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Showing posts from December, 2021

\sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}=\frac{2 \pi}{9\sqrt{3}}+\frac43

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Today we will evaluate the following nice infinte sum First recall the expansion (proved here ) (1) We can rewrite (1) as (2) Proof: If we differentiate both sides of (2) w.r. to x we obtain (3) Proof: Letting       in (3) we obtain

SUM INVOLVING THE CENTRAL BINOMIAL COEFFICIENT

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Here is another interesting (finite) sum involving the central Binomial Coefficient from this Twitter post : Proof: Proof: Proof:

SUMS OF RECIPROCALS OF THE CENTRAL BINOMIAL COEFFICIENTS

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Today we compute sums of reciprocals of the central Binomial Coefficients, namely As we will see, it´s intimately related to the series expansion of the arcsine function. First recall the trigonometric form of the Beta function ( see this post ) if we let   we obtain (1) Proof: Where in (*) we made use of Legendre´s duplication formula proved here Now we multiply (1) by and sum from 1 to we get (2) Proof: Now we integrate (2) to obtain (3) Proof: First we show the L.H.S. of (3) For the R.H.S. Therefore Letting we find that and finally If we now let in (3) we get Setting in (3) Finally letting in (3) we obtain

\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx

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In this short post we will estabilish the following result Where in (**) we used the previous established results ( I and II ): In (*) above we used the following result

\int x/(1+x^2)*1/(tanh(\pi*x/2))dx

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Today´s post we will prove the challenging result posted by @integralsbot here : To this end, we will first establish the following three results: We begin proving two Lemmas: Lemma 1: proof: Lemma 2: Proof: Now, let´s evaluate the first integral where in (*) we applied Lemma 1 Now recall the Laplace transform of the cosine function, a proof can be found in this post : Then In (**) we used the following result proved in this post Therefore, we have that (1) Differentiating (1) w.r. to  gives us Where in the last line we used Lemma 2 , then (2) If we let in (2) we obtain (3) Letting in (3) (4) Now, if we differentiate (4) w.r. to s we obtain (5) And now setting in (5) Proof of     First note that (A.1) Than recall the well known result (a simple proof can be found in this post ) (A.2) We can split the L.H.S. of (A.2) in it´s even and odd terms, namely or

\int_0^\infty \frac{x \ln(1+x^2)}{e^{2 \pi x}+1}\,dx=\frac{19}{24}-\frac{23}{24}\ln 2-\frac{\ln A}{2}

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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