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Fourier Integral
Mehta Chintan B.
D1-14
3rd SEM. Mech. D
Guided By:- Prof. M. S. Suthar
Advanced Engineering Mathematics (2130002)
Fourier Series
• As we know that the fourier series of function f(x) in any interval (-l, l) is given by:
• 𝑓 𝑥 = 𝑎0 + 𝑛=1
∞
𝑎 𝑛 cos
𝑛𝜋𝑥
𝐿
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
𝐿
• Where:-
• 𝑎0 =
1
2𝑙 −𝑙
𝑙
𝑓 𝑡 𝑑𝑡
• 𝑎 𝑛=
1
𝑙 −𝑙
𝑙
𝑓 𝑡 𝑐𝑜𝑠
𝑛𝜋𝑡
𝑙
𝑑𝑡
• 𝑏 𝑛=
1
𝑙 −𝑙
𝑙
𝑓 𝑡 𝑠𝑖𝑛
𝑛𝜋𝑡
𝑙
𝑑𝑡
Fourier Integral
• Let f(x) be a function which is piecewise continuous in every finite interval in
(−∞, ∞) and absolute integral in (−∞, ∞).
• Then 𝑓 𝑥 =
1
𝜋 0
∞
( −∞
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔 𝑡 − 𝑥 𝑑𝑡)𝑑𝜔
• Where :
• 𝜔 =
𝑛𝜋
𝑙
• 𝑙 → ∞
Proof of Fourier Integral
𝑓 𝑥 = 𝑎0 +
𝑛=1
∞
𝑎 𝑛 cos
𝑛𝜋𝑥
𝐿
+ 𝑏 𝑛 sin
𝑛𝜋𝑥
𝐿
𝑓 𝑥 =
1
2𝑙 −𝑙
𝑙
𝑓 𝑡 𝑑𝑡 +
𝑛=1
∞
1
𝑙 −𝑙
𝑙
𝑓 𝑡 𝑐𝑜𝑠
𝑛𝜋𝑡
𝑙
𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
𝑑𝑡 +
𝑛=1
∞
1
𝑙 −𝑙
𝑙
𝑓 𝑡 𝑠𝑖𝑛
𝑛𝜋𝑡
𝑙
𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
𝑑𝑡
=
1
2𝑙 −𝑙
𝑙
𝑓 𝑡 𝑑𝑡 +
𝑛=1
∞
1
𝑙 −𝑙
𝑙
𝑓(𝑡) 𝑐𝑜𝑠
𝑛𝜋𝑡
𝑙
𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
𝑑𝑡 + 𝑠𝑖𝑛
𝑛𝜋𝑡
𝑙
𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
𝑑𝑡
=
1
2𝑙 −𝑙
𝑙
𝑓 𝑡 𝑑𝑡 +
1
𝑙
𝑛=1
∞
−𝑙
𝑙
𝑓 𝑡 𝑐𝑜𝑠
𝑛𝜋
𝑙
𝑡 − 𝑥 𝑑𝑡
• Putting 𝜔 𝑛 =
𝑛𝜋
𝑙
and ∆𝜔 𝑛 = 𝜔 𝑛+1 − 𝜔 𝑛 = 𝑛 + 1
𝜋
𝑙
−
𝜋
𝑙
=
𝜋
𝑙
so
∆𝜔 𝑛
𝜋
=
1
𝑙
𝑓 𝑥 =
1
2𝑙 −𝑙
𝑙
𝑓 𝑡 𝑑𝑡 +
∆𝜔 𝑛
𝜋
𝑛=1
∞
−𝑙
𝑙
𝑓 𝑡 𝑐𝑜𝑠𝜔 𝑛 𝑡 − 𝑥 𝑑𝑡
• As 𝑙 → ∞,
1
𝑙
= 0 and ∆𝜔 𝑛 =
𝜋
𝑙
→ 0, the infinite series in above equation
becomes an integral from 0 𝑡𝑜 ∞
𝑓 𝑥 =
1
𝜋 0
∞
−∞
∞
𝑓 𝑡 cos 𝜔 𝑡 − 𝑥 𝑑𝑡 𝑑𝜔
• Now expanding 𝑐𝑜𝑠𝜔(𝑡 − 𝑥) in above equation.
