The Riemann Transform

Transcription

1 The Riemann Tranform By Armando M. Evangelita Jr. Augut 28, 28 ABSTRACT In hi 859 paper, Bernhard Riemann ued the integral equation f (x ) x dx to develop an explicit formula for etimating the number of prime number le than a given quantity. It i the purpoe of thi preent work to explore ome of the propertie of thi equation.

2 Conider the integral equation given below () F ( ) = f (x ) x dx Formula () i the the integral of f(x) time x for x = to and the reulting function i a function of, ay F() (or the tranform of f(x)). It mut be aume that f(x) i uch that the integral exit (it ha finite value). Example Apply formula () to obtain the tranform of f(x) = e -x. Solution. Subtitute e -x to () F ( ) = e x x dx = Γ ( ) R () <, ine Γ() = e x x dx, R() >, where Γ(S) i the gamma function and R() i the real part of the complex quantity. Unit Step Function (Heaviide Function) The unit tep function or Heaviide function μ(x a) i for x < a, ha a jump ize at x = a (where it i uually conider a undefined), and i for x > a, in a formula: μ ( x a) = { if x < a if x > a a. The tranform of μ(x a) i F () = x μ( x a)dx = x dx = x a a ; here the integration begin at x = a (>) becaue μ(x a) i for x < a. Hence F () = a (a > and > ).

3 Example 2: The Riemann Zeta Function i given by obtain the tranform of n= ζ () = = n= μ ( x n), n =,2,3,4,.... n = R() >, n= n F () = {μ ( x ) + μ (x 2) + μ (x 3) +...}x dx = x + x + x = ( ) = n= n = ζ (), R() >. Example 3: Obtain the tranform of π (x) = μ( x p), where p i a prime number, p = 2, 3, 5, 7, p,. F () = { μ (x p)x dx p } = { μ( x 2) + μ (x 3)+ μ(x 5) + μ (x 7) +... } x dx π ()= ( ) = p p R() >. Dirac Delta Function Conider the function f τ ( x a) = { / τ if a x a+τ otherwie. It integral i I = a+τ τ f τ ( x a)dx = dx =. a We let now let τ become maller and maller and take the limit a τ (τ > ). Thi limit i denoted by δ(x a), that i, δ (x a) = lim f τ ( x a). τ

4 and obtain δ (x a) = { if x = a otherwie and δ (x a)dx =. δ(x a) i called the Dirac delta function or the unit impule function. For a continuou function f(x) one ue the ifting property of δ(x a), f ( x)δ ( x a)dx = f (a). To obtain the tranform of δ(x a), we write and take the tranform f τ ( x a) = τ [ μ( x a) μ (x (a+τ ))] F () = f τ (x a)x dx = [ a (a + τ ) ] = a τ ( + τ a ), a > and R() >. τ Take the limit a τ. By l Hopital rule, the quotient on the right ha the limit /a. Hence, the right ide ha the limit a -( + ). The tranform of δ(x a) define by thi limit i F () = δ (x a)x dx = a (+) a >. Example 4 Obtain the tranform of δ (x n). n= F () = { n= δ (x n)}x dx = n (+) = ζ (+), R()>. n=

5 The Riemann Tranform Many common function like in x, co x, ln x, etc., when applied to formula () don t have finite value. But if the lower limit for () tart at x =, then there are uitable function uch that the integral in () exit. If f(x) i a function defined for all x, it Riemann tranform i the integral of f(x) time x for x = to. It i a function of, ay F(), and i denoted by R(f) ; thu (2) F ( )=R (f )= f (x) x dx. The given function f(x) in (2) i called the invere tranform of F() and i denoted by R - (F); that i, f (x) = R (F ). Example 5 Let f (x) =. Find F(). Solution. From (2) we obtain by integration R(f )= R()= x dx = x = (>). Example 6 Let f (x) = x a, where a i a contant. Find F(). Solution. From (2), R( x a )= x a x dx = a x ( a) = a ( a > ). THEOREM Linearity of the Riemann Tranform The Riemann tranform i a linear operation; that i, for any function f(x) and g(x) whoe tranform exit and any contant a and b the tranform of af(x) + bg(x) exit, and R {af (x) + bg(x)} = af() + bg().

6 Example 7 Find the tranform of coh (alnx) and inh (alnx). Solution. Since coh(a ln x)= 2 (xa + x a ) and inh(aln x)= 2 (xa x a ), we obtain from Example 6 and Theorem, R {coh(aln x)}= 2 (R( xa ) + R( x a )) = 2( a + + a ) = R {inh(aln x)}= 2 (R( xa ) R(x a )) = 2 ( a + a ) = 2 a 2 a 2 a 2. Example 8 Let f (x) = x α i, where i i the imaginary operator (i= ). Find F(). Solution. From Example 6 R( x αi ) = αi = α i + αi + αi = 2 + α 2 + i α 2 + α 2. Example 9 Coine and Sine Derive the formula R {co(α ln x)} = Solution. From Example 8 and Theorem 2 +α 2 and R {in(α ln x)} = α 2 +α 2. x α i = co(α ln x ) + i in(α ln x) R( x αi ) = R(co(α ln x)) + i R(in(α ln x)), thu R {co(α ln x)} = 2 +α 2 and R {in(α ln x)} = α 2 +α 2.

