Mathematical Physics Vol 1

Chapter A. Fractional Calculus: A Survey of Useful Formulas

466

L { f ( t ) } ( s )

f ( t )

( a √ t )

ae a 2 t erfc

1 √ s + a √ s s − a 2 √ s s + a 2 √ s ( s 1

1 √ π t −

( a √ t )

2 t erf

1 √ π t

+ ae a

a 2 t R a √ t 0

τ 2 d τ

1 √ π t −

2 a √ π e −

e

a 2 t erf ( a √ t ) a 2 t R a √ t 0 [ b − a erf ( a ( a √ t )

1 a e

− a 2 )

τ 2 d τ

1 √ s ( s + a 2 ) b 2 − a 2 ( s − a 2 )( 1 √ s ( √ s + a ) 1 ( s + a ) √ s + b

2 a √ π e −

e

√ t )]

( b √ t )

e a 2 t

2 t erfc

− be b

√ s + b )

e a 2 t erfc

e − at erf ( √ b

√ t )

1

− a

√ b

− a

√ t )

( b √ t )

b 2 − a 2

e a 2 t

2 t erfc

[ b

− 1 ]+ e b

a erf ( a

√ s ( s ( 1 − s ) n n + 1 2 ( 1 − s ) n n + 3 2 √ s + 2 a

√ s + b )

− a 2 )(

H 2 n ( √ t )

n ! ( 2 n ) ! √ π t

H 2 n + 1 ( √ t )

n ! ( 2 n + 1 ) ! √ π

√ s

−

at [ I

ae − e − 1

1 ( at )+ I 0 ( at )]

√ s

1 √ s + a √ s + b Γ ( k ) ( s + a ) k ( s + b ) k ( s + a ) √ s + 2 a √ s √ s + 2 a + √ s 1 2 ( s + b ) − 1

a − b

2 ( a + b ) t I

2 t )

0 (

√ π t te − 1

a − b 2 ( a + b ) t [ I

k − 1

1 2 ( a + b ) t I

( a − b

2 e −

, ( k ≥ 0 )

2 t )

k − 1 2

a − b

a − b

2 t )+ I 1 (

2 t )]

0 (

3 2

1 t e − k t e −

at I

1 ( at )

( a − b ) k ( √ s + a + √ s + b ) 2 k √ s + a +( √ s ) − 2 ν √ s √ s + a

1 2 ( a + b ) t I

a − b

, ( k > 0 )

2 t )

k (

1 2 I ν (

1 a ν e −

1 2 t )

, ( ν > − 1 )

1 √ s 2 + a 2 − a 2 ( √ s 2 + a 2 1 √ s 2

J 0 ( at ) I 0 ( at )

− s ) ν

a ν J

, ( ν > − 1 )

ν ( at )

√ s 2 + a 2

√ π Γ ( k ) (

k − 1

1 ( s 2 + a 2 ) k ( √ s 2 + a 2 ( √ s 2 + a 2 + s ) ν √ s 2 − a 2 1 ( s 2 − a 2 ) k 1 s + √ s 2 + a 2 1 ( s + √ s 2 + a 2 ) n 1 s √ s + 1

t 2 a )

( k > 0 )

2 J k

( at )

− 1 2

k t J k ( at )

ka

− s ) k ( k > 0 ) , ( ν > − 1 )

a ν I

ν ( at )

√ π Γ ( k ) ( erf ( √ t )

k − 1

t 2 a )

, (k>0)

2 I k

( at )

− 1 2

J 1 ( at ) at nJ n ( at ) a n t

, n ∈ N

J 1 ( at ) a J n ( at ) a n

1 ( √ s 2 + a 2 ( s + √ s 2 + a 2 ) 1 ( √ s 2 + a 2 ( s + √ s 2 + a 2 ) n

, n ∈ N

, n ∈ N

π s 2 k

k s 2 + k 2 ch

| sin kt | 0 ( 2 √ kt ) 1 √ π t cos2 1 √ π t ch2 1 √ π t sin2 J

1 s e − 1 √ s e −

k / s

√ kt

k / s

√ kt

1 √ s e

k / s

√ kt

1 s √ s e −

k / s

Continued on next page