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As part of one calculation I want to show that the following integral converges absolutely:

$$\int_{0}^{\infty}e^{-x}\frac{1-\cos(x)}{x}dx$$

wihtout calculating its value. Using integral handbooks and/or a computer algebra system like Mathematica I found that the integral converges $ \log(2)/2$. But I do not know how to prove that it converges absolutely.

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Note that $1-\cos x=\displaystyle\int_0^x\sin t\,\mathrm dt\leqslant x$ for every $x\gt0$. Since the integral $\displaystyle\int_0^\infty\mathrm e^{-x}\,\mathrm dx$ converges (absolutely), this is enough to conclude.

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Hint: Using Taylor series, as $x\sim 0$ we have

$$ e^{-x}\frac{1-\cos x}{x} \sim \frac{x}{2} e^{-x}. $$

See related techniques.

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