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

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. –  Aleks Vlasev Oct 20 '12 at 6:26
If you use this technique,the answer is zero. –  Mhenni Benghorbal Oct 20 '12 at 6:32
@ 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. –  Chris's sis Oct 20 '12 at 6:35
@Chris'ssister: Can you show your working? –  wj32 Oct 20 '12 at 6:45

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 $J(0)=0$.
$$\int_0^\infty {{{\cos x - {e^{ - x}}} \over x}dx} = \int_0^\infty {\left( {{s \over {1 + {s^2}}} - {1 \over {1 + s}}} \right)ds} = \mathop {\lim }\limits_{s \to \infty } \log {{\sqrt {{s^2} + 1} } \over {s + 1}} = 0$$
Maybe some words about the Laplace Transform being taken? Where did the $x^{-1}$ term go? –  Pedro Tamaroff Dec 8 '12 at 15:09