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Fractional Calculus - Gauge-institute.org

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<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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