Simplifying special functions via substitution

Question

We have a well-known identity $$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ between beta function and gamma function.

I have a complicated expression involving the terms in the form $\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. I want to simplify by replacing it to $B(a,b)$. How can I do this?

The following is the example that I want to simplify:

(2^(-3 + 4/m)
    E^(-((3 I \[Pi])/m)) (-1 + E^((2 I \[Pi])/m))^3 Gamma[(-2 + m)/
    m] Gamma[(-1 + m)/m] Gamma[
    1/m + (I w3)/
     2] (Beta[(-1 + m)/m, 1/(2 m) + 1/4 I (w1 + w3)] Gamma[
       1/4 (4 - 2/m - 2 I w1 + I w3)] Gamma[(2 + 2 I m w1 - I m w3)/(
       4 m)] - Beta[(-1 + m)/m, 1/(2 m) - 1/4 I (w1 + w3)] Gamma[
       1/4 (4 - 2/m + 2 I w1 - I w3)] Gamma[(2 - 2 I m w1 + I m w3)/(
       4 m)]))/(Gamma[1 - 1/m + (I w3)/2] Gamma[
    1/4 (4 - 2/m + 2 I w1 - I w3)] Gamma[
    1/4 (4 - 2/m - 2 I w1 + I w3)])

(Looking carefully, one can see that the above simplification is possible.)

Answer 1

This may give you what you want.

expr = (2^(-3 + 4/m) E^(-((3 I \[Pi])/m)) (-1 + 
        E^((2 I \[Pi])/m))^3 Gamma[(-2 + m)/m] Gamma[(-1 + m)/
       m] Gamma[
      1/m + (I w3)/
        2] (Beta[(-1 + m)/m, 1/(2 m) + 1/4 I (w1 + w3)] Gamma[
         1/4 (4 - 2/m - 2 I w1 + I w3)] Gamma[(2 + 2 I m w1 - 
            I m w3)/(4 m)] - 
       Beta[(-1 + m)/m, 1/(2 m) - 1/4 I (w1 + w3)] Gamma[
         1/4 (4 - 2/m + 2 I w1 - I w3)] Gamma[(2 - 2 I m w1 + 
            I m w3)/(4 m)]))/(Gamma[1 - 1/m + (I w3)/2] Gamma[
      1/4 (4 - 2/m + 2 I w1 - I w3)] Gamma[
      1/4 (4 - 2/m - 2 I w1 + I w3)]) // ExpandAll

expr //. Gamma[a_ + b_] -> (Gamma[a] Gamma[b])/Beta[a, b]

% // Simplify

You still get a complicated expression, but I think all the terms involving Gammas are single values rather than sums.

I initially used ExpandAll on your expression so that there was no factor with the sum values in the Gamma arguments. My second statement would not work if there were.