Tabella di alcuni integrali utili nel calcolo dei coefficienti di Fourier Il

Tabella di alcuni integrali utili nel calcolo dei coefficienti di Fourier
Il parametro k e’ intero e positivo; il parametro a reale.
π
Z
Z
Z
= −
x cos(kx) dx
=
x2 sin(kx) dx
=
x2 cos(kx) dx
=
x3 sin(kx) dx
=
x3 cos(kx) dx
=
x4 sin(kx) dx
=
x4 cos(kx) dx
=
x5 sin(kx) dx
=
x5 cos(kx) dx
=
eax sin(kx) dx
=
eax cos(kx) dx
=
0
π
0
π
0
Z π
0
Z
π
0
Z π
0
Z
π
0
Z π
0
Z
π
0
Z π
Z
0
π
0
Z π
Z
0
π
2
eax sin(kx) dx
=
0
Z
0
π
(−1)k π
k
−1 + (−1)k
k2
(−1)k 2 − k 2 π 2 − 2
k3
2(−1)k π
k2
(−1)k π k 2 π 2 − 6
−
k3
3(−1)k k 2 π 2 − 2 + 6
k4
k
24 − (−1) π 4 k 4 − 12π 2 k 2 + 24
k5
k
2 2
4(−1) π k π − 6
k4
(−1)k π π 4 k 4 − 20π 2 k 2 + 120
−
k5
5(−1)k π 4 k 4 − 12π 2 k 2 + 24 − 120
k6
k aπ
k − (−1) e k
a2 + k 2
(−1)k aeaπ − a
a2 + k 2
√ [k2 /(4a)]
e
π
k
k − 2iaπ
√
√
√
−2Erf
+ Erf
+
4 a
2 a
2 a
2iπa + k
√
Erf
2 a
√ [k2 /(4a)]
π
k − 2iaπ
2iπa + k
ie
√
√
√
Erf
− Erf
4 a
2 a
2 a
x sin(kx) dx
2
eax cos(kx) dx
=
1
π/2
Z
Z
=
(−1)k
(2k + 1)2
x cos[(2k + 1)x] dx
=
(−1)k (2k + 1)π − 2
2(2k + 1)2
x2 sin[(2k + 1)x] dx
=
4(−1)k (2k + 1)π − 8
4(2k + 1)3
0
π/2
Z
Z
x sin[(2k + 1)x] dx
0
π/2
0
π/2
2
x cos[(2k + 1)x] dx
=
(−1)k (2πk + π)2 − 8
4(2k + 1)3
0
3
x sin[(2k + 1)x] dx
Z
k
π/2
Z
=
0
π/2
(2πk + π)2 − 8
3(−1)
4(2k + 1)4
k
3
x cos[(2k + 1)x] dx
=
(−1) (2k + 1)π (2πk + π)2 − 24 + 48
8(2k + 1)4
0
2(−1)k (2k + 1)π
π/2
Z
x4 sin[(2k + 1)x] dx
=
(−1)k π 4 k +
π/2
x4 cos[(2k + 1)x] dx
=
Z
Z
Z
x5 sin[(2k + 1)x] dx
=
0
π/2
5
x cos[(2k + 1)x] dx
=
eax sin[(2k + 1)x] dx
=
eax cos[(2k + 1)x] dx
=
0
π/2
1 2
2
π 2 − 6 + 24
− 3(2πk + π)2 + 24
1 4
2
− 3(2πk + π)2 + 24
(2k + 1)6
1
(−1)k (2k
32
+ 1)π (2πk + π)4 − 80(2πk + π)2 + 1920 − 120
(2k + 1)6
aπ
0
π/2
Z
1 4
2
5(−1)k π 4 k +
π/2
k+
(2k + 1)5
0
Z
(2k + 1)5
0
Z
(−1)k e 2 a + 2k + 1
a2 + (2k + 1)2
aπ
0
π/2
2
eax
sin[(2k + 1)x] dx
=
0
(−1)k e 2 (2k + 1) − a
a2 + (2k + 1)2
√ (2k+1)2
π
2k + 1
−iπa + 2k + 1
√ −2Erf
√
√
e 4a
+ Erf
+
4 a
2 a
2 a
iπa + 2k + 1
√
2 a
√ (2k+1)2
π
−iπa + 2k + 1
iπa + 2k + 1
√ Erf
√
√
ie 4a
− Erf
4 a
2 a
2 a
+Erf
Z
π/2
2
eax
cos[(2k + 1)x] dx
0
=
Gli integrali dei tipi
Z π/2
f (x) sin[(2k)x]dx ,
Z
π/2
f (x) cos[(2k)x]dx ,
0
0
si riportano immediatamente ad integrali su [0, π] del tipo considerato nella
tabella precedente.
2