MY MATEX NOTES

In today´s post We will compute these wonderfull integrals: From last post we know (1) (2) (3) Lemma 1 Proof: Therefore, from Lemma 1 and from (1) and (2) we obtain Lemma 2 Therefore from Lemma 2 and (2) and (3) we obtain

In this post we will compute the following three integrals: Consider the following three integrals (1) (2) (3) Differentiating the three w.r. to z we obtain (4) (5) (6) Now recall Binet´s Integral representation for the Digamma function (7) Making the change of variable and multiplying (7) by 4z we get (8) Multiplying (3) by (9) Integrating (8) w.r. to z   Letting Where we used that Proved here . For the second integral, integrating from 0 to z Letting Where we used that Proved here . For the last integral, We Multiply both sides of (7) by to get Now let And now let to obtain Integrating from 0 to z Letting

In this post we will proof the following two integrals related to the derivative of the Riemann Zeta function at Recall Kummer´s fourier expansion for LogGamma    (1) Integrating (1) from 0 to (2) Letting in (2) Where we used that  Proved here , and that For the second integral, lets first evalute the following integral                                                Then, differentiating both sides w.r. to s Setting s=2 Setting Where we again made use of the fact that