>>34
f(x)：周期2L　⇔　f(x) = f(x+2L)
∫[-L,L]f(x)dx = ∫[-L+2nL,L+2nL]f(x)dx, n=1,2,3,・・・
　: f(x)が周期関数なので,∫範囲を全て2Lだけずれても同じ。

∫[-L,L]f(x)dx
=∫[-L,-L+C]f(x)dx + ∫[-L+C,L]f(x)dx
=∫[-L+2L,-L+C+2L]f(x)dx + ∫[-L+C,L]f(x)dx
=∫[L,L+C]f(x)dx + ∫[-L+C,L]f(x)dx
=∫[-L+C,L+C]f(x)dx