On sépare l'intégrale en deux morceaux,
\[
\int_{-L}^Lf(x)\,dx
=
\int_{-L}^0f(x)\,dx
+\int_{0}^Lf(x)\,dx \,.
\]
Avec le changement de variable \(x=\varphi(u):= -u\),
\(\varphi'(u)=-1\),
\[\begin{aligned}
\int_{-L}^0f(x)\,dx=\int_{L}^0f(-u)(-1)\,du
&=\int_{0}^Lf(-u)\,du\,.
\end{aligned}\]
- Si \(f\) est paire, alors
\[
\int_{0}^Lf(-u)\,du
=\int_{0}^Lf(u)\,du
\]
et donc
\[
\int_{-L}^Lf(x)\,dx
=
\int_0^Lf(u)\,du
+\int_{0}^Lf(x)\,dx
=2\int_0^Lf(x)\,dx \,.
\]
- Si \(f\) est impaire,
\[
\int_0^Lf(-u)\,du
=\int_0^L (-f(u))\,du
=-\int_0^Lf(u)\,du\,,
\]
et donc
\[
\int_{-L}^Lf(x)\,dx
=
-\int_0^Lf(x)\,dx
+\int_0^Lf(x)\,dx
=0\,.
\]