Can I apply Fundamental Theorem of Calculus to $\frac{d}{d a}\int_a^{\infty}(x-a)f(x)dx$? I want to find the derivative:
$$\frac{d}{d a}\int_a^{\infty}(x-a)f(x)dx$$
What I have tried is by starting to rewrite the integral:
$$\int_a^{\infty}(x-a)f(x)dx=\int_{\infty}^{a}af(x)dx-\int_{\infty}^{a}xf(x)dx$$
How can I use Fundamental Theorem of Calculus find an expression for the first derivative $d/da$
I might be able to use the technique discussed here 
Is there a fundamental theorem of calculus for improper integrals? but I haven't yet found a proper way?
 A: Just write it as$$a\int_\infty^af(x)\,\mathrm dx-\int_\infty^axf(x)\,\mathrm dx.$$Now use the fundamental theorem of Calculus for improper integrals and the rule of the derivative of a product of functions.
A: $$\int_{a}^{+\infty}(x-a)f(x)\,dx\stackrel{x\mapsto a+z}{=}\int_{0}^{+\infty}z f(z+a)\,dz \tag{1}$$
implies
$$\frac{d}{da}\int_{a}^{+\infty}(x-a)f(x)\,dx=\frac{d}{da}\int_{0}^{+\infty}z f(z+a)\,dz =\int_{0}^{+\infty}zf'(z+a)\,dz\tag{2}$$
and by integration by parts
$$\int_{0}^{+\infty}zf'(z+a)\,dz=\lim_{z\to +\infty}zf(z+a)-\int_{0}^{+\infty}f(z+a)\,dz.\tag{3}$$
Of course we are tacitly assuming that all the involved integrals/limits make sense.
A: Mostly ab ovo: With 
$$F(x):=\int_x^\infty(t-x)f(t)\,\mathrm dt,$$
assume that


*

*the improper integral converges for all $x\approx a$

*$f$ is bounded near $a$, say $|f(x)|<M$ for all $x\approx a$.


Then for $h\approx 0$,
$$\begin{align}
\frac{F(a+h)-F(a)}h&=\frac1h\left(\int_{a+h}^\infty(t-a-h)f(t)\,\mathrm dt-\int_a^\infty(t-a)f(t)\,\mathrm dt\right)\\
&=\frac1h\left(\int_a^\infty(-h)f(t)\,\mathrm dt-\int_a^{a+h}(t-a-h)f(t)\,\mathrm dt\right)\\
&=-\int_a^\infty f(t)\,\mathrm dt-\frac1h\int_a^{a+h}(t-a-h)f(t)\,\mathrm dt.
\end{align}$$
For the second summand we find
$$\left|\frac1h\int_a^{a+h}(t-a-h)f(t)\,\mathrm dt\right|\le \frac1{|h|}\int_0^{|h|}tM\,\mathrm dt=\frac12|h|M $$
and therefore 
$$F'(a)=\lim_{h\to0}\frac{F(a+h)-F(a)}h=-\int_a^\infty f(t)\,\mathrm dt. $$
