let-n-Z-shov-that-0-sin-x-n-ln-x-x-dx-pi-2n-2-where-is-the-Euler-Mascheroni-constan-

Question Number 157003 by MathSh last updated on 18/Oct/21

let n \in Z^{+}

shov that \underset{0}{\int^{\infty}} \frac{\sin (x^{- n}) \ln (x)}{x} dx = \frac{π γ}{{2 n}^{2}}

where γ is the Euler - Mascheroni constan

Answered by mindispower last updated on 18/Oct/21

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

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

= \frac{1}{n^{2}} \int_{0}^{\infty} \frac{s i n (t)}{t} l n (t)

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

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

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

Δ = \frac{1}{n^{2}} f^{'} (- 1) = \frac{π}{2 n^{2}} \lim_{a \to 0} (c o s (\frac{a}{2} π) Γ (a) + Γ^{'} (a) a)

Γ^{'} (a) = Ψ (a) Γ (a)

= \lim_{a \to 0} \frac{π}{2 n^{2}} (Γ (a) + Ψ (a) Γ (a) a)

Ψ (1 + a) = Ψ (a) + \frac{1}{a} \Rightarrow a Ψ (a) = a Ψ (1 + a) - 1

= \lim_{a \to 0} \frac{π}{2 n^{2}} (Γ (a) + a Ψ (1 + a) Γ (a) - Γ (a))

= \lim_{a \to 0} . \frac{π}{2 n^{2}} (Γ (1 + a) Ψ (1 + a)) = \frac{π}{2 n^{2}} Ψ (1) = \frac{π γ}{2 n^{2}}

Commented by MathSh last updated on 18/Oct/21

Perfect dear Ser, thank you

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