Can you take the derivative of a function inside of an integral? I was trying to prove the differentiation property of the fourier transform:
$$x(t) = 1/(2\pi)\int_{-\infty}^\infty X(jw)e^{jwt}dw$$ Then take the derivative of both sides:
$$dx(t)/dt = 1/(2\pi)\int_{-\infty}^{\infty} \frac{d(X(jw)e^{jwt}}{dt}dw = 1/(2\pi)\int_{-\infty}^{\infty} X(jw)*jw*e^{jwt}dw$$ However, I'm not sure about the rules for this. Is it legal for me to take the derivative of a function depending on a variable "w" when it is in the integral? Am I allowed to manipulate functions dependent on the variable of integration?
 A: Here are sufficient conditions for differentiating under an improper integral.
Let $I$ be a compact subset of $\mathbb{R}$ and let $t\in I$.  If $f(t,\omega)$ and $\frac{\partial f(t,\omega)}{\partial t}$ are continuous for all $t\in I$ and $\omega\in \mathbb{R}$, and if $\int_{-\infty}^\infty f(t,\omega)\,d\omega$ converges for some $t_0\in I$ and $\int_{-\infty}^\infty \frac{\partial f(t,\omega)}{\partial t}\,d\omega$ converges uniformly for all $t\in I$, then 
$$\frac{d}{dt}\int_{-\infty}^\infty f(t,\omega)\,d\omega=\int_{-\infty}^\infty \frac{\partial f(t,\omega)}{\partial t}\,d\omega \tag 1$$
for $t\in I$.
So, let $f(t,\omega)=X(j\omega)e^{j\omega t}$.
Note that the equality in $(1)$ is not true in general.  Here are two examples, which shows that differentiation under the integral sign fails. 

EXAMPLE $1$
For a first example, the integral $I(x)$ as given by 
$$I(x)=\int_{-\infty}^\infty\frac{\sin(xt)}{t}\,dt$$
converges uniformly for all $|x|\ge \delta>0$.  But the integral of the derivative with respect to $x$, $\int_{-\infty}^\infty \cos(xt)\,dt$ diverges for all $x$.

EXAMPLE $2$
As another example, let $J(x)$ be the integral given by 
$$J(x)=\int_0^\infty x^3e^{-x^2t}\,dt$$
Obviously, $J(x)=x$ for all $x$ and hence $J'(x)=1$.  However, 
$$\int_0^\infty (3x^2-2x^4t)e^{-x^2t}\,dt=\begin{cases}1&,x\ne 0\\\\0&,x=0\end{cases}$$
Thus, formal differentiation under the integral sign leads to an incorrect result for $x=0$ even though all integrals involved are absolutely convergent.

