integration - How to integrate $intlimits_0^infty e^{-a x^2}cos(b x) dx$ where $a>0$

How to integrate

$$\int\limits_0^\infty e^{-a x^2}\cos(b x) dx$$

where $a>0$

The real problem is this integral

$$\lim\limits_{\alpha\rightarrow 2}\int\limits_0^\infty e^{-a x^\alpha}\cos(b x) dx$$

I tried integration by parts and then the change of variable $z=x^2$ but it does not work.

Answer

Using Euler's identity, we get:

$$
\int\limits_0^\infty e^{-a x^2}\cos(b x) dx=Re \left( \int\limits_0^\infty e^{-a x^2} e^{ibx} dx \right)
$$

$$
\int\limits_0^\infty e^{-a x^2} e^{ibx} dx = \int\limits_0^\infty e^{-a x^2+ibx} dx
$$

Let's forget about imaginary unit and take $ib=\beta$ for simplicity:

$$
-ax^2+\beta x=-a (x^2-\frac{\beta}{a}x+\frac{\beta^2}{4a^2})+\frac{\beta^2}{4a}=-a(x-\frac{\beta}{2a})^2+\frac{\beta^2}{4a}
$$

$$
\int\limits_0^\infty e^{-a x^2+\beta x} dx=e^{\frac{\beta^2}{4a}} \int\limits_0^\infty e^{-a(x-\frac{\beta}{2a})^2} dx
$$

I believe you will not have trouble with the rest.

Hints:

$dx=d(x-\frac{\beta}{2a})$

$\beta^2=-b^2$.