𝑓 𝑥 =
1
𝜋 0
∞
(
−∞
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡 𝑐𝑜𝑠𝜔𝑥 + 𝑠𝑖𝑛𝜔𝑡 𝑠𝑖𝑛𝜔𝑥) 𝑑𝜔
=
1
𝜋 0
∞
−∞
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 +
1
𝜋 0
∞
−∞
∞
𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
=
0
∞
𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 +
0
∞
𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
• Where:
• 𝐴 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡
• B 𝜔 =
1
𝜋 −∞
∞
𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
Fourier cosine integrals
• When 𝑓(𝑥) is an even function:
• 𝐴 𝜔 =
2
𝜋 0
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 and B 𝜔 = 0
• So the fourier integrals of an even function is given by:
• 𝑓(𝑥) = 0
∞
𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔
Fourier sin integral
• When 𝑓(𝑥) is an odd function:
• 𝐴 𝜔 = 0 and B 𝜔 =
2
𝜋 0
∞
𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
• So the fourier integral of odd function is given by:
• 𝑓(𝑥) = 0
∞
𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
Fourier cosine sum
• Find the fourier cosine integral of 𝒇 𝒙 = 𝒆−𝒌𝒙, where 𝒙 > 𝟎, 𝒌 > 𝟎 hence show that
𝟎
∞ 𝒄𝒐𝒔𝝎𝒙
𝒂 𝟐+𝝎 𝟐 𝒅𝝎 =
𝝅
𝟐𝒂
𝒆−𝒂𝒙
The fourier cosine integral of 𝑓 𝑥 is given by:
𝑓 𝑥 =
0
∞
𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔
𝐴 𝜔 =
2
𝜋 0
∞
𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡
=
2
𝜋 0
∞
𝑒−𝑘𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡
=
2
𝜋
𝑒−𝑘𝑡
𝑘2 + 𝜔2
(−𝑘𝑐𝑜𝑠𝜔𝑡 + 𝜔𝑠𝑖𝑛𝜔𝑡 (𝑓𝑟𝑜𝑚 0 𝑡𝑜∞)
=
2
𝜋
(
𝑎
𝑎2 + 𝜔2)
• Hence:
𝑓 𝑥 =
2𝑎
𝜋 0
∞
1
𝑎2 + 𝜔2
𝑐𝑜𝑠𝜔𝑥𝑑𝜔
0
∞
𝑐𝑜𝑠𝜔𝑥
𝑎2 + 𝜔2
𝑑𝜔 =
𝜋
2𝑎
𝑓(𝑥)
=
𝜋
2𝑎
𝑒−𝑎𝑥 (x > 0, 𝑎 > 0)
Fourier sine integral sum
• Find the sine integral of 𝑓 𝑥 = 𝑒−𝑏𝑥
, hence show that
𝜋
2
𝑒−𝑏𝑥
=
0
∞ 𝜔𝑠𝑖𝑛𝜔𝑥
𝑏2+𝜔2 𝑑𝜔
The fourier sine integral of 𝑓 𝑥 is given by:
𝑓(𝑥) =
0
∞
𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
𝐵 𝜔 =
2
𝜋 0
∞
𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
=
2
𝜋 0
∞
𝑒−𝜔𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
