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Let $n$ be an integer $\geq 2$ and let $p_n(x)=\displaystyle \frac{1}{n!}(x^n(1-x)^n)^{(n)}$ be a Legendre polynomial. Let $k $ be a positive integer.

I would like know if there is a closed formula (such as a hypergeometric function in $n$ and in $k$ ) to $$ I_{n,k}=\displaystyle \int_{0}^1 p_n(x)x^n(1-x)^k \log(x) \,dx$$

thanks for your help.

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I don't have a general expression as a function of $n$, but there is one in terms of harmonic numbers $H_k$ as a function of $k\in\mathbb{R}^+$ for given $n$; for example

$$I_{2,k}=\tfrac{1}{2}\Gamma(k+1)\left(\frac{2 H_{k+3}-3}{ \Gamma(k+4)}+\frac{6\left(25-12 H_{k+5}\right)}{\Gamma (k+6)}\right)$$ $$I_{3,k}=3 \Gamma (k+1) \left(\frac{12 H_{k+5}-25}{\Gamma (k+6)}+\frac{30 \left(49-20 H_{k+7}\right)}{\Gamma (k+8)}\right)$$ $$I_{4,k}=\tfrac{3}{4} \Gamma (k+1) \left(\frac{25-12 H_{k+5}}{\Gamma (k+6)}+\frac{180 \left(20 H_{k+7}-49\right)}{\Gamma (k+8)}+\frac{840 \left(761-280 H_{k+9}\right)}{\Gamma (k+10)}\right)$$ $$I_{5,k}=\tfrac{45}{2} \Gamma (k+1) \left(\frac{147-60 H_{k+7}}{\Gamma (k+8)}+\frac{56 \left(280 H_{k+9}-761\right)}{\Gamma (k+10)}-\frac{504 \left(2520 H_{k+11}-7381\right)}{\Gamma (k+12)}\right)$$

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  • $\begingroup$ ok, thank you for your help mr Carlo $\endgroup$ – mamiladi May 29 at 1:36

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