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I want to evaluate $$\int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx}$$ I run this integral on Maple, It does converge. How we get a closed form? Is that related to polylogs? $\operatorname{Li}_{5}\left(\frac{1}{2}\right)$

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A very nice question (+1) –  Chris's sis Dec 28 '12 at 14:01

2 Answers 2

up vote 8 down vote accepted

$$\begin{eqnarray*} \int_{0}^{1}{\frac{{{\ln }^{2}}\left( 1-x \right){{\ln }^{2}}\left( 1+x \right)}{1+x}dx} &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \int_0^1 dx\, (1-x)^s(1+x)^{t-1} \right) \right|_{s=t=0} \\ &=& \left. \left(\frac{\partial^2}{\partial s^2}\frac{\partial^2}{\partial t^2} \,\frac{{}_2F_1(1-t,1;s+2;-1)}{s+1} \right) \right|_{s=t=0} \end{eqnarray*}$$ Here we've used Euler's integral representation for the hypergeometric function.

Addendum: Using the series representation for the hypergeometric function we can take the derivatives. After a little work we find the integral can be written as $$\sum_{k=2}^\infty \frac{(-1)^k}{k+1} \left(H_k^2 - H_k^{(2)}\right) \left(H_{k+1}^2 + H_{k+1}^{(2)}\right),$$ where $H_k$ and $H_k^{(n)}$ are the harmonic and generalized harmonic numbers, respectively. This sum has bad convergence behavior, the terms go like $(-1)^k(\log k)^4/k$ $(k\to\infty)$. Since the sum is alternating we could accelerate it using the Euler transform, for example.

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Thx! But there should be a nicer form : ) –  Ryan Dec 28 '12 at 2:36
    
@Ryan: If it exists someone will post it here. I would be surprised, but I have been before. :-) –  user26872 Dec 28 '12 at 3:02
    
@oen: I went exactly that way, but it didn't lead me to a nice form. Chris. (+1) –  Chris's sis Dec 28 '12 at 9:32
    
@Chris'ssister : Sure buddy : ) –  Ryan Dec 28 '12 at 10:25
    
@Chris'ssister I m sry guys, Actually one of the integral got a missing lnx term :( You can see here (lots of integral and papers etc) : tieba.baidu.com/… –  Ryan Jan 30 '13 at 2:29

The result is -4〖Li〗_5 (1/2)+4ζ(3) 〖Log〗^2 2-(2π^2)/9 〖Log〗^3 2- π^2/3 ζ(3)-π^4/20 Log2+ 7/30 〖Log〗^5 2+ 63/8 ζ(5) =0,418709830998418751408037243448…

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Do you mean:$$-4\operatorname{Li}_5(\frac12)+4\zeta(3)\log^2(2)-\frac{2\pi^2}9 \log^3(2)-\frac{\pi^2}{3}\zeta(3)-\frac{\pi^4}{20}\log2+\frac7{30}\log^5(2)+ \frac{63}8\zeta(5)$$ –  columbus8myhw Sep 21 at 20:32
    
Also: That seems to be correct, numerically. How did you get this? –  columbus8myhw Sep 21 at 20:42

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