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

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Maple says $-\gamma/2$. – AD. Oct 13 '12 at 12:30
@Chris'ssister: I like your question here. They are really like simple or difficult puzzles. Thanks for sharing them here. – Babak S. Oct 13 '12 at 13:11
@Babak Sorouh: I'm really glad to read these words. Thank you! – Chris's sis the artist Oct 13 '12 at 13:14

3 Answers 3

up vote 21 down vote accepted

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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your way seems so simple! Thanks! (+1) – Chris's sis the artist Oct 13 '12 at 19:30
@Chris'ssister: You are welcome. – Mhenni Benghorbal Oct 13 '12 at 19:43
+1. It's quite simple and elegant. – Felix Marin Oct 30 '14 at 20:00
@FelixMarin: Thanks for the comment. I really appreciate it. – Mhenni Benghorbal Dec 26 '14 at 19:16

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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thanks for the answer (+1) – Chris's sis the artist Oct 13 '12 at 12:35
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. – robjohn Oct 25 '14 at 13:02

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.

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