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