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Evaluate $$\int_{0}^{\infty} \frac{\cos x - e^{-x}}{x} \ dx$$

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    If you evaluate this at 1 you get that C is your integral, so you can't get rid of it.2012-10-20
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    If you use this [technique](http://math.stackexchange.com/questions/212101/evaluate-int-0-infty-frac-cos-x-e-x2x-dx/212238#212238),the answer is zero.2012-10-20
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    @ Mhenni Benghorbal: that's true! Thanks. However, I try to evaluate it by using differentiation under the integral sign. It would be interesting to know how to get rid of $C$ and finish it up this way.2012-10-20
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    @Chris'ssister: Can you show your working?2012-10-20

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To use your technique: $$J(a)=\int_0^\infty\frac{\cos x-e^{-x}}{x}e^{-ax}dx$$ then $$dJ/da=\int_0^\infty(\cos x e^{-ax} -e^{-(1+a)x})dx=a/(a^2+1)-1/(a+1)$$ hence $J(a)=\frac12\log(a^2+1)-\log(a+1)+C$. $C$ can be determined by $J(a)\to 0$ as $a\to\infty$, hence $C=0$ and $J(0)=0$.

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$$\int_0^\infty {{{\cos x - {e^{ - x}}} \over x}dx} = \int_0^\infty {\left( {{s \over {1 + {s^2}}} - {1 \over {1 + s}}} \right)ds} = \mathop {\lim }\limits_{s \to \infty } \log {{\sqrt {{s^2} + 1} } \over {s + 1}} = 0$$

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    Maybe some words about the Laplace Transform being taken? Where did the $x^{-1}$ term go?2012-12-08