$\int f' g d \lambda = - \int f g' d \lambda$ I know that this result is quite elementary, but nevertheless, I don't think that the result is trivial. So, let $f,g \in C^1 (\mathbb R)$ with $f, g, f', g' \in L^2 (\mathbb R)$.
Why is it true that $\displaystyle\int_{\mathbb R} f'g\,d\lambda = - \int_{\mathbb R} fg'\,d\lambda\;?$ 
Thanks for your help!
 A: You are right that this is usually treated "in the literature" (i.e., in textbooks and papers) as trivial, although it is not quite that.
First, note (by Cauchy Schwarz) that $F := f \cdot g \in L^1$ is continuously differentiable with $F'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \in L^1$. It is well-known (see for instance here Absolute continuity of the Lebesgue integral) that this ($F'$ being an integrable function) implies for $\epsilon > 0$ that there is $\delta > 0$ satisfying $\int_A |F'(x)|dx < \epsilon$ as soon as $\lambda(A) < \delta$ (here, $\lambda$ is the Lebesgue measure). Therefore, if $|x-y| < \delta$ and (without loss of generality) $x \leq y$, then $|F(x) - F(y)| \leq \int_x^y |F'(t)| dt \leq \epsilon$.
In other words, $F$ is uniformly continuous. Since $F$ is also integrable, this implies (see here: Limit at infinity of a uniformly continuous integrable function) that $F$ vanishes at infinity, that is, $F(x) \to 0$ as $x \to \infty$ or $x \to -\infty$.
Now, we can apply the fundamental theorem of calculus to deduce
$$
\int_{\Bbb{R}} F'(t) dt
= \lim_n \int_{-n}^n F'(t) d t
= \lim_n \big[ F(n) - F(-n) \big]
= 0.
$$
If you recall that $F'(t) = f'(t) g(t) + f(t) g'(t)$, this implies the claim.