=
2
𝜋
𝑒−𝑏𝑡
𝑏2 + 𝜔2
(−𝑏𝑠𝑖𝑛𝜔𝑡 − 𝜔𝑐𝑜𝑠𝜔𝑡) (𝑓𝑟𝑜𝑚 0 𝑡𝑜 ∞)
=
2
𝜋
(
𝜔
𝑏2 + 𝜔2
)
• Hence:
𝑓 𝑥 =
2
𝜋 0
∞
𝜔𝑠𝑖𝑛𝜔𝑥
𝑏2 + 𝜔2
𝑑𝜔
0
∞
𝜔𝑠𝑖𝑛𝜔𝑥
𝑏2 + 𝜔2
𝑑𝜔 =
𝜋
2
𝑓 𝑥
0
∞
𝜔𝑠𝑖𝑛𝜔𝑥
𝑏2 + 𝜔2
𝑑𝜔 =
𝜋
2
𝑒−𝑏𝑥(x > 0, 𝑏 > 0)
References
• Advanced engineering mathematics ofTATA McGraw Hill
• https://www.wikipedia.org>wiki>fourier_integral
• https://mathonline.wikidot.com
ThankYou

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Fourier integral of Fourier series

  • 1. Gandhinagar Institute ofTechnology Fourier Integral Mehta Chintan B. D1-14 3rd SEM. Mech. D Guided By:- Prof. M. S. Suthar Advanced Engineering Mathematics (2130002)
  • 2. Fourier Series • As we know that the fourier series of function f(x) in any interval (-l, l) is given by: • 𝑓 𝑥 = 𝑎0 + 𝑛=1 ∞ 𝑎 𝑛 cos 𝑛𝜋𝑥 𝐿 + 𝑏 𝑛 sin 𝑛𝜋𝑥 𝐿 • Where:- • 𝑎0 = 1 2𝑙 −𝑙 𝑙 𝑓 𝑡 𝑑𝑡 • 𝑎 𝑛= 1 𝑙 −𝑙 𝑙 𝑓 𝑡 𝑐𝑜𝑠 𝑛𝜋𝑡 𝑙 𝑑𝑡 • 𝑏 𝑛= 1 𝑙 −𝑙 𝑙 𝑓 𝑡 𝑠𝑖𝑛 𝑛𝜋𝑡 𝑙 𝑑𝑡
  • 3. Fourier Integral • Let f(x) be a function which is piecewise continuous in every finite interval in (−∞, ∞) and absolute integral in (−∞, ∞). • Then 𝑓 𝑥 = 1 𝜋 0 ∞ ( −∞ ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔 𝑡 − 𝑥 𝑑𝑡)𝑑𝜔 • Where : • 𝜔 = 𝑛𝜋 𝑙 • 𝑙 → ∞
  • 4. Proof of Fourier Integral 𝑓 𝑥 = 𝑎0 + 𝑛=1 ∞ 𝑎 𝑛 cos 𝑛𝜋𝑥 𝐿 + 𝑏 𝑛 sin 𝑛𝜋𝑥 𝐿 𝑓 𝑥 = 1 2𝑙 −𝑙 𝑙 𝑓 𝑡 𝑑𝑡 + 𝑛=1 ∞ 1 𝑙 −𝑙 𝑙 𝑓 𝑡 𝑐𝑜𝑠 𝑛𝜋𝑡 𝑙 𝑐𝑜𝑠 𝑛𝜋𝑥 𝑙 𝑑𝑡 + 𝑛=1 ∞ 1 𝑙 −𝑙 𝑙 𝑓 𝑡 𝑠𝑖𝑛 𝑛𝜋𝑡 𝑙 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 𝑑𝑡 = 1 2𝑙 −𝑙 𝑙 𝑓 𝑡 𝑑𝑡 + 𝑛=1 ∞ 1 𝑙 −𝑙 𝑙 𝑓(𝑡) 𝑐𝑜𝑠 𝑛𝜋𝑡 𝑙 𝑐𝑜𝑠 𝑛𝜋𝑥 𝑙 𝑑𝑡 + 𝑠𝑖𝑛 𝑛𝜋𝑡 𝑙 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑙 𝑑𝑡 = 1 2𝑙 −𝑙 𝑙 𝑓 𝑡 𝑑𝑡 + 1 𝑙 𝑛=1 ∞ −𝑙 𝑙 𝑓 𝑡 𝑐𝑜𝑠 𝑛𝜋 𝑙 𝑡 − 𝑥 𝑑𝑡
  • 5. • Putting 𝜔 𝑛 = 𝑛𝜋 𝑙 and ∆𝜔 𝑛 = 𝜔 𝑛+1 − 𝜔 𝑛 = 𝑛 + 1 𝜋 𝑙 − 𝜋 𝑙 = 𝜋 𝑙 so ∆𝜔 𝑛 𝜋 = 1 𝑙 𝑓 𝑥 = 1 2𝑙 −𝑙 𝑙 𝑓 𝑡 𝑑𝑡 + ∆𝜔 𝑛 𝜋 𝑛=1 ∞ −𝑙 𝑙 𝑓 𝑡 𝑐𝑜𝑠𝜔 𝑛 𝑡 − 𝑥 𝑑𝑡 • As 𝑙 → ∞, 1 𝑙 = 0 and ∆𝜔 𝑛 = 𝜋 𝑙 → 0, the infinite series in above equation becomes an integral from 0 𝑡𝑜 ∞ 𝑓 𝑥 = 1 𝜋 0 ∞ −∞ ∞ 𝑓 𝑡 cos 𝜔 𝑡 − 𝑥 𝑑𝑡 𝑑𝜔 • Now expanding 𝑐𝑜𝑠𝜔(𝑡 − 𝑥) in above equation.
