How do I integrate this using the Fourier transform? 
The integral is:$$I=\displaystyle\int_{0}^{\infty}e^{-ax^2}\cos(bx)\,dx$$

What I tried:
I was trying it with Fourier transform, so I have here:
$$f(t)=e^{-at^2}\cos(bt)$$
I know that: $$\mathcal{F}\left(e^{-ax^2} \right)=\dfrac{1}{\sqrt{2a}}e^{-\frac{\omega^2}{4a}}$$
Then used frequency modulation property: $$\cos(bt)=\dfrac{1}{2}\left(e^{-jbt}+e^{jbt}\right)$$
That gives me:
$$\mathcal{F}\left[f(t)\right]=\dfrac{1}{2\sqrt{2a}}\left[e^{\dfrac{-(\omega+b)^2}{4a}}+e^{\dfrac{-(\omega-b)^2}{4a}}\right]$$
For the integral I have to evaluate at $\omega=0$:
$$\mathcal{F}\bigg{|}_{\omega=0}=\dfrac{1}{\sqrt{2a}}\left[e^{-\frac{b^2}{4a}}\right]$$
This is what i got but the answer is:
$$I=\sqrt{\dfrac{\pi}{4a}}\left[e^{-\frac{b^2}{4a}}\right]$$
I don't know from where this $\sqrt{\dfrac{\pi}{2}}$ term is coming?
 A: A differentiation trick works well for this problem.
\begin{align}
            F(b) & = \int_{0}^{\infty}e^{-ax^2}\cos(bx)dx \\
    F'(b) & = -\int_{0}^{\infty}e^{-ax^2}x\sin(bx)dx \\
     & = \frac{1}{2a}\int_{0}^{\infty}\left(\frac{d}{dx}e^{-ax^2}\right)\sin(bx)dx \\
     & = -\frac{1}{2a}\int_{0}^{\infty}e^{-ax^2}\frac{d}{dx}\sin(bx)dx \\
     & = -\frac{b}{2a}\int_{0}^{\infty}e^{-ax^2}\cos(bx)dx \\
     & = -\frac{b}{2a}F(b).
\end{align}
Therefore, there is a constant $C$ such that
$$
      F(b) = Ce^{-b^2/4a}
$$
The constant is $F(0)=C$, which is
\begin{align}
        F(0) & =\int_{0}^{\infty}e^{-ax^2}dx \\
  & = \frac{1}{2}\int_{-\infty}^{\infty}e^{-ax^2}dx \\
  & = \frac{1}{2}\left[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-a(x^2+y^2)}dxdy\right]^{1/2} \\
  & = \frac{1}{2}\left[\int_{0}^{2\pi}\int_{0}^{\infty}e^{-ar^2}rdrd\theta\right]^{1/2} \\
  & = \frac{\sqrt{2\pi}}{2}\left[\frac{1}{2a}\int_{0}^{\infty}e^{-ar^2}(2ar)dr\right]^{1/2} \\
  & = \frac{\sqrt{2\pi}}{2}\frac{1}{\sqrt{2a}}
\end{align}
Therefore,
$$
      \int_{0}^{\infty}e^{-ax^2}\cos(bx)dx = F(b)= F(0)e^{-b^2/4a}=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-b^2/4a}.
$$
