for $f\in C^2(\mathbb{R})$, finding the derivative of $\frac{d}{dt}\int_0^\infty f(x+t)\cdot xdx$ 
Let $f\in C^2(\mathbb{R})$,
(a) Prove that
$$\frac{\mathrm{d}}{\mathrm{d}t}\int_0^\infty f(x+t)\cdot x\mathbb{d}x=-\int_0^\infty f(x)\mathrm{d}x$$
(b) Prove that
$$ \iint_{(0,\infty)\times(0,\infty)}f(x+y)\mathrm{d}x\mathrm{d}y = \int_0^\infty f(x)\cdot x\mathrm{d}x$$

I'm actually really lost on both sections without much clue where to start, so sorry about my own attempt being short, but for (a), I've tried differentiating under the integral sign, which got me nowhere, and for (b) I've tried using (a) and saying that:
$$ \iint_{(0,\infty)\times(0,\infty)}f(x+y)\mathrm{d}x\mathrm{d}y = \int_0^\infty \int_0^\infty f_y(x)\mathrm{d}x\mathrm{d}y$$
$$ = \int_0^\infty-\int_0^\infty\frac{\mathrm{d}}{\mathrm{d}y}f_y(x)\cdot x\mathrm{d}x\mathrm{d}y = -\int_0^\infty x\int_0^\infty\frac{\mathrm{d}}{\mathrm{d}y}f_y(x)\mathrm{d}y\mathrm{d}x$$
where I have no idea how to proceed.
 A: I proceed very naively
on the first question,
since that's the math I know.
If
$g(t)
=\int_0^\infty f(x+t)xdx
$,
$\begin{array}\\
g(t+h)-g(t)
&=\int_0^\infty f(x+t+h)x\,dx-\int_0^\infty f(x+t)x\, dx\\
&=\int_0^\infty (f(x+t+h)-f(x+t))\, x\,dx\\
\text{so}\\
\dfrac{g(t+h)-g(t)}{h}
&=\int_0^\infty\dfrac{f(x+t+h)-f(x+t)}{h}x\,dx\\
\text{so}\\
\lim_{h \to 0}\dfrac{g(t+h)-g(t)}{h}
&=\lim_{h \to 0}\int_0^\infty\dfrac{f(x+t+h)-f(x+t)}{h}x\,dx\\
&=\int_0^\infty\lim_{h \to 0}\dfrac{f(x+t+h)-f(x+t)}{h}x\,dx\\
&=\int_0^\infty f'(x+t)x\,dx\\
&=(xf(x))|_0^{\infty}-\int_0^\infty f(x+t)dx
\qquad\text{(integrating by parts)}\\
&=-\int_0^\infty f(x+t)dx
\qquad\text{(assuming }\lim_{x \to \infty} x\, f(x) = 0)\\
\end{array}
$
I'll leave all the justifications
for the taking of limits
and reversing limit and integration
to those more knowledgable.
Of course,
that is probably the point
of the question.
A: Just for the fun of it,
here is a proof of
the discrete version of
$\iint_{(0,\infty)\times(0,\infty)}f(x+y)\mathrm{d}x\mathrm{d}y = \int_0^\infty f(x)\cdot x\mathrm{d}x
$.
$\begin{array}\\
\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a(i+j)
&=\sum_{n=0}^{\infty}\sum_{j=0}^{n}a(n)\\
&=\sum_{n=0}^{\infty}a(n)\sum_{j=0}^{n}1\\
&=\sum_{n=0}^{\infty}(n+1)a(n)\\
\end{array}
$
