MY MATEX NOTES

In this blog entry we will evaluate the beautiful series below involving the square of the central binomial coefficient Click here to see the proof We used the previous result proved here

Today we will prove the following result that appears in this Twitter post Click here to see the proof.

Today we will show the following result that appears in this twitter post We will start by proving two lemmas involving the Beta and Digamma functions. Lemma 1: (1) Recall the Beta function (2) If we differentiate (2) w.r. to s we obtain Lemma 2: (3) Recall the Beta function (4) If we differentiate (4) w.r. to a we obtain Now let´s evaluate the integral: For the proof of the special values of the Digamma function see this post . Plugging the values of J and K back in the original integral we obtain:

   Today we will show the following result that appears in this post from @integralsbot Let Then: And         We then get: By the reflection formula Letting we obtain that By the functional equation of the Gamma function We obtain for instance that

Lets proof the following result: The Cauchy Schlömilch transformation (proof can be found here ) The following results hold: Proof: Consider Equating Real and Imaginary parts We obtain the desired results

Today we will show the following result, a integral representation for the reciprocal of the Beta Function For , , and To this end We will use a contour integral. The function that we will integrate is: Note that this function pocesses three branch points. We will integrate along the contour below. The dot lines represent the branch cuts. Since is analytic inside this contour, by Cauchy´s theorem the  contour integral equals zero, i.e. (1) Lets first show that the integral along vanishes as Choosing the integral over becomes Taking the limit as in the last expression The other two integrals along and also vanishes as for , and For the integral along the segment we have For the integral along let since the radius of the semi-circle is 1 Putting all together back in (1) we obtain: And (2) Now recall the reflection formual of the gamma function (3) From (3) we have that (4) Plugging (4) in (2) we Obtain