Fractional Calculus - Gauge-institute.org

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<strong>Gauge</strong> Institute Journal, Volume 5, No 1, February 2009H. Vic Dannon2.2 Direct Derivation of the Fundamental TheoremProof:( −1) d dDF2 ( x ) =f ( x )−12( dx ( − a)) ( dx ( − a))d 1−1= ( x −u) 2f( u)du( − ) Γ()∫( dx a)12u=xu=a−121 d−1= ( x −u) 2f( u)duΓ()dx∫12u=xu=aBy Leibnitz Rule for Differentiation of Integrals, [Spieg, p. 95],ddxu= ψ( x) u=ψ( x)∫ Gxudu ( , ) = ∫ ∂xGxudu ( , ) + ϕ'( xGx ) ( , ϕ( x)) −ψ'( xGx ) ( , ψ( x)).u= ϕ( x) u=ϕ( x)If we take−12Gxu (, ) = ( x− u) fu (), then Gxx (, ) is not defined. Toremove the singularity, we integrate by partsu= x u=xu=x−1 1 1⎡⎤( x − u) 2f( u) du = 2( x u) 2f( u) 2 ( x u) 2⎢ − ⎥ − − f '( u)du⎣⎦∫ ∫u=au= a u=aTherefore,u=x1 12 2=−2( x −a) f( a) −2 ( x −u) f '( u)du∫u=a10
<strong>Gauge</strong> Institute Journal, Volume 5, No 1, February 2009H. Vic Dannon⎧u x(1)1 d=⎫−1 1DF2( x) =− ⎪2( x a) 2f ( a) 2 ( x u) 2⎨ − + − f '( u)du ⎪⎬Γ()1 dx∫2⎪⎩u=a⎪⎭⎧1u=x⎫1 d1−=− ⎪( x a) 2f( a) 2( x u) 2⎨ − + − f '( u)du⎪⎬Γ()1dx∫2⎪⎩u=a⎪⎭If we take Gxu (, ) = 2( x− u) 2fu (), then by Leibnitz rule1⎧1u=x⎫1 1−−=− ⎪( x a) 2f( a) ( x u) 2⎨ − + − f '( u)du⎪⎬Γ()1∫ (H )2⎪⎩u=a⎪⎭Integrating by parts with respect to uu= x u=xu=x−1 1 3⎡ − ⎤( )2'( ) ( )2( ) 1−x − u f u du = x u f u ( x u) 2⎢ − ⎥ + − f( u)du⎣⎦∫ ∫2u=au= a u=aSubstituting into (H ),−u=x1 32 1−2=−( x − a) f( a) + ( x −u) f( u)duu=x( − ) 11()1 22 u=aSince Γ− (1+ 1) =−1Γ− (1),∫2u=a1−32 2DF ( x) =− ( x −u) f ( u)duΓ∫ .2 2 2( −1)1−3DF2( x) = ( x −u) 2f ( u)duΓ− ( )∫112u=xu=a= Dfx2().11
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