$\int f' g d \lambda = - \int f g' d \lambda$ 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!

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.
