Fourier Integral. Dr Mansoor Alshehri. King Saud University. MATH204-Differential Equations Center of Excellence in Learning and Teaching 1 / 22

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1 Dr Mansoor Alshehri King Saud University MATH4-Differential Equations Center of Excellence in Learning and Teaching /

2 Fourier Cosine and Sine Series Integrals The Complex Form of Fourier Integral MATH4-Differential Equations Center of Excellence in Learning and Teaching /

3 Formula of Fourier Integral The Fourier Integral of f(x) defined on the interval (, ) is given by f(x) = π where and A() cos(x) d + π A() = B() = Formula () can be written as f(x) = π f(t) cos(t) dt, f(t) sin(t) dt. B() sin(x) d, () f(t) cos (t x) dtd. () MATH4-Differential Equations Center of Excellence in Learning and Teaching 3 /

4 Theorem If f is absolutely integrable ( ) f(x) dx <, and f, f are piecewise continuous on every finite intreval, then Fourier integral of f converges to f(x) at a point of continuity and converges to at a point of discontinuity. f(x + ) + f(x ) MATH4-Differential Equations Center of Excellence in Learning and Teaching 4 /

5 Example () Express the function f(x) = as a Fourier integral. Hence evaluate value of Solution Since sin d. f(x) = π = π {, x, x >, sin cos x d and deduce the f(t) cos (t x)dtd cos (t x)dtd MATH4-Differential Equations Center of Excellence in Learning and Teaching 5 /

6 Hence = π = π = π = π Fourier Integral sin (t x) d sin ( x) sin ( x) d sin ( + x) + sin ( x) d sin cos x d. { sin cos x π d =, x <, x >, At x = ±, f(x) is discontinuous and the integral has the value ( π + ) = π 4. Now by setting x =, we have sin d = π. MATH4-Differential Equations Center of Excellence in Learning and Teaching 6 /

7 Example () Compute the Fourier integral of the function, < x < π, π < x < f(x) =, < x < π, π < x <. MATH4-Differential Equations Center of Excellence in Learning and Teaching 7 /

8 Solution We have f(x) = π = π = π + π = cos (t x)dtd + π π sin (t x) d + π π sin x + sin (π + x) d sin (π x) + sin x d ( cos π) sin(x)d. π sin (t x) cos (t x)dtd π d MATH4-Differential Equations Center of Excellence in Learning and Teaching 8 /

9 This Fourier integral converges at the discontinuities points π,, π respectively to f(( π) + ) + f(( π) ) f( + ) + f( ) f(π + ) + f(π ) =, =. =, MATH4-Differential Equations Center of Excellence in Learning and Teaching 9 /

10 Example (3) Consider the function, x < f(x) = x, x <, x. Sketch the graph of f, find the Fourier integral and deduce the value of sin d. Solution y (, ) x MATH4-Differential Equations Center of Excellence in Learning and Teaching /

11 () A() = = = f(t) cos(t)dt ( t) cos(t)dt }{{} by parts [ ] sin(t) ( t) = [ ] cos(t) sin( ) cos() cos( ) = sin() MATH4-Differential Equations Center of Excellence in Learning and Teaching /

12 () B() = = = = f(t) sin(t)dt ( t) sin(t)dt }{{} by parts [ cos(t) ] ( t) cos( ) sin() sin( ) [ ] sin(t) = cos() sin() MATH4-Differential Equations Center of Excellence in Learning and Teaching /

13 Thus, = π = π f(x + ) + f(x ) At x =, we have A() cos(x)d + B() sin(x)d π [ ( sin cos(x) cos + sin ) sin(x) ] d, hence, f( + ) + f( ) = + π = = = π sin d sin d, MATH4-Differential Equations Center of Excellence in Learning and Teaching 3 /

14 Exercises Find the Fourier integral for the following functions 3 f(x) = {, x < e x, x >., < x <, < x < f(x) =, < x <, x >. f(x) = { C, x, x >, where C is a constant such that C. Deduce the value of the sin α integral α dα. MATH4-Differential Equations Center of Excellence in Learning and Teaching 4 /

15 Fourier Cosine and Sine Series Integrals Fourier Cosine and Sine Series Integrals The Fourier sine integral is given by f(x) = π where C() = The Fourier cosine integral is given by f(x) = π where D() = C() sin(x)d, f(t) sin(t)dt. D() cos(x)d, (3) f(t) cos(t)dt. MATH4-Differential Equations Center of Excellence in Learning and Teaching 5 /

16 Fourier Cosine and Sine Series Integrals Example Compute the Fourier integral of the function { sin x, x π f(x) =, x π, and deduce that cos π + cos ( π ) d = π. Solution We observe that the function f is even on the interval (, ). So It has a Fourier cosine integral given by (3), that is f(x) = π D() cos(x)d, MATH4-Differential Equations Center of Excellence in Learning and Teaching 6 /

17 where Thus D() = = = = f(x) = π π Fourier Integral f(t) cos(t)dt = Fourier Cosine and Sine Series Integrals π sin t cos(t)dt sin t( ) + sin t( + ) dt cos t( ) π cos t( + ) ( ) ( + ) [cos π + ]. [cos π + ] cos(x)d. (4) Since f is continuous on the whole interval (, ), the above integral converges to the given function f(x). Setting x = π/ in (4), we get ( ) π [cos π + ] cos d = π. MATH4-Differential Equations Center of Excellence in Learning and Teaching 7 / π

18 Fourier Cosine and Sine Series Integrals Exercises Find the Fourier sine and Fourier cosine integral for the following functions f(x) = { x, < x, x >, x, x f(x) = x +, < x <, x. MATH4-Differential Equations Center of Excellence in Learning and Teaching 8 /

19 The Complex Form of Fourier Integral The Complex Form of Fourier Integral The complex form of Fourier integral is given by f(x) = π β()e ix d, where β() = f(t)e it dt MATH4-Differential Equations Center of Excellence in Learning and Teaching 9 /

20 The Complex Form of Fourier Integral Example Find the complex form of the Fourier integral for the function { e f(x) = x, x, x > Solution We have Hence β() = f(t)e it dt = e (i+)t dt = (i + ) e(i+)t = (e (i+) e (i+)) (i + ) = i [ + e (i+) e (i+)]. f(x) = i [ π + e (i+) e (i+)] e ix d. MATH4-Differential Equations Center of Excellence in Learning and Teaching /

21 The Complex Form of Fourier Integral Exercise Find the complex form of the Fourier integral for the function {, x < f(x) = e x, x >. MATH4-Differential Equations Center of Excellence in Learning and Teaching /

22 The Complex Form of Fourier Integral Acknowledgment This project was supported by King Saud University, Center of Excellence in Learning and Teaching. MATH4-Differential Equations Center of Excellence in Learning and Teaching /