Faculty of Engineering and Technology.
Department of Applied Science & Humanities.
Subject: Mathematics – II (303191151)
Semester: 2nd Sem. B.Tech Programme (All Branches)
Lecture Notes: Unit – 4 Fourier Integral
Introduction:
We have learnt Fourier series for periodic functions. There exist many practical problems in
engineering which involve non-periodic functions. We can solve such problems on the basis of Fourier series
technique by converting non-periodic functions in terms of sine and cosine functions. This conversion will lead to
the extension of Fourier series to Fourier integral.
Fourier Integral:
Let 𝑓(𝑥) be a function which is piecewise continuous in every finite interval (−∞, ∞) and absolutely
integrable in (−∞, ∞), then Fourier integral is given by following formula.
Formula:
❖ Fourier Integral:
∞
𝑓(𝑥) = ∫ [𝐴(𝜆) cos 𝜆𝑥 + 𝐵(𝜆) sin 𝜆𝑥] 𝑑𝜆
0
1 ∞ 1 ∞
𝑤ℎ𝑒𝑟𝑒 𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣, 𝐵(𝜆) = ∫ 𝑓(𝑣) sin 𝜆𝑣 𝑑𝑣
𝜋 −∞ 𝜋 −∞
❖ Fourier Cosine Integral:
∞
2 ∞
𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆, 𝑤ℎ𝑒𝑟𝑒 𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
0 𝜋 0
❖ Fourier Sine Integral:
∞
2 ∞
𝑓(𝑥) = ∫ 𝐵(𝜆) sin 𝜆𝑥 𝑑𝜆, 𝑤ℎ𝑒𝑟𝑒 𝐵(𝜆) = ∫ 𝑓(𝑣) sin 𝜆𝑣 𝑑𝑣
0 𝜋 0
❖ Existence of Fourier Integral:
The Fourier integral of function 𝑓(𝑥) exists if the area under the function is finite. This means that the
function must be absolutely integrable. If 𝑥 is point of continuity of 𝑓(𝑥), then the Fourier integral expansion
𝑓(𝑥+0)+𝑓(𝑥−0)
of 𝑓(𝑥) holds. If 𝑥 is a point of discontinuity, then 𝑓(𝑥) is replaced by .
2
Examples
𝟏 − 𝒙𝟐 𝒊𝒇 |𝒙| ≤ 𝟏
1. Find the Fourier Integral representation of the function 𝒇(𝒙) = {
𝟎 𝒊𝒇 |𝒙| > 𝟏
Solution:
∞
𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆
0
1 | P a g e Lecture Notes -Unit-4 Fourier Integral Mathematics – II (303191151)
, Faculty of Engineering and Technology.
Department of Applied Science & Humanities.
2 ∞
𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 ∞
= ∫ (1 − 𝑣 2 ) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 2)
sin 𝜆𝑣 cos 𝜆𝑣 sin 𝜆𝑣 1
= |(1 − 𝑣 − (−2𝑣) (− ) + (−2) (− 3 )|
𝜋 𝜆 𝜆2 𝜆 0
2 2 cos 𝜆 2 sin 𝜆
= (− + ) [∵ sin 0 = 0]
𝜋 𝜆2 𝜆3
4 sin 𝜆 − 𝜆 cos 𝜆
= ( )
𝜋 𝜆3
4 ∞ sin 𝜆 − 𝜆 cos 𝜆
𝑓(𝑥) = ∫ ( ) cos 𝜆𝑥 𝑑𝜆
𝜋 0 𝜆3
2. Find the Fourier cosine and sine Integral of 𝒇(𝒙) = 𝒆−𝒌𝒙 , (𝒙 > 𝟎, 𝒌 > 𝟎)
Solution: (a)
2 ∞
𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 ∞ −𝑘𝑣
= ∫ 𝑒 cos 𝜆𝑣 𝑑𝑣
𝜋 0
∞
2 𝑒 −𝑘𝑣
= [ 2 (−𝑘 cos 𝜆𝑣 + 𝜆 sin 𝜆𝑣)]
𝜋 𝑘 + 𝜆2 0
2 𝑘
= [0 + 2 ]
𝜋 𝑘 + 𝜆2
2𝑘
=
𝜋(𝑘 2+ 𝜆2 )
We obtain the Fourier cosine integral representation
∞
𝑖. 𝑒. 𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆
0
2𝑘 ∞ cos 𝜆𝑥
𝑓(𝑥) = 𝑒 −𝑘𝑥 = ∫ 𝑑𝜆, (𝑥 > 0, 𝑘 > 0)
𝜋 0 (𝑘 2 + 𝜆2 )
From this, we see that
∞
cos 𝜆𝑥 𝜋 −𝑘𝑥
∫ 𝑑𝜆 = 𝑒 , (𝑥 > 0, 𝑘 > 0)
0 (𝑘 2 + 𝜆2 ) 2𝑘
(b) Similarly, we have
2 | P a g e Lecture Notes -Unit-4 Fourier Integral Mathematics – II (303191151)
Department of Applied Science & Humanities.
