由f(-x) = f(x),f(x)為偶函數(shù),可知其Fourier級數(shù)不含正弦項(xiàng).
對正整數(shù)n,∫{-π,π} f(x)cos(nx)dx = 2π·∫{0,π} e^x·cos(nx)dx.
e^x·cos(nx)的一個原函數(shù)為e^x·(cos(nx)+n·sin(nx))/(1+n²).
于是∫{-π,π} f(x)cos(nx)dx = 2π·(e^π·cos(nπ)-1)/(1+n²) = 2π·(e^π·(-1)^n-1)/(1+n²).
于是對正整數(shù)n,cos(nx)的系數(shù)為2(e^π·(-1)^n-1)/(1+n²).
常數(shù)項(xiàng) = ∫{-π,π} f(x)dx/(2π) = ∫{0,π} e^xdx = e^π-1.
f(x)的Fourier級數(shù)為e^π-1+2·∑{1 ≤ n} (e^π·(-1)^n-1)·cos(nx)/(1+n²).