∫_0 ^1 sin(logx)dx


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Question Number 103198 by Dwaipayan Shikari last updated on 13/Jul/20

$\int_{0}^{1} s i n (l o g x) d x$
Commented by Dwaipayan Shikari last updated on 13/Jul/20

$s i n x = \frac{e^{i x} - e^{- i x}}{2 i}$ $s i n (l o g x) = \frac{x^{i} - x^{- i}}{2 i}$ $\int_{0}^{1} \frac{x^{i} - x^{- i}}{2 i} d x = \frac{1}{2 i} \int x^{i} - x^{- i} d x = \frac{1}{2 i} . {[(\frac{x^{i + 1}}{i + 1} - \frac{x^{- i + 1}}{1 - i})]}_{0}^{1}$ $= \frac{1}{2 i} (\frac{1}{i + 1} - \frac{1}{1 - i}) = - \frac{1}{2}$ $Is it true ? ? ? ? ? ? ?$
	Answered by OlafThorendsen last updated on 13/Jul/20

	$I = \int_{0}^{1} \sin (\ln x) d x$ $x = e^{u} \Rightarrow d x = e^{u} d u$ $I = \int_{- \infty}^{0} \sin u . e^{u} d u$ $I = \int_{- \infty}^{0} \frac{e^{i u} - e^{- i u}}{2 i} e^{u} d u$ $I = \frac{1}{2 i} \int_{- \infty}^{0} (e^{(1 + i) u} - e^{(1 - i) u}) d u$ $I = \frac{1}{2 i} {[\frac{e^{(1 + i) u}}{1 + i} - \frac{e^{(1 - i) u}}{1 - i}]}_{- \infty}^{0}$ $I = \frac{1}{2 i} [\frac{1}{1 + i} - \frac{1}{1 - i}]$ $I = \frac{1}{2 i} [\frac{1 - i}{1 - i^{2}} - \frac{1 + i}{1 - i^{2}}]$ $I = \frac{1}{2 i} (\frac{- 2 i}{2}) = - \frac{1}{2}$
	Answered by Smail last updated on 13/Jul/20

	$B y p a r t s$ $u = s i n (l n x) \Rightarrow u^{'} = \frac{1}{x} c o s (l n x)$ $v^{'} = 1 \Rightarrow v = x$ $\int_{0}^{1} s i n (l n x) d x = {[x s i n (l n x)]}_{0}^{1} - \int_{0}^{1} c o s (l n x) d x$ $u = c o s (l n x) \Rightarrow u^{'} = - \frac{1}{x} s i n (l n x)$ $v^{'} = 1 \Rightarrow v = x$ $\int_{0}^{1} s i n (l n x) d x = - {[x c o s (l n x)]}_{0}^{1} - \int_{0}^{1} s i n (l n x) d x$ $2 \int_{0}^{1} s i n (l n x) d x = - c o s (l n 1)$ $\int_{0}^{1} s i n (l n x) d x = \frac{- 1}{2}$
	Answered by mathmax by abdo last updated on 13/Jul/20

	$A = \int_{0}^{1} \sin (lnx) dx changement lnx = - t give$ $A = - \int_{0}^{\infty} \sin (- t) (- e^{- t}) dt = - \int_{0}^{\infty} e^{- t} sint dt$ $= - Im (\int_{0}^{\infty} e^{- t + it} dt) and \int_{0}^{\infty} e^{(- 1 + i) t} dt = {[\frac{1}{- 1 + i} e^{(- 1 + i) t}]}_{0}^{\infty}$ $= - \frac{1}{- 1 + i} = \frac{1}{1 - i} = \frac{1 + i}{2} \Rightarrow A = - \frac{1}{2}$
	Answered by Aziztisffola last updated on 13/Jul/20
	$let x=e^t ⇒dx=e^t dt ∫_0 ^1 sin(logx)dx=∫_(−∞) ^( 0) e^t sin(t)dt=∫_0 ^(+∞) −e^t sin(t)dt With Laplace transform: L{−e^t sin(t)}= L{−sin(t)}(s−1)=((−1)/((s−1)^2 +1)) then ∫_0 ^( +∞) −e^(−st) e^t sin(t)dt=((−1)/((s−1)^2 +1)) let s=0 ⇒ ∫_0 ^( +∞) −e^t sin(t)dt=((−1)/((−1)^2 +1))=((−1)/2) Hence ∫_(−∞) ^( 0) e^t sin(t)dt=((−1)/2) ∴ ∫_0 ^1 sin(logx)dx= ((−1)/2)$
	$let x = e^{t} \Rightarrow dx = e^{t} dt$ $\int_{0}^{1} s i n (l o g x) d x = \int_{- \infty}^{0} e^{t} \sin (t) dt = \int_{0}^{+ \infty} - e^{t} \sin (t) dt$ $With Laplace transform :$ $L {- e^{t} \sin (t)} = L {- \sin (t)} (s - 1) = \frac{- 1}{{(s - 1)}^{2} + 1}$ $then \int_{0}^{+ \infty} - e^{- st} e^{t} \sin (t) dt = \frac{- 1}{{(s - 1)}^{2} + 1}$ $let s = 0 \Rightarrow \int_{0}^{+ \infty} - e^{t} \sin (t) dt = \frac{- 1}{{(- 1)}^{2} + 1} = \frac{- 1}{2}$ $Hence \int_{- \infty}^{0} e^{t} \sin (t) dt = \frac{- 1}{2}$ $∴ \int_{0}^{1} s i n (l o g x) d x = \frac{- 1}{2}$
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