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Question Number 190371 by TUN last updated on 02/Apr/23

	Answered by qaz last updated on 02/Apr/23
	$∫_0 ^1 sin (ln(1/x))((x^b −x^a )/(lnx))dx=∫_(−∞) ^0 sin (−u)∙((e^(ub) −e^(ua) )/u)e^u du =∫_0 ^∞ ((sin u)/u)[e^(−u(a+1)) −e^(−u(b+1)) ]du =∫_0 ^∞ L{e^(−u(a+1)) sin u−e^(−u(b+1)) sin u}∙L^(−1) {u^(−1) }du =∫_0 ^∞ [(1/((u+a+1)^2 +1))−(1/((u+b+1)^2 +1))]du =[arctan (u+a+1)−arctan (u+b+1)]_0 ^∞ =arctan ((a−b)/(1+(u+a+1)(u+b+1)))∣_0 ^∞ =arctan ((b−a)/(1+(a+1)(b+1)))$
	$\int_{0}^{1} \sin (l n \frac{1}{x}) \frac{x^{b} - x^{a}}{l n x} d x = \int_{- \infty}^{0} \sin (- u) \cdot \frac{e^{u b} - e^{u a}}{u} e^{u} d u$ $= \int_{0}^{\infty} \frac{\sin u}{u} [e^{- u (a + 1)} - e^{- u (b + 1)}] d u$ $= \int_{0}^{\infty} L {e^{- u (a + 1)} \sin u - e^{- u (b + 1)} \sin u} \cdot L^{- 1} {u^{- 1}} d u$ $= \int_{0}^{\infty} [\frac{1}{{(u + a + 1)}^{2} + 1} - \frac{1}{{(u + b + 1)}^{2} + 1}] d u$ $= {[\arctan (u + a + 1) - \arctan (u + b + 1)]}_{0}^{\infty}$ $= \arctan \frac{a - b}{1 + (u + a + 1) (u + b + 1)} ∣_{0}^{\infty} = \arctan \frac{b - a}{1 + (a + 1) (b + 1)}$
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