https://ameblo.jp/titchmarsh/entry-12430842489.html
の結果より
∫[-∞→∞] exp(-x²/2) {e^(-uxi)}dx=√(2π) exp(-u²/2)
∫[-∞→∞] x exp(-x²/2) {e^(-uxi)}dx=-i√(2π) uexp(-u²/2)
∫[-∞→∞] x² exp(-x²/2) {e^(-uxi)}dx=√(2π) (1-u²)exp(-u²/2)
であることがわかった．
これを使って
F(u)=∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx
の値を求めてみよう．
∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx
=∫[-∞→∞] x³ exp(-x²/2) {cos(ux)-isin(ux)}dx
=∫[-∞→∞] x³ exp(-x²/2) {cos(ux)}dx
-∫[-∞→∞] x³ exp(-x²/2) {isin(ux)}dx
=-i∫[-∞→∞] x³ exp(-x²/2) {sin(ux)}dx
=-i∫[-∞→∞] x exp(-x²/2) {x²sin(ux)}dx
=i∫[-∞→∞]{exp(-x²/2)}’ {x²sin(ux)}dx
=i{exp(-x²/2)}{x²sin(ux)}[-∞→∞]
-i∫[-∞→∞]{exp(-x²/2)}{x²sin(ux)}’dx
=-i∫[-∞→∞]{ exp(-x²/2)}{x²sin(ux)}’dx
=-i∫[-∞→∞]{ exp(-x²/2)}{2xsin(ux)+ux²cos(ux)}dx
=-2i∫[-∞→∞]{xexp(-x²/2)}{sin(ux)dx
-iu∫[-∞→∞]{x²exp(-x²/2)}{cos(ux)}dx….（＊）
ここで，
∫[-∞→∞] x exp(-x²/2) {e^(-uxi)}dx=-i√(2π) uexp(-u²/2)
より
∫[-∞→∞] x exp(-x²/2) {cos(ux)-isin(ux)}dx
=-i√(2π) uexp(-u²/2)
-∫[-∞→∞] x exp(-x²/2) {isin(ux)}dx=-i√(2π) u exp(-u²/2)
∫[-∞→∞] x exp(-x²/2) {sin(ux)}dx=√(2π) u exp(-u²/2)
であるので，（＊）の前項は，
-2i√(2π) uexp(-u²/2)
また，
∫[-∞→∞] x² exp(-x²/2) {e^(-uxi)}dx
=√(2π) (1-u²)exp(-u²/2)
より
∫[-∞→∞] x² exp(-x²/2) {cos(ux)+isin(ux)}dx
=√(2π) (1-u²)exp(-u²/2)
∫[-∞→∞] x² exp(-x²/2) {cos(ux)}dx
=√(2π) (1-u²)exp(-u²/2)
だから，後項は，-iu√(2π) (1-u²)exp(-u²/2)
以上より
∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx
=-2i√(2π) uexp(-u²/2)-i√(2π) u(1-u²)exp(-u²/2)
=-2i√(2π) uexp(-u²/2)-i√(2π)uexp(-u²/2)+i√(2π)u³exp(-u²/2)
=-3i√(2π) uexp(-u²/2)+i√(2π)u³exp(-u²/2)
=-i√(2π)(3u-u³)exp(-u²/2)
よって
F(u)=∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx
=-i√(2π)(3u-u³)exp(-u²/2)
フーリエ変換の定義式を
F(u)=1/√(2π)∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx
とすると
F(u)=-i(3u-u³)exp(-u²/2)
＜別解＞
∫[-∞→∞] x² exp(-x²/2) {e^(-uxi)}dx=√(2π) (1-u²)exp(-u²/2)
の両辺をu で微分すると
-i∫[-∞→∞] x³ exp(-x²/2) {e^(-uxi)}dx=√(2π) {(-2u)exp(-u²/2)+(1-u²)(-u)exp(-u²/2)}
=-√(2π) {2u+u(1-u²)}exp(-u²/2)
よって
∫[-∞→∞] x² exp(-x²/2) {e^(-uxi)}dx=-i√(2π)(3u-u³)exp(-u²/2)
が得られる．