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Let $P_{n,k}$ be the $k$-th number of the Pascal's triangle on the $n$-th row, that is $P_{n,k} = \binom {n-1}{k-1}$

For example, $(P_{3,1},P_{3,2},P_{3,3})=(1,2,1)$.

I have noticed through random computations that the following integrals seems to be all equal to $\pi$ (up to a rational).

So I strongly believe that:

$$\forall n\geqslant2, \quad \exists r_n\in \mathbb{Q}, \quad r_n\pi = \displaystyle\int_{\mathbb{R}^+} \frac{x^2}{\sqrt{\displaystyle\sum_{k=1}^{n+1}P_{2n,k}\cdot x^{2k-2}}} \ \mathrm{d}x.$$

My question is:

Is there a way to find an express $r_n$ for all $n$ ? Is this a known sequence ?

The first values are :

$$r_2=\frac 14, r_3=\frac 18, r_4=\frac 1{16},r_5=\frac 5{256}.$$

Ant
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E. Joseph
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1 Answers1

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Suggestion/hint: If you apply the binomial formula to $$ (1 + x^2)^{2n} $$ I suspect that you'll be able to substantially simplify your denominator. And then letting $x = \tan^2 u$ looks as if it'll reduce the integral to something not too bad.

John Hughes
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