7 THEOREM 2 -Shifting Theorem If f(x) ha the tranform F() (where > k for ome k), then x a f(x) ha the tranform F( a) (where a > k). In formula, or, if we take the invere on both ide R {x a f (x )} = F( a) x a f (x) = R {F ( a)}. PROOF We obtain F( a) by replacing with a in the integral in (), o that F ( a) = x ( a) f ( x)dx = x [ x a f (x )]dx = R {x a f (x)}. Example From Example 9 and the -Shifting theorem one can obtain the Riemann tranform for R {x a co(α ln x )} = a ( a) 2 + α 2 and R{x a in(α ln x)} = α ( a) 2 + α 2. Exitence and Uniquene of Riemann Tranform A function f(x) ha a Riemann tranform if it doe not grow too fat, ay, if for all x and ome contant M and k it atifie (3) f (x ) Mx k. THEOREM 3 Exitence Theorem for Riemann Tranform If f(x) i defined and piecewie continuou on every finite interval on x and atifie (3) for all x and ome contant M and k, then the Riemann tranform R(f) exit for all > k.

8 PROOF Since f(x) i piecewie continuou, x -- f(x) i integrable over any finite interval on the x-axi, R(f ) = f (x )x f (x) x dx M x k x dx = M k. Uniquene. If the Riemann tranform of a given function exit, it i uniquely determined and if two continuou function have the ame tranform, they are completely identical. Tranform of Derivative and Integral THEOREM 4 Riemann Tranform of Derivative The tranform of the firt and econd derivative of f(x) atify (4) R(f ') = (+)F(+) f () (5) R(f '' ) = (+2)(+)F (+2) (+)f () f '() Formula (4) hold if f(x) i continuou for all x and atifie (3) and f (x) i piecewie continuou on every finite interval for x. Formula (5) hold if f and f are continuou for all x and atify (3) and f i piecewie continuou on every finite interval for x. PROOF Uing integration by part on formula (4) R(f ) = f '(x )x dx = [ f ( x)x ] + (+) f (x)x 2 dx = f () + (+)F(+). The proof of (5) now follow by applying integration by part twice on it, that i R(f '') = f ''(x )x dx = [ f '(x )x ] + (+) f '( x)x 2 dx = f '() + (+) [ f (x )x 2 + (+2) f (x)x 3 dx ] = f '() (+)f () + (+2)(+)F(+2).

9 Repeatedly uing integration by part a in the proof of (5) and uing induction, we obtain the following Theorem. THEOREM 5 Riemann Tranform of the Derivative f (n) of Any Order Let f, f,, f (n-) be continuou for all x and atify (2). Furthermore, let f (n) be piecewie continuou on every finite interval for x. Then the tranform of f (n) atifie R(f (n) ) = (+n)(+n ) (+)F (+n) (+n )(+n 2) f () (+n 2)(+n 3) f '() f (n ) (). Example Let f(x) = x 2. Then f() =, f (x) = 2x, f () = 2, f (x) = 2. Obtain R{f}, R{f }, and R{f }. Solution. R {f } = F () = 2, F (+) =, F (+2) =. Hence, by formula (4) and (5), R(f ') = (+) = 2 and R(f '') = (+2)(+) (+) 2 = 2. THEOREM 6 Riemann Tranform of Integral Let F() denote the tranform of a function f(x) which i piecewie continuou for x and atifie formula (3). Then, for >, > k, and x >, (6) R { x } { f (τ )d τ = x F ( ), thu f ( τ )d τ = R } F( ). PROOF Let the integral in (6) be g(x) then g (x) = f(x). Since g() = (the integral from to i zero), R {f (x )} = R{g'( x)} = (+)G(+) g() = (+)G(+) = F(), replace by, ([ ] + )G([ ] + ) = F( ) = G() = F( ). Diviion by and interchange of the left and right ide give the firt formula in (6), from which the econd follow.