  • 6. 𝑓 𝑥 = 1 𝜋 0 ∞ ( −∞ ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡 𝑐𝑜𝑠𝜔𝑥 + 𝑠𝑖𝑛𝜔𝑡 𝑠𝑖𝑛𝜔𝑥) 𝑑𝜔 = 1 𝜋 0 ∞ −∞ ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 + 1 𝜋 0 ∞ −∞ ∞ 𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 𝑠𝑖𝑛𝜔𝑥𝑑𝜔 = 0 ∞ 𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 + 0 ∞ 𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔 • Where: • 𝐴 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 • B 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
  • 7. Fourier cosine integrals • When 𝑓(𝑥) is an even function: • 𝐴 𝜔 = 2 𝜋 0 ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 and B 𝜔 = 0 • So the fourier integrals of an even function is given by: • 𝑓(𝑥) = 0 ∞ 𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔
  • 8. Fourier sin integral • When 𝑓(𝑥) is an odd function: • 𝐴 𝜔 = 0 and B 𝜔 = 2 𝜋 0 ∞ 𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 • So the fourier integral of odd function is given by: • 𝑓(𝑥) = 0 ∞ 𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
  • 9. Fourier cosine sum • Find the fourier cosine integral of 𝒇 𝒙 = 𝒆−𝒌𝒙, where 𝒙 > 𝟎, 𝒌 > 𝟎 hence show that 𝟎 ∞ 𝒄𝒐𝒔𝝎𝒙 𝒂 𝟐+𝝎 𝟐 𝒅𝝎 = 𝝅 𝟐𝒂 𝒆−𝒂𝒙 The fourier cosine integral of 𝑓 𝑥 is given by: 𝑓 𝑥 = 0 ∞ 𝐴 𝜔 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 𝐴 𝜔 = 2 𝜋 0 ∞ 𝑓 𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 = 2 𝜋 0 ∞ 𝑒−𝑘𝑡 𝑐𝑜𝑠𝜔𝑡𝑑𝑡 = 2 𝜋 𝑒−𝑘𝑡 𝑘2 + 𝜔2 (−𝑘𝑐𝑜𝑠𝜔𝑡 + 𝜔𝑠𝑖𝑛𝜔𝑡 (𝑓𝑟𝑜𝑚 0 𝑡𝑜∞) = 2 𝜋 ( 𝑎 𝑎2 + 𝜔2)
  • 10. • Hence: 𝑓 𝑥 = 2𝑎 𝜋 0 ∞ 1 𝑎2 + 𝜔2 𝑐𝑜𝑠𝜔𝑥𝑑𝜔 0 ∞ 𝑐𝑜𝑠𝜔𝑥 𝑎2 + 𝜔2 𝑑𝜔 = 𝜋 2𝑎 𝑓(𝑥) = 𝜋 2𝑎 𝑒−𝑎𝑥 (x > 0, 𝑎 > 0)
  • 11. Fourier sine integral sum • Find the sine integral of 𝑓 𝑥 = 𝑒−𝑏𝑥 , hence show that 𝜋 2 𝑒−𝑏𝑥 = 0 ∞ 𝜔𝑠𝑖𝑛𝜔𝑥 𝑏2+𝜔2 𝑑𝜔 The fourier sine integral of 𝑓 𝑥 is given by: 𝑓(𝑥) = 0 ∞ 𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥𝑑𝜔
  • 12. 𝐵 𝜔 = 2 𝜋 0 ∞ 𝑓 𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 = 2 𝜋 0 ∞ 𝑒−𝜔𝑡 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 = 2 𝜋 𝑒−𝑏𝑡 𝑏2 + 𝜔2 (−𝑏𝑠𝑖𝑛𝜔𝑡 − 𝜔𝑐𝑜𝑠𝜔𝑡) (𝑓𝑟𝑜𝑚 0 𝑡𝑜 ∞) = 2 𝜋 ( 𝜔 𝑏2 + 𝜔2 )
  • 13. • Hence: 𝑓 𝑥 = 2 𝜋 0 ∞ 𝜔𝑠𝑖𝑛𝜔𝑥 𝑏2 + 𝜔2 𝑑𝜔 0 ∞ 𝜔𝑠𝑖𝑛𝜔𝑥 𝑏2 + 𝜔2 𝑑𝜔 = 𝜋 2 𝑓 𝑥 0 ∞ 𝜔𝑠𝑖𝑛𝜔𝑥 𝑏2 + 𝜔2 𝑑𝜔 = 𝜋 2 𝑒−𝑏𝑥(x > 0, 𝑏 > 0)
  • 14. References • Advanced engineering mathematics ofTATA McGraw Hill • https://www.wikipedia.org>wiki>fourier_integral • https://mathonline.wikidot.com