proof of Gauss’ digamma theorem

proof of Gauss’ digamma theorem

Proof.

The proof follows the

argument

given in

[

1

]

, which in turn derives from that given in

[

2

]

.

The first formula is the

logarithmic derivative

of

Γ

⁢

(

x

+

n

)

=

(

x

+

n

-

1

)

⁢

(

x

+

n

-

2

)

⁢

⋯

⁢

x

⁢

Γ

⁢

(

x

)

By the partial fraction decomposition satisfied by the

ψ

function

,

ψ

⁢

(

p

q

)

+

γ

=

∑

n

=

0

∞

(

1

n

+

1

-

q

p

+

n

⁢

q

)

=

lim

t

→

1

-

⁡

∑

n

=

0

∞

(

1

n

+

1

-

q

p

+

n

⁢

q

)

⁢

t

p

+

n

⁢

q

using Abel’s limit theorem.

Now,

∑

n

=

0

∞

(

1

n

+

1

-

q

p

+

n

⁢

q

)

⁢

t

p

+

n

⁢

q

=

∑

n

=

0

∞

t

p

+

n

⁢

q

n

+

1

-

∑

n

=

0

∞

q

⁢

t

p

+

n

⁢

q

p

+

n

⁢

q

=

t

p

-

q

⁢

∑

n

=

0

∞

t

(

n

+

1

)

⁢

q

n

+

1

-

q

⁢

∑

n

=

0

∞

t

p

+

n

⁢

q

p

+

n

⁢

q

Since

-

ln

⁡

(

1

-

t

)

=

∑

n

=

1

∞

t

n

n

the first term is

-

t

p

-

q

⁢

ln

⁡

(

1

-

t

q

)

Using the algorithm for extracting every

q

th

term of a series (

http://planetmath.org/ExtractingEveryNthTermOfASeries

), the second term is

∑

n

=

0

q

-

1

ω

-

n

⁢

p

⁢

ln

⁡

(

1

-

ω

n

⁢

t

)

and therefore

∑

n

=

0

∞

(

1

n

+

1

-

q

p

+

n

⁢

q

)

⁢

t

p

+

n

⁢

q

=

-

t

p

-

q

⁢

ln

⁡

(

1

-

t

q

)

+

∑

n

=

0

q

-

1

ω

-

n

⁢

p

⁢

ln

⁡

(

1

-

ω

n

⁢

t

)

=

-

t

p

-

q

⁢

ln

⁡

1

-

t

q

1

-

t

-

(

t

p

-

1

-

1

)

⁢

ln

⁡

(

1

-

t

)

+

∑

n

=

1

q

-

1

ω

-

n

⁢

p

⁢

ln

⁡

(

1

-

ω

n

⁢

t

)

Let

t

→

1

-

to get

ψ

⁢

(

p

q

)

=

-

γ

-

ln

⁡

q

+

∑

n

=

1

q

-

1

ω

-

n

⁢

p

⁢

ln

⁡

(

1

-

ω

n

)

Replace

p

by

q

-

p

and add the two expressions to obtain

ψ

⁢

(

p

q

)

+

ψ

⁢

(

q

-

p

q

)

=

-

2

⁢

γ

-

2

⁢

ln

⁡

q

+

2

⁢

∑

n

=

1

q

-

1

cos

⁡

(

2

⁢

π

⁢

n

⁢

p

q

)

⁢

ln

⁡

(

1

-

ω

n

)

The left side is real, so it is equal to the

real part

of the right side. But

ℜ

(

ln

(

1

-

ω

n

)

)

=

ln

|

1

-

ω

n

|

1

/

2

=

ln

|

(

1

-

cos

2

⁢

π

⁢

n

q

)

2

+

sin

2

2

⁢

π

⁢

n

q

|

1

/

2

=

1

2

ln

(

2

-

2

cos

2

⁢

π

⁢

n

q

)

and so

ψ

⁢

(

p

q

)

+

ψ

⁢

(

q

-

p

q

)

=

-

2

⁢

γ

-

2

⁢

ln

⁡

q

+

∑

n

=

1

q

-

1

cos

⁡

(

2

⁢

π

⁢

n

⁢

p

q

)

⁢

ln

⁡

(

2

-

2

⁢

cos

⁡

2

⁢

π

⁢

n

q

)

(1)

But

ψ

⁢

(

x

)

-

ψ

⁢

(

1

-

x

)

=

d

d

⁢

x

⁢

ln

⁡

(

Γ

⁢

(

x

)

⁢

Γ

⁢

(

1

-

x

)

)

=

-

π

⁢

cot

⁡

π

⁢

x

by the

Euler reflection formula

and thus

ψ

⁢

(

p

q

)

-

ψ

⁢

(

q

-

p

q

)

=

-

π

⁢

cot

⁡

π

⁢

p

q

(2)

Add equations (

1

) and (

2

) to get

ψ

⁢

(

p

q

)

=

-

γ

-

π

2

⁢

cot

⁡

π

⁢

p

q

-

ln

⁡

q

+

1

2

⁢

∑

n

=

1

q

-

1

cos

⁡

2

⁢

π

⁢

n

⁢

p

q

⁢

ln

⁡

(

2

-

2

⁢

cos

⁡

2

⁢

π

⁢

n

q

)

=

-

γ

-

π

2

⁢

cot

⁡

π

⁢

p

q

-

ln

⁡

q

+

∑

n

=

1

q

-

1

cos

⁡

2

⁢

π

⁢

n

⁢

p

q

⁢

ln

⁡

(

2

⁢

sin

⁡

π

⁢

n

q

)

where the last equality holds since

ln

(

2

-

2

cos

(

2

θ

)

)

=

ln

(

2

-

2

(

1

-

2

sin

2

θ

)

=

ln

(

4

sin

2

θ

)

=

2

ln

(

2

sin

θ

)

References

1

G.E. Andrews, R. Askey, R. Roy,

Special Functions

, Cambridge University Press, 2001.

2

J.L. Jensen [1915-1916], An elementary exposition of the theory of the

gamma function

,

Ann. Math.

17

, 124-166.

Title

proof of Gauss’ digamma theorem

Canonical name

ProofOfGaussDigammaTheorem

Date of creation

2013-03-22 16:24:07

Last modified on

2013-03-22 16:24:07

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Proof

Classification

msc 30D30

Classification

msc 33B15

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