10 Example 2 Let f(x) = x. Obtain the tranform of x g(x ) = τ d τ = G(). Solution. F () = R {x } =, F( ) = 2, then G() = ( 2). Differentiation and Integration of Tranform Differentiation of Tranform Given a function f(x), the derivative F () = df/d of the tranform F() = R(f) can be obtained by differentiating F() under the integral ign with repect to. Thu, if F ( ) = f (x ) x dx, then F ' () = ln x f ( x) x dx. Conequently, if R(f) = F(), then R {ln x f (x )} = F ' () and R {F ' ()} = ln x f (x ), where the econd formula i obtained by applying on both ide of the firt formula. In thi way, differentiation of a function correpond to the multiplication of the function by -lnx. Example 3 Obtain the tranform of ln x in(α ln x ) and ln x co(α ln x). Solution. R {ln x in(α ln x )} = d α d { 2 + α } = 2 R {ln x co(α ln x)} = d d { 2 + α } 2 2α ( 2 + α 2 ) 2 = ( 2 + α 2 ) 2 2 ( 2 + α 2 ) 2 = 2 α 2 ( 2 + α 2 ) 2.

11 Integration of Tranform Given a function f(x), and the limit of f(x)/lnx, a x approache from the right, exit, then for > k, R { f ( x) ln } x = F(σ )dσ hence R { F(σ )dσ } = f (x) ln x. In thi way, integration of the tranform of a function f(x) correpond to the diviion of f(x) by lnx. From the definition it follow that F(σ )dσ = [ x σ f (x)dx ]dσ = f (x ) [ x σ dσ ] dx x. Integration of x - σ with repect to σ give x - σ /(-ln x). Hence the integral over σ on the right equal x - /ln x. Therefore, F(σ )dσ = x f ( x) ln x { dx = R f (x) ln x } ( > k ). Example 4: Find the invere tranform of ln ( + α 2 Solution. Denote the given tranform by F(). It derivative i 2 ) ( = ln 2 + α 2 ). 2 Taking the invere tranform, we obtain F '() = d d [ln(2 +α 2 ) ln 2 ] = { R 2 F '() = R 2 +α 2 Hence the invere f(x) of F() i α = 2co(α ln x) 2 = ln x f (x). } f (x) = 2 ( co(α ln x )). ln x

12 Alternatively, if we let G() = α 2 2, then g(x ) = R {G} = 2[ co(α ln x )]. From thi and uing { the integral of tranform we get, R ln 2 + α 2 } R { = G()d 2 } = g(x ) ln x = 2 [ co(α ln x )]. ln x The Riemann Tranform and the Laplace Tranform The Laplace tranform i the integral of f(y) time e -y from y = to where f(y) i defined for all y. It i denoted by L{f}, (7) L{f } = f ( y )e y dy. The Riemann tranform i given below (8) R {f } = f (x ) x dx. Replace x = e y ( or y = lnx) in formula (8) and ince x = to, y = (ln) to (ln). f (x ) x dx = f (e y ) e y y d (e y ) = f ( y) e y dy, which i formula (7). The Bilateral Laplace Tranform Formula (7) i uually called the Unilateral Laplace tranform ince the integral i evaluated from to. The integral below i known a the Bilateral Laplace tranform becaue the integral i taken from - to, (9) B {f } = f ( y)e y dy.

13 Now, conider the integral equation () f (x ) x dx, Replace x = e y ( or y =lnx) in formula (4) and ince x = to, y = - to, thu f (x ) e x dx = f (e y ) e y y d (e y ) = f ( y) e y dy, which i (9). Riemann Tranform: General Formula Formula F ( )=R {f ( x)}= f (x ) x dx f ( x) = R (F ()) Name Definition of Tranform Invere Tranform R {af (x) + bg(x)} = ar {f (x)} + br{g(x )} Linearity R {x a f ( x)} = F ( a) R {F ( a)} = x a f (x ) -Shifting Theorem R(f ') = (+)F(+) f () Differentiation of Function R(f '' ) = (+2)(+)F (+2) (+)f () f '() x R { f (τ )d τ } = F ( ), Integration of Function R {ln x f (x)} = F '() ln x } = R { f ( x) F(σ )dσ Differentiation of Tranform Integration of Tranform

14 Table: Some Riemann Tranform f(x) F() = R{f(x)} 2 x 3 x a a 4 x α i αi 5 co(α ln x) 2 + α 2 6 in(α ln x ) α 2 + α 2 7 coh(a ln x) 2 a 2 8 inh(aln x) a 2 a 2 9 x b co(α ln x) b ( b) 2 + α 2 x b in(α ln x ) α ( b) 2 + α 2 δ (x a) a (+) 2 n= δ (x n) n= n + 3 δ (x p) p p 4 μ (x a) a 5 μ (x n) n= n= n 6 μ (x p) p p = ζ (+) p (+) = ζ () 7 2 ln x [ co(α ln x )] ln( 2 + α 2 p 2 )

15 8 arctan in(αln x) ln x α 9 2 ln x [ coh(aln x)] ln( 2 a 2 2 ln x (xb x a ) ) 2 ln ( a b ) REFERENCE Riemann, Bernhard (859). On the Number of Prime Number le than a Given Quantity. pp. 5-7.