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This is a question regarding the answer presented here.

In order to make this post self-contained, I am wondering if someone can explain why the sum $$ \sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\frac{1}{k+2} \tag{1} $$ can be re-written as $$ \int_{0}^{1}x\, Q_n(x)\,dx \tag{2} $$ where in the above $Q_n(x)$ is the shifted Legendre polynomial that satisfy $$ Q_n(x)=P_n(2x-1)=\sum_{k=0}^{n}(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}x^k \tag{3} $$ with $P_n(x)$ the ordinary Legendre polynomials.

I am aware of how to re-write a sum as an integral, but I have never encountered an example of the form of eq.(1) and I find passing from eq.(1) to eq.(2) a bit confusing.

  • Simply perform term-wise integration by swapping the integral and the finite sum, and just use the fact that with $k\geq 0$, $$\int_{0}^{1}x^{k+1},\mathrm{d}x=\frac{x^{k+2}}{k+2}\bigg]_{0}^{1}=\frac{1}{k+2}.$$ That’s all there is to it. – KStarGamer Sep 04 '23 at 17:57
  • @KStarGamer thanks for the comment, however, that was not my issue. I guess what I wrote was not clear enough. I was able to go from eq.(2) to eq.(1) without issues, which is what you explained. My question was how to go from eq.(1) to eq.(2) directly. Is this clearer now? –  Sep 04 '23 at 19:02
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    I’m unsure why you cannot just reverse the process. Write $\frac{1}{k+2}$ as $\int_{0}^{1}x^{k+1},\mathrm{d}x$ and once again swap the integral and finite sum. – KStarGamer Sep 04 '23 at 22:50
  • @KStarGamer yes, thanks for pointing this out. It was only after I wrote my comment that I realized your hint! –  Sep 05 '23 at 17:46
  • No problem! Glad I could help :) – KStarGamer Sep 05 '23 at 20:12

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