a) = - \frac{π}{2} \frac{- Γ^{'} (1)}{1^{2}} = \frac{π Γ^{'} (1)}{2} = - \frac{π}{2} γ

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

Commented by M±th+et+s last updated on 21/Jun/20

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