■ある三角関数の積分(その5)

[4]ある三角関数の積分

 (定理2)

  ∫(0,∞)sin^n(ax)/x^ndx=a^(n-1)π/2-a^(n-1)π/(m-1)!2^(n-1)Σ(-1)^(r-1)(n-1,r-1)(n-2r)^(n-1)

  (n-1,r-1)は2項係数n-1Cr-1,r=0~[(n-1)/2],[]はガウス記号

が成り立ちます.

 

 この定理を用いると

  ∫(0,∞)sin(ax)/xdx=π/2

  ∫(0,∞)sin^2(ax)/x^2dx=aπ/2

  ∫(0,∞)sin^3(ax)/x^3dx=3a^2π/8

  ∫(0,∞)sin^4(ax)/x^4dx=a^3π/3

  ∫(0,∞)sin^5(ax)/x^5dx=115a^4π/384

  ∫(0,∞)sin^6(ax)/x^6dx=11a^5π/40

  ∫(0,∞)sin^7(ax)/x^7dx=5887a^6π/23040

  ∫(0,∞)sin^8(ax)/x^8dx=151a^7π/630

  ∫(0,∞)sin^9(ax)/x^9dx=259732a^8π/1146880

  ∫(0,∞)sin^10(ax)/x^10=15619a^9π/72576

  ∫(0,∞)sin^11(ax)/x^11dx=381773117a^10π/1857945600

  ∫(0,∞)sin^12(ax)/x^12dx=655177a^11π/3326400

  ∫(0,∞)sin^13(ax)/x^13dx=20646903199a^12π/108999475200

  ∫(0,∞)sin^14(ax)/x^14dx=27085381a^13π/148262400

  ∫(0,∞)sin^15(ax)/x^15dx=467168310097a^14π/2645053931520

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n=2kのとき、オイラー数Ak^(n)

Ak^(n)=Σ(-1)^(r)(n+1,r)(k+1-r)^(n)

を用いて

  ∫(0,∞)sin^2k(ax)/x^2kdx=a^2k-1π/2・Ak-1^(2k-1)/(2k-1)!

と書ける

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