Subject: Mathematics – II (303191151)
Semester: 2nd Sem. B.Tech Programme (All Branches)
Lecture Notes: Unit – 4 Fourier Integral
Introduction:
We have learnt Fourier series for periodic functions. There exist many practical problems in
engineering which involve non-periodic functions. We can solve such problems on the basis of Fourier series
technique by converting non-periodic functions in terms of sine and cosine functions. This conversion will lead to
the extension of Fourier series to Fourier integral.
Fourier Integral:
Let 𝑓(𝑥) be a function which is piecewise continuous in every finite interval (−∞, ∞) and absolutely
integrable in (−∞, ∞), then Fourier integral is given by following formula.
Formula:
❖ Fourier Integral:
∞
𝑓(𝑥) = ∫ [𝐴(𝜆) cos 𝜆𝑥 + 𝐵(𝜆) sin 𝜆𝑥] 𝑑𝜆
0
1 ∞ 1 ∞
𝑤ℎ𝑒𝑟𝑒 𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣, 𝐵(𝜆) = ∫ 𝑓(𝑣) sin 𝜆𝑣 𝑑𝑣
𝜋 −∞ 𝜋 −∞
❖ Fourier Cosine Integral:
∞
2 ∞
𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆, 𝑤ℎ𝑒𝑟𝑒 𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
0 𝜋 0
❖ Fourier Sine Integral:
∞
2 ∞
𝑓(𝑥) = ∫ 𝐵(𝜆) sin 𝜆𝑥 𝑑𝜆, 𝑤ℎ𝑒𝑟𝑒 𝐵(𝜆) = ∫ 𝑓(𝑣) sin 𝜆𝑣 𝑑𝑣
0 𝜋 0
❖ Existence of Fourier Integral:
The Fourier integral of function 𝑓(𝑥) exists if the area under the function is finite. This means that the
function must be absolutely integrable. If 𝑥 is point of continuity of 𝑓(𝑥), then the Fourier integral expansion
𝑓(𝑥+0)+𝑓(𝑥−0)
of 𝑓(𝑥) holds. If 𝑥 is a point of discontinuity, then 𝑓(𝑥) is replaced by .
2
Examples
𝟏 − 𝒙𝟐 𝒊𝒇 |𝒙| ≤ 𝟏
1. Find the Fourier Integral representation of the function 𝒇(𝒙) = {
𝟎 𝒊𝒇 |𝒙| > 𝟏
Solution:
∞
𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆
0
1 | P a g e Lecture Notes -Unit-4 Fourier Integral Mathematics – II (303191151)
, Faculty of Engineering and Technology.
Department of Applied Science & Humanities.
2 ∞
𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 ∞
= ∫ (1 − 𝑣 2 ) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 2)
sin 𝜆𝑣 cos 𝜆𝑣 sin 𝜆𝑣 1
= |(1 − 𝑣 − (−2𝑣) (− ) + (−2) (− 3 )|
𝜋 𝜆 𝜆2 𝜆 0
2 2 cos 𝜆 2 sin 𝜆
= (− + ) [∵ sin 0 = 0]
𝜋 𝜆2 𝜆3
4 sin 𝜆 − 𝜆 cos 𝜆
= ( )
𝜋 𝜆3
4 ∞ sin 𝜆 − 𝜆 cos 𝜆
𝑓(𝑥) = ∫ ( ) cos 𝜆𝑥 𝑑𝜆
𝜋 0 𝜆3
2. Find the Fourier cosine and sine Integral of 𝒇(𝒙) = 𝒆−𝒌𝒙 , (𝒙 > 𝟎, 𝒌 > 𝟎)
Solution: (a)
2 ∞
𝐴(𝜆) = ∫ 𝑓(𝑣) cos 𝜆𝑣 𝑑𝑣
𝜋 0
2 ∞ −𝑘𝑣
= ∫ 𝑒 cos 𝜆𝑣 𝑑𝑣
𝜋 0
∞
2 𝑒 −𝑘𝑣
= [ 2 (−𝑘 cos 𝜆𝑣 + 𝜆 sin 𝜆𝑣)]
𝜋 𝑘 + 𝜆2 0
2 𝑘
= [0 + 2 ]
𝜋 𝑘 + 𝜆2
2𝑘
=
𝜋(𝑘 2+ 𝜆2 )
We obtain the Fourier cosine integral representation
∞
𝑖. 𝑒. 𝑓(𝑥) = ∫ 𝐴(𝜆) cos 𝜆𝑥 𝑑𝜆
0
2𝑘 ∞ cos 𝜆𝑥
𝑓(𝑥) = 𝑒 −𝑘𝑥 = ∫ 𝑑𝜆, (𝑥 > 0, 𝑘 > 0)
𝜋 0 (𝑘 2 + 𝜆2 )
From this, we see that
∞
cos 𝜆𝑥 𝜋 −𝑘𝑥
∫ 𝑑𝜆 = 𝑒 , (𝑥 > 0, 𝑘 > 0)
0 (𝑘 2 + 𝜆2 ) 2𝑘
(b) Similarly, we have
2 | P a g e Lecture Notes -Unit-4 Fourier Integral Mathematics – II (303191151)
