Aug 22, 2018 - 2] Gradshteyn, I. S. and Ryzhik, I. M., Tables of Integrals, Series and Products, Academic. Press, 2000. 2. INFINITE PRODUCT FOR GAMMA ...
Innite Product for Gamma Function and Innite Series for Log Gamma Function August 22, 2018 It is the spirit that quickeneth; the esh proteth nothing: the words that I speak unto you, they are spirit, and they are life. - John 6:63.
ABSTRACT. I derive an innite product for gamma function and innite series for log gamma function.
1. INTRODUCTION In this paper, I derive the formulas below:
(k + z) ( z ) = [ { (1 + k)
log (z)+ z 1= and
n
1 n log (k + z )k z k ( 1) n+1k ( k ) [ (1+ k) k ]
n =0
ez
+
1+
=0
n
n =0
k =0
k +z
1
(
1+ k
]
1)k( nk )
1
/(n+1)
} ,
that allowed me to write the identity
§ ª ª
e =
n =0
n
ª ¨ ª ª © k =0
/ ¡ 1 + k) +k ¤¥¥¥¥ (1)k( nk )«ªªª n ¢ ¢ ¢ ¢(2 ¢ ¥ ¬ ¢ ¢ (1+ k )1+k ¥ ¥ ª ª £ ¦ 1
1 2
( +1)
=( √1 ) Å( √8 ) Å 2 Å 53 Å 5 Å 2 Å 5 Å√ 5 Å( 2 Å 5 ) Å & ) ( 3 Å 7 Å 21 ) 3 Å 7 3 3 ( 2 1
1
/1
4
/2
2
√
1
/3
14
6
13
7
1
/4
√
3
40
10
15
14
1
/5
2. INFINITE PRODUCT REPRESENTATION FOR GAMMA FUNCTION AND INFINITE SERIES FOR LOG GAMMA FUNCTION.
Theorem 2.1. For 0< z < 1, then
(k + z) (z )= [ (1+ k) {
log (z)+ z 1=
n =0
and z 1
e
n
1 n log (k + z )k z k ( 1) n +1 k ( k ) [ (1 + k) k ]
(2.1)
=0
n
n =0
+
1+
k +z 1+k
k =0
(
]
1)k( nk )
1
/(n+1)
} ,
(2.2)
where log (z) denotes the log gamma function, log(z) denotes the natural logarithm function, ez denotes the exponential function and (z ) denotes the gamma function.
Proof. In [1, page 18], we have (u)=
n =0
n
1 n k n +1 k (1) ( k )log(u + k ).
(2.3)
=0
Integrating both sides of (2.3) from 1 at z with respect to u, we nd z
+
1
(u)d u =
n =0
n
z 1 n k ( 1) n +1 k ( k )+ log(u + k)du. 1
=0
1
(2.4)
2
INFINITE PRODUCT FOR GAMMA FUNCTION AND INFINITE SERIES FOR LOG GAMMA FUNCTION
I know [2, page 651, formula 6.461] that z
+
(u)d u = log (z ).
1
(2.5)
On the other hand, I calculate that z
+
1
log(u + k)du =1 z (1 + k)log(1 + k) +(k + z)log(k + z)
=1 z log(1+ k)
1+ k
(2.6)
(k + z )k z . + log(k + z)k z =1 z +log[ (1+ k) k ] +
+
1+
From (2.4) at (2.6), it follows that
n
1 (k + z)k z (1)k( nk ){1 z + log[ (1 n + 1 + k) k ]} n k n (k + z)k z 1 (1)k( nk )log[ (1+ =1 z + n +1 k) k ] k n n 1 (k + z )k z . Òlog (z)+ z 1= n +1 (1)k( nk )log[ (1 + k) k ] n k
log (z)=
=0
+
1+
=0
+
(2.7)
1+
=0
=0
=0
+
1+
=0
The exponentiation of (2.7) give me z 1
e ( z ) =
which are the desired results.
n
n =0
k =0
{
n
(k + z)k z ( k ) [ (1 + k) k ] } +
k 1)
(
1
/(n+1)
1+
,
¡
Example 2.2. Set z = / in the Theorem above and encounter 1
2
§ ª ª
e =
n =0
n
ª ¨ ª ª © k =0
/ ¡ 1 + k) +k ¤¥¥¥ (1)k( nk )«ªªª n ¢ 2 ¢ ( ¢ ¢ ¥ ¬ ¢ ¥ ¢ ¥ ¢ 1+ k ¥ ª ª ¦ £ (1+ k ) 1
1 2
( +1)
=( √1 ) Å( √8 ) Å 2 Å 53 Å 5 Å 2 Å 5 Å√ 5 Å( 2 Å 5 ) Å & ) ( 3 Å 7 Å 21 ) 3 Å 7 3 3 ( 2 1
/1
1
/2
4
√
2
6
1
/3
14
13
7
3
√
1
/4
40
10
15
14
1
/5
REFERENCES [1] Guillera, Jesús and Sondow, Jonathan, Double Integrals and Innite Products for Some Classical Constants via Analytical Continuations of Lerch's Transcendent, arXiv:math/0506319v3 [math.NT], 5 Aug 2006. [2] Gradshteyn, I. S. and Ryzhik, I. M., Tables of Integrals, Series and Products, Academic Press, 2000.
