$\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx$ Prove that: $\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx=\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx$
This is my proof:
$\forall \ t \in \mathbb{R}$ (the improper integral coverge absolutely $\forall \ t \in \mathbb{R}$) I consider: $g(t)=\int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx$. 
Let $h \ne 0$ 
$\left| \frac{g(t+h)-g(t)}{h}-\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx \right|\le\int_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx$
For the main value theorem and since $\int_{-\infty}^{\infty} e^{-x^2} dx=\sqrt{\pi}$
$\int_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx=\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}$
$\cos(2tx)$ is derivable in $\bar{x}$ then fixed a $\epsilon>0 \ $ if $\ 0<|h|<\delta$:
$\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}<\sqrt{\pi} \ \epsilon$
It is correct?
There are other ways?
UPDATE
probably the proof is incorrect, because when I use the mean value theorem $x$ depends also from $h$ and hence the continuity of $x(h)$ is not obvious, then I can't guarantee the derivability of $\cos(2 t x(h))$ in $x$ for $h \rightarrow 0$. Am I right?
 A: You way looks good.  Here's an alternate way: evaluate both integrals, and see that the derivative of one equals the other.
For example,
$$\begin{align}\int_{-\infty}^{\infty} dx \, e^{-x^2} \, \cos{2 t x} &= \Re{\left [\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i 2 t x} \right ]}\\ &= e^{-t^2}\Re{\left [\int_{-\infty}^{\infty} dx \, e^{-(x-i t)^2} \right ]}\\ &= \sqrt{\pi}\, e^{-t^2}\end{align}$$
The derivative of this with respect to $t$ is
$$\frac{d}{dt} \int_{-\infty}^{\infty} dx \, e^{-x^2} \, \cos{2 t x} = -2 \sqrt{\pi} t e^{-t^2}$$
Now try with taking the derivative inside the integral:
$$\begin{align}-2 \int_{-\infty}^{\infty} dx \,x \, e^{-x^2} \,\sin{2 t x} &= -2 \Im{\left [\int_{-\infty}^{\infty} dx\,x \, e^{-x^2} e^{i 2 t x} \right ]}\\ &= -2 e^{-t^2}\Im{\left [\int_{-\infty}^{\infty} dx\,x \, e^{-(x-i t)^2} \right ]}\\ &= -2 e^{-t^2}\Im{\left [\int_{-\infty}^{\infty} dx\,(x+i t) e^{-x^2} \right ]} \\ &= -2 t \sqrt{\pi} e^{-t^2} \end{align}$$
QED
A: There is much simpler way using the Leibniz integral rule, which allows you to swap the integral and the derivation:
\begin{align}
\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx
&= \int_{-\infty}^{\infty} \left(\frac{\partial}{\partial t} e^{-x^2} \cos(2tx) \right) dx \\
&=\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx
\end{align}

Update: As far as I know, the Leibniz rule can be used for integral $\frac{\partial}{\partial t}\int_a^b f(x,t)dx$ with possibly infinite limits $a, b$ as well, under some additional conditions. In particular, Theorem 10.3. of this article describes it:

  
*
  
*there are upper bounds $|f(x,t)|\leq A(x)$ and $|\frac{\partial}{\partial t} f (x,t)| \leq B(x)$, both being independent of $t$, such that a $\int_a^b A(x) dx$ and $\int_a^b B(x) dx$ converge.
  

In our case we can pick $A(x)=e^{-x^2}$ and $B(x)=2|x| e^{-x^2}$.
See also Using Leibniz Integral Rule on infinite region.
