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Example text
9) cannot be satisfied by any ordinary functions in classical mathematics. Hence, the delta function is not a function in the classical sense. However, it can be treated as a function in the generalized sense, and in fact, δ(x) is called a generalized function or distribution. The concept of the delta function is clear and simple in modern mathematics. It is very useful in physics and engineering. Physically, the delta function represents a point mass, that is a particle of unit mass located at the origin.
6 (General Parseval’s Relation). If F {f (x)} = F (k) and F {g(x)} = G(k) then ∞ ∞ f (x) g(x)dx = −∞ PROOF F (k) G(k)dk. 25) −∞ We proceed formally to obtain ∞ ∞ F (k) G(k)dk = −∞ −∞ = 1 2π 1 dk · 2π ∞ e © 2007 by Taylor & Francis Group, LLC g(x)dx −∞ ∞ g(x) dx −∞ −∞ ∞ ∞ f (y) dy −∞ e−ikx g(x) dx f (y) dy −∞ ∞ −∞ ∞ = ∞ −iky eik(x−y) dk −∞ ∞ δ(x − y)f (y) dy = −∞ f (x)g(x) dx. 20). We now use an indirect method to obtain the Fourier transform of sgn(x), that is, 2 1 . 8), ik F {sgn(x)} = 2 , π or 2 1 · .
18) ∞ 1 = cosec2 (x). 14) to derive ∞ π 1 = 2 2 (n + b ) b n=−∞ = π b ∞ exp(−2|n|πb) n=−∞ ∞ ∞ exp(−2nπb) + n=0 exp(2nπb) n=1 which is, by writing r = exp(−2πb), = π b π = b ∞ ∞ rn + n=0 1+r 1−r © 2007 by Taylor & Francis Group, LLC n=1 = 1 r n = π coth(πb). 19) π 1 2 b exp(−b|k|). 14) that ∞ (n2 n=−∞ π 1 + e−2πb 1 = . 2 +b ) b (1 − e−2πb ) Or, ∞ 2 1 1 π 1 + e−2πb + . = (n2 + b2 ) b2 b (1 − e−2πb ) n=1 It turns out that ∞ π 1 = 2 + b2 ) (n 2b n=1 1 + e−2πb 1 − (1 − e−2πb ) πb π 2 (1 + e−x ) 2 − , (2πb = x) x (1 − e−x ) x π 2 x (1 + e−x ) − 2 (1 − e−x ) = 2 x (1 − e−x ) = = π x 2 x3 1 2 x− − x2 2!