prove-that-0-sin-3-x-ln-x-x-dx-pi-8-2-ln-3-m-n-

Question Number 154478 by mnjuly1970 last updated on 18/Sep/21

You can't use 'macro parameter character #' in math mode

\int_{0}^{\infty} \frac{{s i n}^{3} (x) . l n (x)}{x} d x \overset{?}{=} \frac{π}{8} (- 2 γ + l n (3)) \dots . . m . n

Answered by ARUNG_Brandon_MBU last updated on 18/Sep/21

\int_{0}^{\infty} \frac{\sin^{3} x \cdot \ln x}{x} d x

2 i \sin x = (e^{i x} - e^{- i x})

- 8 i \sin^{3} x = e^{3 i x} - 3 e^{i x} + 3 e^{- i x} - e^{- 3 i x}

\sin^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3 x

Ω (α) = - \int_{0}^{\infty} \frac{\sin^{3} x}{x^{α}} = \frac{1}{4} \int_{0}^{\infty} (\frac{\sin 3 x}{x^{α}} - \frac{3 \sin x}{x^{α}}) d x = \frac{1}{4} [\frac{π 3^{α - 1} - 3 π}{2 Γ (α) \sin (\frac{π}{2} α)}]

Ω^{'} (α) = \frac{1}{8} [\frac{π 3^{α - 1} \ln 3 (Γ (α) \sin (\frac{π}{2} α)) - (π 3^{α - 1} - 3 π) (\frac{π}{2} \cos (\frac{π}{2} α) Γ (α) + \sin (\frac{π}{2} α) Γ^{'} (α))}{Γ^{2} (α) \sin^{2} (\frac{π}{2} α)}]

Ω^{'} (1) = \int_{0}^{\infty} \frac{\sin^{3} x \cdot \ln x}{x} d x = \frac{1}{8} [\frac{π \ln 3 + 2 π (Γ^{'} (1))}{Γ^{2} (1) \sin^{2} (\frac{π}{2})}] = \frac{1}{8} (π \ln 3 - 2 π γ) = \frac{π}{8} (\ln 3 - 2 γ)

Commented by mnjuly1970 last updated on 18/Sep/21

t h a n k y o u s o m u c h s i r A r u n g . .

Commented by Ar Brandon last updated on 18/Sep/21

My pleasure, Sir ��