MY MATEX NOTES

In this entry we prove the follwoing result, an integral I came across today in this video Click here for the proof.

                    In today´s post we want to prove this beautiful integral representation of the Stieltjes constants The Stieltjes constants are given by the limit We start by proving the following result: (1) Proof: With the aid of (1) we now prove the following result (2) Proof:  With the result (2), we can now prove our goal integral:

               In this post We will proof the following integrals related to the Catalan´s Constant . The proofs are based on the following paper: Representations of Catalan’s constant, David Bradley, 2001 (1) Proof: (2) Proof: (3) Proof: (4) Proof: By equation (A.7) of the appendix we have Than, we can rewrite our itegral as Equating (3) and (4) we conclude that (5) Letting in (5) we obtain (6) We can Integrate by part (6) to get And from (6) we conclude that (7) Now, let in (7) Ans from (7) we conclude that (8) Appendix (A.1) Proof: Recall the addition formulas of sine and cosine (see here ) (A.2) (A.3) Examples: Another two examples that will be useful in the next proof (A.4) Proof: Recall the double angle formulas (see here ): (A.5) (A.6) Than (A.7) Proof: Reference Representations of Catalan’s constant, David Bradley, 2001

In this post we will prove the beautiful results below related to the moments of the Loggamma function in the interval between zero and  1/2. First recall Kummer´s fourier series for the Loggamma function (1) Then multiply both sides of (1) by and integrate from zero and 1/2: Where We used that And     For the second integral, multiply both sides of (1) by and integrate from 0 to 1/2 Collecting all the results and putting together we obtain Appendix Derivative of the Dirichlet Eta function Then : We used that: And in terms of Now, recall the functional equation of the Riemann Zeta function Differentiating w.r. to we obtain: Let Solving for we obtain Special Values of the Digamma Function Recall the functional equation of the Digamma function for we have for for