How to integrate $\int\limits_0^\infty e^{-a x^2}\cos(b x) dx$ 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.
 A: I will consider the integral without the limit process. $$I(a,b)=\int\limits_0^\infty e^{-a x^2}\cos(b x) dx=\sum_{n=0}^{\infty}\frac{(-1)^n.b^{2n}}{(2n)!}\int_0^{\infty}x^{2n}e^{-ax^2}dx$$ by expanding the cosine. The integrals in the above sum are the familiar Gaussian Integrals defined by $$I_m=\int_0^{\infty}x^me^{-ax^2}dx$$ for non negative integral m. One may note that $I_0=\frac{1}{2}\sqrt{\frac{\pi}{a}}$ and that $I_{2n}=(-1)^n\frac{d^n}{da^n}I_0$.  It follows that $$I(a,b)=\frac{\sqrt\pi}{2}\sum_{n=0}^{\infty}\frac{b^{2n}}{(2n)!}\frac{d^n}{da^n}a^{-\frac{1}{2}}.$$ It is not difficult to see that $$\frac{d^n}{da^n}a^{-\frac{1}{2}}=\frac{-1^n}{\sqrt a}\frac{1.2...(2n-1)}{2^n}\frac{1}{a^n}.$$ Substituting the above in the expression for $I(a,b)$ one has $$I(a,b)=\sqrt{\frac{\pi}{4a}}\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{b^2}{4a}\right)^n$$ $$\implies I(a,b)=\sqrt{\frac{\pi}{4a}} e^{-\frac{b^2}{4a}} .$$
A: 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: 
$dx=d(x-\frac{\beta}{2a})$
$\beta^2=-b^2$.
