Can I apply Fundamental Theorem of Calculus to $\frac{d}{d a}\int_a^{\infty}(x-a)f(x)dx$?

Question

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?

Answer 1

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.

Answer 2

$$\int_{a}^{+\infty}(x-a)f(x)\,dx\stackrel{x\mapsto a+z}{=}\int_{0}^{+\infty}z f(z+a)\,dz \tag{1}$$ 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.

Answer 3

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 $$F'(a)=\lim_{h\to0}\frac{F(a+h)-F(a)}h=-\int_a^\infty f(t)\,\mathrm dt. $$