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I was looking over one of Coffey's papers where is shows the following series, but with no evaluation.

I am just wondering if anyone would know how to evaluate this series:

$$\sum_{n=1}^{\infty}(-1)^{n}\left[1+\frac{2}{n+1}\right]\binom{x}{n+2}=\frac{1}{2}(2-x)(x-1)-x\left(1-2\gamma+x-2\psi(x+1)\right)$$

It is related to the derivation of the integral $$\int_{0}^{1}\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)^{2}dx$$.

It is in his paper entitled, "certain log integrals, zeta values, and the Stieltjes constant".

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Assuming $x>0$ and starting with the identity $$\sum_{n=1}^{+\infty}(-1)^{n}\binom{x}{n}=-1$$ it is quite easy to deduce from $\binom{x}{n}=\binom{x-1}{n}+\binom{x-1}{n-1}$ that: $$\sum_{n=1}^{+\infty}(-1)^{n}\binom{x}{n+2}=-\frac{1}{2}(x-1)(x-2),\tag{1}$$ so we just need an expression for: $$S=\sum_{n=1}^{+\infty}\frac{(-1)^n}{n+1}\binom{x}{n+2}=\int_{0}^{1}\sum_{n=1}^{+\infty}(-y)^n\binom{x}{n+2}\,dy.\tag{2}$$ In virtue of the binomial theorem, we just have: $$ S = \int_{0}^{1}\frac{(1-y)^x-1+xy}{y^2}dy+\int_{0}^{1}\frac{x(1-x)}{2}dy,\tag{3} $$ and up to integration by parts: $$ \int_{0}^{1}\frac{(1-y)^x-1+xy}{y^2}dy = \left.-\frac{(1-y)^x-1+xy}{y}\right|_{0}^{1}+x\cdot\int_{0}^{1}\frac{1-(1-y)^{x-1}}{y}dy,$$ so: $$ S = x\cdot\int_{0}^{1}\frac{1-(1-y)^{x-1}}{y}dy+\left(1+\frac{x}{2}\right)(1-x),\tag{4} $$ but the integral in the RHS is just $\psi(x)+\gamma$ due to the Euler formula for harmonic numbers: $$ \gamma + \psi(x+1) = \int_{0}^{1}\frac{1-y^x}{1-y}\,dy.$$

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  • $\begingroup$ Probably I messed some computations somewhere in the middle but I hope the main ideas are clearly detectable :) $\endgroup$ – Jack D'Aurizio Mar 5 '14 at 14:01
  • $\begingroup$ Very nice, JD. I had not thought of beginning with that identity. Sorry I didn't award the bounty sooner. It was my first time doing it. This is stupid, but may I ask how you went from the first to the second line. That is, how you went from $$\sum_{n=1}^{\infty}(-1)^{n}\binom{x}{n}=-1$$ to $$\sum_{n=1}^{\infty}(-1)^{n}\binom{x}{n+2}=-1/2(x-1)(x-2)$$. I see you claimed it is easy to deduce from $$\binom{x-1}{n}+\binom{x-1}{n-1}=\binom{x}{n}$$, but I still can not see what you mean. $\endgroup$ – Cody Mar 8 '14 at 12:34
  • $\begingroup$ Just re-index the sum: $$\sum_{n=1}^{+\infty}(-1)^n\binom{x}{n+2}=\sum_{n=3}^{+\infty}(-1)^n\binom{x}{n}=-1+\binom{x}{1}-\binom{x}{2}.$$ $\endgroup$ – Jack D'Aurizio Mar 8 '14 at 14:44
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    $\begingroup$ Yes, I see it. DUH!!. I was just coming back to tell you to disregard my previous comment. Thank you very much. It is a nice solution. That was Parseval's identity you began with. I think I am half asleep yet :) $\endgroup$ – Cody Mar 8 '14 at 14:56

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