1
$\begingroup$

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.)

$\endgroup$
2
  • 1
    $\begingroup$ Should the B[(-1 + m)/m, 1/(2 m) - 1/4 I (w1 + w3)]read Beta[(-1 + m)/m, 1/(2 m) - 1/4 I (w1 + w3)]? $\endgroup$
    – Bob Hanlon
    Oct 2 '21 at 3:31
  • $\begingroup$ @BobHanlon Yes you are right. I modified the question. $\endgroup$
    – eigenvalue
    Oct 3 '21 at 2:22
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks. Although not completely general, I think it is a nice trick! $\endgroup$
    – eigenvalue
    Oct 3 '21 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.