(Project)study of fourier integrals
- 2. -:CONTENT:- Fourier integrals Fourier cosine integral Fourier sine integral Problem Conclusion
- 3. -: Fourier integrals :- Fourier integral is an extension of Fourier series in non-periodic functions. Hear integration is used instead of Summation in a Fourier series. The Fourier integrals of a function f(x) is given by. 𝑓 𝑥 = 0 ∞ 𝐴 𝜔 cos 𝜔𝑥 + 𝐵 𝜔 𝑠𝑖𝑛𝜔𝑥 𝑑𝜔 Where, 𝐴 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑢 cos 𝜔𝑢 𝑑𝑢 𝐵 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑢 𝑠𝑖𝑛𝜔𝑢 𝑑𝑢 In this there are three type of integral. I. Fourier integral II. Fourier cosine integral III. Fourier sine integral
- 4. -:Fourier cosine integral:- Suppose f(x) is an even function . As we know cos𝜔x is an even function and sin𝜔𝑥 is an odd function .There fore f(x) cos𝜔𝑥 is an Even function & f(x) sin𝜔𝑥 is an odd function. Now, 𝐴 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑢 𝑐𝑜𝑠𝜔𝑢 𝑑𝑢 = 2 𝜋 0 ∞ 𝑓 𝑢 𝑐𝑜𝑠𝜔𝑢 𝑑𝑢 𝐵 𝜔 = 1 𝜋 −∞ ∞ 𝑓(𝑢) sin 𝜔𝑢 𝑑𝑢 = 0 Fourier cosine integral represented by 𝑓 𝑥 = 0 ∞ 𝐴(𝜔) cos 𝜔𝑥 𝑑𝜔
- 5. -:Fourier sine integral:- Suppose f(x) is an even function. Now sin 𝜔𝑥 is an odd function then f(x) cos𝜔𝑥 is also odd function and f(x) sin 𝜔𝑥 is an even function. Now, 𝐴 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑢 cos 𝜔𝑢 𝑑𝑢 = 0 𝐵 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑢 sin 𝜔𝑢 𝑑𝑢 = 2 𝜋 0 ∞ 𝑓 𝑢 sin 𝜔𝑢 𝑑𝑢 Fourier sine integral represented as 𝑓 𝑥 = 0 ∞ 𝐵(𝜔) sin 𝜔𝑥 𝑑𝜔
- 6. Problem:- Find the Fourier cosine and Fourier sine integral of 𝑓 𝑥 = 𝑒−𝑘𝑥 where x>0 and k>0. Ans:- Fourier cosine integral of f(x) is given by 𝑓 𝑥 = 0 ∞ 𝐴(𝜔) cos 𝜔𝑥 𝑑𝜔 … . . (1) Where, 𝐴 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑥 cos 𝜔𝑥 𝑑𝑥 = 1 𝜋 −∞ ∞ 𝑒−𝑘𝑥 cos 𝜔𝑥 𝑑𝑥 Since f(x) is even so the integration is even = 2 𝜋 0 ∞ 𝑒−𝑘𝑥 cos 𝜔𝑥 𝑑𝑥 Now, by integration by parts = 2 𝜋 −𝑘 𝑘2+𝜔2 𝑒−𝑘𝑥 −𝜔 𝑘 sin 𝜔𝑥 + cos 𝜔𝑥 ∞ 0 = 2 𝜋 0 + 𝑘 𝑘2+𝜔2 = 2𝑘 𝜋 𝑘2+𝜔2 By substituting 𝐴 𝜔 into (1) we obtain the Fourier cosine integral 𝑓 𝑥 = 2𝑘 𝜋 0 ∞ cos 𝜔𝑥 𝑘2 + 𝜔2 𝑑𝜔
- 7. Fourier sine integral of f(x) is given by 𝑓 𝑥 = 0 ∞ 𝐵(𝜔) sin 𝜔𝑥 𝑑𝜔 … … (2) Where, 𝐵 𝜔 = 1 𝜋 −∞ ∞ 𝑓 𝑥 sin 𝜔𝑥 𝑑𝑥 Since f(x) is odd the integral is even = 2 𝜋 0 ∞ 𝑒−𝑘𝑥 sin 𝜔𝑥 𝑑𝑥 Now, by integration by parts = 2 𝜋 −𝜔 𝑘2 + 𝜔2 𝑒−𝑘𝑥 𝑘 𝜔 sin 𝜔𝑥 + cos 𝜔𝑥 ∞ 0 = 2 𝜋 0 + 𝜔 𝑘2 + 𝜔2 = 2 𝜋 𝜔 𝑘2 + 𝜔2 By substituting B 𝜔 into (2) we obtain the Fourier cosine integral 𝑓 𝑥 = 2 𝜋 0 ∞ 𝜔 sin 𝜔𝑥 𝑘2 + 𝜔2 𝑑𝜔
- 8. -:Conclusion:- Many problems involve functions that are non –periodic and are of interest on the whole x-axis to find Fourier series of such function we use Fourier integrals.
- 9. THANK YOU