1. Introduction 

One of the fundamental results of Fourier analysis is the Poisson Summation Formula, which can be written [1] 

where a  > 0 , ab= 2p,  and

is the Fourier Transform of f. Equation (1) is valid under relatively weak conditions. For example f(X)=O [(1+ ½X½^C) ^-1, for X Î R and for some C > 0. It is also known that if

is the Cosine Transform of a function f then we have the Poisson Summation Cosine Formula [2] 

where a  > 0 , ab= 2p. 

In this work using the Poisson Summation Formula we generalize an exponential formula of Ramanujan and arrive at a new Mellin Summation Formula. This formula is like (4) but now the part of Fourier Cosine Transform is replaced by the Mellin Transform. We also give several applications. 

Definition [2] 

The Mellin Transform of a function Y is defined to be

For the applications we will need the following 

2. The Theorems 

Let a  > 0 and the function f  be odd and analytic in the uper half plane Im(z) > 0 and continuous in Im(z) ]]> ³ 0 and let there exist C,N > 0 and 0 £ b < p such that

for every z in Im(z) ³ 0, then

where ab= 2p and . 

The Main Theorem. (The Mellin Summation Formula-MSF) 

Let a, b > 0 and ab= 2p, then

where f (x)= Re [(My) (ix) f (-x)], with and f, y, x real, f (o)= 0, provided that all sums

converge.

To prove the MSF we use a Theorem which appeared first in [4]. Here we give a complete proof and the conditions under which this Theorem holds. 

Theorem 2. [4] 

for every z with Im(z) ³ 0, C, l, d > 0 constantst, with the condition |z|= x + N + 1/2, N sufficiently large natural number. Then for x > 0 the integral  

converges absolutely, the series

converges in the Abel sense and

then the series 

converges and we can drop the limit in (10). 

3. The Proofs of Theorems 

The proof of Theorem 1 is in [3]. For the proof of Theorem 2 we need a Lemma. 

Lemma 

Then the Mellin Transform of  Y can be extended analytically into a meromorphic function in the half plane Re (z) < x with simple poles at the points z= -m for m Î z with m > -x, m ³ 0.

Proof 

Let 0 < a < r. Then if z is not an integer

Thus the function is meromorphic in C whose only poles are z= -m with residue . 

And for the derivative we have

This completes the proof of the Lemma. 

Now let the function Y be as in Lemma and let x Î R be such that (11) holds. We define the function

Then from the Lemma g is meromorphic in Im(z) > 0 and continuous in Im(z) ³ 0 with simple poles at z= i(m + x), where m Î z , m ³ 0, m + x > 0 and

and thus

So, if the integral ò ^¥_{- ¥} f(t) (MY)(x + it)dt exists and also

the series converges and we have 

The condition m > -x not needed if x > 0. 

Having in mind the above and the Lemma we can proceed to the following 

From (9) the integral 

ò ^¥_{- ¥} f (t) (MY)(x + it)e^ita dt

is absolutely convergent for a > 0. Also if 0 < a £ d we will have

for R_N ® ¥ when 0 £ q £ p. 

Hence for every a > 0 one as

from which (10) follows. 

If f (x) = Re [(MY)(ix) f (-ix)], with f, y, x real, then f (x) is an even function. The reason is that we can write according Theorem 2 

From the above we have 

also 

where ab= 2p, and . This completes the proof. 

Notes 

1. In the same way one can prove a formula of Ramajunan [5] (which we give here a more general form) 

If ab= 2p, then

where

2. The use of Theorem 2 for finding Self Reciprocal functions. 

From [2] we have that whenever f_c (x) = f (x) then

and

where , f_e (s) being an even function. Using Theorem 2 with Y(x)= e^-x we have

which is more convenient to calculate. 

Similar expansions hold for the Fourier sine transform and Hankel transforms. 

4. Applications 

1. Let and f(k) = kn^k, then Y(x)= e^-x and (MY)(S) = G(S). Hence if ab= 2p

2. Let and f(k) = k, then we have . Hence if a,n > 0, noting that 

we have from the MSF.

Thus we arrive at

3. From the relations [7]

and 

from the Main Theorem we get

whenever . 

4. Also we have 

5. If  a,n > 0 and , denotes the Polylogarithm function then

Proof 

Let and f(k)= k, then if   we have  

In the same way we have

where   is the Riemann Zeta function. 

]]> Or, if ab= 2p then

6. A consequence of Jacobi’s triple identity. Note that the calculations of sums in Application 5 depend on finding the values 

where we have set

To find C₁ (a) we differentiate X (n, a) twice with respect to v to get

 

]]>  

 

 

where q_j (0, q)= q_j (q). 

Note use Theorem 1 with along with Jacobi’s triple identity [6] to arrive at

But it is known from tables that

where k_b is the solution of (note that when b Î Q^*+ then the k_b are algebraic numbers) and

 

]]>

Hence we have 

Proposition 2.1