Find $f'(t)$, where $f(t)=\int_{-\infty}^{\infty}e^{-x^2}\cos(xt)dx$ 
Let $f(t)=\int_{-\infty}^{\infty}e^{-x^2}\cos(xt)dx.$ Fine $f'(t).$

I thought about two different approach. I think my first solution is accurate, but I would like to work on the second solution and have a precise argument based on epsilon, I appreciate if you help me. 
$\mathbf{My\: frist\: solution:}$ 
Add $i\int_{-\infty}^{\infty}e^{-x^2}\sin(xt)dx$, so we have
$$f(t)=\Re{(\int_{-\infty}^{\infty}e^{-x^2+ixt}}dx)=\Re(e^{-\frac{t^2}{4}}\int^{\infty}_{-\infty}e^{-(x-\frac{it}{2})^2}dx).$$
Hence $$f(t)=\sqrt{\pi}\exp{\{-\frac{t^2}{4}}\}$$
and so 
$$f'(t)=-\frac{\sqrt{\pi }t}{2}\exp{\{-\frac{t^2}{4}}\}$$.
$\mathbf{Second\: solution}$
By Leibniz integral rule, I know 
$$f'(t)=-2\int_{0}^{\infty}xe^{-x^2}\sin{(xt)}dx$$.
Define $g(t)=-2\int_{0}^{\infty}xe^{-x^2}\sin{(xt)}dx$
Given $\epsilon$, we need to find $h$ sufficiency small such that we have
$$|\frac{f(t+h)-f(t)}{h}-g(t)|\leq \epsilon$$. 
By definition of $f$ and $g$ we have have the following
$$|\frac{f(t+h)-f(t)}{h}-g(t)|=2|\int^{\infty}_0e^{-x^2}\big{(}\frac{\cos(x(t+h))-\cos(tx)}{h}+x\sin(xt))dx|$$
AS $$\frac{d}{dt}(\cos{xt})=-x\sin{(xt)}$$
for the $\epsilon$, there exsist $h$ small enough such that
$$|\frac{\cos{(xt+xh)}-\cos{xt}}{h}+x\sin{(xt)}|\leq \epsilon$$
Hence
$$2|\int^{\infty}_0e^{-x^2}\big{(}\frac{\cos(x(t+h))-\cos(tx)}{h}+x\sin(xt))dx|\leq 2\epsilon\int_0^{\infty}e^{-x^2}dx=\sqrt{\pi}\epsilon,$$
which shows that $f'(t)=-2\int_{0}^{\infty}xe^{-x^2}\sin{(xt)}dx.$
 A: Replace $\cos(xt)$ by $\frac{e^{ixt}+e^{-ixt}}{2}$ and look at
$$h(a) = \int_{-\infty}^\infty e^{-x^2}e^{2ax} dx $$
In your second part you proved that $h'(a) = \int_{-\infty}^\infty 2x e^{-x^2}e^{2ax} dx$. Another way is to integrate $\int_0^b \int_{-\infty}^\infty 2x e^{-x^2}e^{2ax} dxda= \int_{-\infty}^\infty \int_0^b 2x e^{-x^2}e^{2ax}da dx= h(a)-h(0)$, and since the integrand is continuous it means $h'(b) = \lim_{h \to 0} \frac{1}{h}\int_b^{b+h} \int_{-\infty}^\infty 2x e^{-x^2}e^{2ax} dxda= \int_{-\infty}^\infty 2x e^{-x^2}e^{2bx} dx$.
Integrating by parts $$h'(a) =\int_{-\infty}^\infty 2x e^{-x^2}e^{2ax} dx=\int_{-\infty}^\infty e^{-x^2}2a e^{2ax} dx= 2a h(a)$$ Thus $$\log h(a) = \log h(0)+\int_0^a \frac{h'(b)}{h(b)} db = \log h(0)+\int_0^a 2b db = \log h(0) + a^2$$ and $ h(a) = e^{a^2} h(0)=e^{a^2}\sqrt{\pi}$.
About the first part,  note in real analysis $\int_{-\infty}^\infty e^{-x^2} dx =\int_{-\infty}^\infty e^{-(x-a)^2} dx$ is correct only for $a \in \mathbb{R}$. 
In complex analysis it is correct for $a \in \mathbb{C}$ when using the Cauchy integral theorem. 
Another method is for $a \in \mathbb{R}$, $h(a) =e^{a^2} \int_{-\infty}^\infty e^{-(x-a)^2} dx=e^{a^2} \int_{-\infty}^\infty e^{-x^2} dx= \sqrt{\pi} e^{a^2}$.
Since $h(a)-\sqrt{\pi} e^{a^2}$ is analytic for $a \in \mathbb{C}$ and non-constant analytic functions have only isolated zeros and $h(a)-\sqrt{\pi} e^{a^2} = 0$ for $a \in \mathbb{R}$, then $h(a)-\sqrt{\pi} e^{a^2} = 0$ for $a \in \mathbb{C}$. 
A: Let $f(t)$ be given by the integral
$$f(t)=\int_{-\infty}^\infty e^{-x^2}\cos(xt)\,dx\tag1$$
Differentiating $(1)$ reveals
$$f'(t)=-\int_{-\infty}^\infty xe^{-x^2}\sin(xt)\,dx\tag2$$
Integrating by parts the integral on the right-hand side of $(2)$ with $u=\sin(xt)$ and $v=-\frac12e^{-x^2}$ yields
$$\begin{align}
f'(t)&=- \frac12t\int_{-\infty}^\infty e^{-x^2}\cos(xt)\,dx\\\\
&=-\frac12 tf(t)\tag3
\end{align}$$
Hence $f(t)$ satisfies the ODS 
$$f'(t)+\frac12 tf(t)=0\tag4$$
subject to the initial condition $f(0)=\sqrt \pi$.  It is straightforward to solve $(4)$ subject to the initial condition.  Proceeding we find that
$$f(t)=\sqrt \pi e^{-t^2/4} \tag5$$
Differentiating $(5)$, we obtain
$$f'(t)=-\frac{\sqrt \pi}2 te^{-t^2/2}$$
