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3. Let f be a non-periodic function. similar to Fourier series, we can represent the function by Fourier integral as follows. f(t) = 1 π Z ∞ 0  coswt Z ∞ −∞ f(v) coswv dv + sinwt Z ∞ ∞ f(v) sinwv dv dw = 1 π Z ∞ 0 Z ∞ −∞ f(v)[coswt coswv + sinwtsinwv] dv dw = 1 π Z ∞ 0 Z ∞ −∞ f(v) cosw(v − t)dv dw. (10) Now cosw(v − t) is an even function of w and so we can write f(t) = 1 2π Z ∞ −∞ Z ∞ −∞ f(v) cosw(v − t)dv dw. (11) Also as sinw(v − t) is an odd function of w 0 = i 2π Z ∞ −∞ Z ∞ −∞ f(v) sinw(v − t)dv dw. (12) Adding the above two equations gives f(t) = 1 2π Z ∞ −∞ Z ∞ −∞ f(v)[cosw(v − t) + isinw(v − t)] dv dw = 1 2π Z ∞ −∞ Z ∞ −∞ f(v)e i(v−t)w dv dw = 1 2π Z ∞ −∞ e −itw Z ∞ −∞ f(v)e iwv dv dw. (13) Let F(w) = Z ∞ −∞ f(v)e iwv dv (14) then f(t) = 1 2π Z ∞ −∞ F(w)e −iwt dw. (15) 3 Define a Fourier transform pair as follows The Fourier transform of f(t) is F(f) = F(w) = Z ∞ −∞ f(t)e iwt dt (16) and the inverse Fourier transform is f(t) = F −1 (f) = 1 2π Z ∞ −∞ F(w)e −iwt dw. (17) (a) The convolution f ∗ g of two functions f and g is defined by (f ∗ g)(x) = Z ∞ −∞ f(u)g(x − u)du = Z ∞ −∞ f(x − u)g(u)du. (18) Show that F(f ∗ g) = F(f)F(g). (19) (b) Show that F  d n f dxn  = (−iw)F  d n−1 f dxn−1  . (20) (c) Consider the Laplace’s equation in two dimensions uxx + uyy = 0 (21) in the half plane y ≥ 0 subject to the boundary condition u(x, 0) = f(x) (−∞ < x < ∞) and the condition u(x, y) → 0 as p x 2 + y 2 → ∞. Find the solution by using a Fourier transformation with respect to x.

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