The integral $$\int_0^\infty\!e^{-x^2}\cos{tx}\,dx \;:=\; w(t)$$ is a generalisation of the Gaussian integral $w(0) = \frac{\sqrt{\pi}}{2}$ . For evaluating it we first form its derivative which may be done by differentiating under the integral sign: $$w'(t) \;=\; \int_0^\infty\!e^{-x^2}(-x)\sin{tx}\,dx \;=\; \frac{1}{2}\int_0^\infty\!e^{-x^2}(-2x)\sin{tx}\,dx$$ Using integration by parts this yields $$w'(t) \;=\; \frac{1}{2}\!\sijoitus{x=0}{\quad\infty}\!e^{-x^2}\sin{tx}-\frac{t}{2}\int_0^\infty\!e^{-x^2}\cos{tx}\,dx \,=\, \frac{1}{2}(0-0)-\frac{t}{2}\int_0^\infty\!e^{-x^2}\cos{tx}\,dx \,=\, -\frac{t}{2}w(t).$$ Thus $w(t)$ satisfies the linear differential equation $$\frac{dw}{dt} \;=\; -\frac{1}{2}tw,$$ where one can separate the variables and integrate: $$\int\!\frac{dw}{w} \;=\; -\frac{1}{2}\int\!t\,dt.$$ So, $\ln{w} \,=\, -\frac{1}{4}t^2+\ln{C}$ , i.e. $w = w(t) = Ce^{-\frac{1}{4}t^2}$ , and since there is the initial condition $w(0) = \frac{\sqrt{\pi}}{2}$ , we obtain the result $$w(t) \;=\; \frac{\sqrt{\pi}}{2}e^{-\frac{1}{4}t^2}.$$