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Evaluate

$\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx$

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    @Babak Sorouh: I'm really glad to read these words. Thank you!2012-10-13

4 Answers 4

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Related problems: (I), (II). Recalling the Mellin transform of a function $f$

$ F(s) = \int_{0}^{\infty} x^{s-1}f(x) \,dx .$

Then we consider the more general integral

$ F(s) = \int_{0}^{\infty} x^{s-1}\left(\cos x - e^{-x^2}\right) \, dx \,. $

The value of the integral in our problem follows by taking the limit as $s\to 0 $ in the above integral. Evaluating the above integral gives

$ F(s) = \Gamma \left( s \right) \cos \left( \frac{\pi \,s}{2} \right) - \frac{1}{2}\, \Gamma \left(\frac{s}{2} \right) \,.$

Taking the limit as $s \to 0 \,,$ we get the desired result

$ \int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx = -\frac{\gamma}{2}\,. $

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    @MhenniBenghorbal It would help if you indicate that the limit $s\to 0$ is explained in your link https://math.stackexchange.com/questions/713464/integration-with-exponential-constant/713570#7135702017-06-11
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The result is

$ \int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \, dx = -\frac{\gamma}{2},$

where $\gamma$ is the Euler-Mascheroni constant.

Some direct calculations are available, but I prefer to consider it as a difference of some sort of log-singularities. you can find a slightly general method in this line of approach to calculate integrals of this form in my blog posting.

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    It would be nice to see more of the behind-the-scenes work here. Perhaps an application of the centers that are mentioned in your blog.2014-10-25
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Contour integration along the contour $[0,R]\cup Re^{i\pi/2[0,1]}\cup i[R,0]$ says that $ \int_0^\infty\frac{e^{ix}}{x^\alpha}\mathrm{d}x=e^{i\pi(1-\alpha)/2}\int_0^\infty\frac{e^{-x}}{x^\alpha}\mathrm{d}x\tag{1} $ since the integral along the curve vanishes as $R\to\infty$. Thus, $ \begin{align} \int_0^\infty\frac{e^{ix}-e^{-x^2}}{x^\alpha}\mathrm{d}x &=e^{i\pi(1-\alpha)/2}\Gamma(1-\alpha)-\frac12\Gamma\left(\frac{1-\alpha}2\right)\\ &=\frac{e^{i\pi(1-\alpha)/2}\Gamma(2-\alpha)-\Gamma\left(\frac{3-\alpha}2\right)}{1-\alpha}\tag{2} \end{align} $ Take the limit of $(2)$ as $\alpha\to1^-$ using L'Hospital and the fact that $\Gamma'(1)=-\gamma$: $ \begin{align} \int_0^\infty\frac{e^{ix}-e^{-x^2}}{x}\mathrm{d}x &=\frac{-\frac{i\pi}2+\gamma-\frac\gamma2}{-1}\\[4pt] &=-\frac\gamma2+\frac{i\pi}2\tag{3} \end{align} $ Therefore, we have both $ \boxed{\bbox[5pt]{\displaystyle\int_0^\infty\frac{\cos(x)-e^{-x^2}}{x}\mathrm{d}x=-\frac\gamma2}}\tag{4} $ and $ \int_0^\infty\frac{\sin(x)}{x}\mathrm{d}x=\frac\pi2\tag{5} $ where $\gamma$ is the Euler-Mascheroni Constant.