Derivative of $\int_{1}^{\infty}\frac{\cos(xt)}{t^{2}}dt$ I just wanna know if my intuition is right, I have the integral:
$$
F(x)=\int_{1}^{\infty}\frac{\cos(tx)}{t^{2}}dt
$$
I want to find $F'(x)$. I claim the answer to be $0$, heres why:
$$
\frac d{dx}\int_{\alpha(x)}^{\beta(x)}f(t)dt = \beta'(x)\cdot f(\beta(x)) - \alpha'(x)\cdot f(\alpha(x))
$$
Then what we have here essentially is $\alpha'(x)=\frac{d}{dx}0=0$.
Now I either solved it correctly, or did something stupid. Hope someone can confirm for me.
 A: $F'(x)$ certainly cannot be $0$ everywhere, since $F(0)=1$ is easily computed, and $F(x)\to 0$ for $x\to\infty$ since the ever closer neigboring wiggles of the cosine tend to cancel each other out the more there are of them.
The error in your reasoning is that you have forgotten that your $f$ is a function of both $x$ and $t$. What you really have is
$$\frac d{dx}\int_{\alpha(x)}^{\beta(x)}f(t,x)\,dt = \beta'(x)\cdot f(\beta(x),x) - \alpha'(x)\cdot f(\alpha(x),x)+\int_{\alpha(x)}^{\beta(x)} \Bigl[\frac{d}{dx} f(t, x)\Bigr] \, dt$$

Here's a heuristic estimate of $F(x)$  when $x\gg0$. Perhaps that can tell you enough about the general shape of the derivative for your purposes:
If $x$ is large, the integrand consists of a lot of narrow positive and negative peaks of the cosine numerator. The with of each half-period is $\frac{\pi}{x}$. If we flip the sign of every other half-period and slide each of them to cover the interval $t\in[1,1+\frac\pi x]$, the entire integral is the area between every other of the resulting curves -- and therefore tends to simply half of integral of the first half-period.
Within the first period the $\frac1{t^2}$ factor is eventually just $1$, so we can calculate
$$ F(x) \approx \frac12 \int_{1}^{1+\pi/x} \cos(xt)\,dt = \frac{\sin(x+\pi)-\sin(x)}{2x} 
= \frac{-\sin(x)}{x} $$
A: A formal differentiation under the integral suggests that
$$
F'(x) = -\int_1^\infty {\sin(xt)\over t} dt.
$$
One problem with a direct attack on this, using dominated convergence, for example, is that this integral doesn't converge absolutely. Taking a step back,  rewrite $F$ using one integration by parts:
$$
F(x) = -{\sin x\over x}+{2\over x}\int_1^\infty {\sin(xt)\over t^3} dt.
$$
Differentiation of this integral
$$
G(x):=\int_1^\infty {\sin(xt)\over t^3} dt.
$$
can be justified by dominated comvergence and the result is
$$
G'(x)=\int_1^\infty {\cos(xt)\over t^2} dt=F(x).
$$
Thus, because
$$
F(x) =-{\sin x\over x}+{2\over x}G(x)
$$
you have
$$
\eqalign{
F'(x) 
&= {\sin x-x\cos x\over x^2} + {2\over x}F(x) -{2\over x^2}G(x)\cr
&={\sin x-x\cos x\over x^2}+ {2\over x}F(x) -{2\over x^2}\left[{x\over 2}F(x)+{\sin x\over 2} \right]\cr
&= -{\cos x\over x}+{1\over x}F(x).
}
$$
Finally, again integrating by parts,
$$
F(x) :=\int_1^\infty{\cos(xt)\over t^2} dt= \cos x-x\int_1^\infty{\sin (xt)\over t}dt
$$
so
$$
\eqalign{
F'(x) 
&= -{\cos x\over x}+{1\over x}F(x)=-{\cos x\over x}+{1\over x}\left[ \cos x-x\int_1^\infty{\sin (xt)\over t}dt\right]\cr
&=-\int_1^\infty{\sin(xt)\over t} dt,\cr
}
$$
as suspected
