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In evaluating integrals like (link to another example)

$$I=\int_0^1\frac{\log(x) \log^2(1-x)dx}{x}$$

one can make the substitution $x=\sin^2(\theta)$ to obtain

$$I=16\int_0^\frac{\pi}{2}\frac{\log(\sin(x)) \log^2(\cos(x)) \cos(x)dx}{\sin(x)}$$ which is the partial derivative of the beta function at $(x=0,y=1)$:

$$I=\lim_{x \rightarrow 0^+}\partial_y^2\partial_x B(x,1)$$ because $$B(x,y)=2 \int_0^\frac{\pi}{2} \cos^{2x-1}(x)\sin^{2y-1}(x)dx$$ $$\partial^2_y\partial_xB(x,y)=16\int_0^\frac{\pi}{2} \log(\cos(x))^2\log(\sin(x)) \cos^{2x-1}(x)\sin^{2y-1}(x)dx.$$

My problem: $I$ is convergent, and taking the partial derivatives of $B(x,y)$ in the way above does yeild $2I$, however $B(x,y)$ is only defined for $\Re(x), \Re(y) >0$. So, how does one:

  1. Go about proving that the limit exists (directly using the partial derivatives of the beta function, without reference to $I$)? (Extra points for intuition, as I don't see how the partial derivative can exist at a place where the function doesn't).
  2. Find the limit in this case?
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Meow
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  • Maybe some integral representations require $\Re(x), \Re(y) >0, ;$but the Beta function is defined for most $(x,y);$ e.g. by $$ B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\cdot $$ Special considerations are needed only if $x+y$ is a non-positive integer: If $x$ and $y$ are no integers then $B=0$. Since $B(x,y)$ is symmetric, let $x \le y$. If $y > 0$ use the Pochhammer symbol to get $B(x,y) = \Gamma(y)/(x)_y$, otherwise the result is undefined. – gammatester Dec 13 '13 at 08:00
  • Are you sure about the relation $2I = \lim_{x \rightarrow 0^+}\partial_y^2\partial_x B(x,1);$? Using Maple I get $$\partial_y^2\partial_x B(x,y) = -\Psi(2,x+y) B(x,y)-2 \Psi(1,x+y) \Big(\Psi(y)-\Psi(x+y)\Big) B(x,y) +\Big(\Psi(x)-\Psi(x+y)\Big) \Big(\Psi(1,y)-\Psi(1,x+y)\Big) B(x,y) +\Big(\Psi(x)-\Psi(x+y)\Big) \Big(\Psi(y)-\Psi(x+y)\Big)^2 B(x,y)$$ and for the limit $$\lim_{x \rightarrow 0^+}\partial_y^2\partial_x B(x,1) = -\frac{\pi^4}{180} \approx -0.5411616.$$ The value -0.5411616 is the numerical result from Maple or Wolfram Alpha. – gammatester Dec 16 '13 at 09:49
  • @gammatester http://www.wolframalpha.com/input/?i=integrate+log%28x%29log%5E2%281-x%29%2Fx+from+0+to+1. It seems that the integral is equal to the limit, although the factor of $2$ before $2I$ was incorrect. – Meow Dec 16 '13 at 16:45
  • @gammatester Thank you, although my question boils down to finding a nicer form for that limit and (even if the prior isn't possible) to prove that the limit equals $\frac{-1}{2}\zeta(4).$ – Meow Dec 16 '13 at 16:52

1 Answers1

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[Initial apology: this is just a bunch of considerations too long to fit in a comment] Now we have a proof.

$$ I = -\int_{0}^{+\infty} x\cdot\log^2(1-e^{-x})\,dx = \int_{0}^{+\infty}\frac{x^2}{e^{x}-1}\log(1-e^{-x})\,dx,$$ $$ I = -\int_{0}^{+\infty}\frac{x^2}{e^{x}-1}\sum_{k=1}^{+\infty}\frac{e^{-kx}}{k}dx = -\int_{0}^{+\infty}x^2 e^{-x}\sum_{k=1}^{+\infty}H_k e^{-kx}dx,$$ $$ I = -\sum_{k=1}^{+\infty}H_k \int_{0}^{+\infty}x^2 e^{-(k+1)x}dx = -\sum_{k=1}^{+\infty}\frac{2H_k}{(k+1)^3}.$$

Interestingly, integrating by parts in another way we get:

$$ I = \int_{0}^{1}\operatorname{Li}_2(x)\left(\frac{\log(1-x)}{x}-\frac{\log x}{1-x}\right)dx = -\frac{1}{2}\operatorname{Li}_2^2(1)-\int_{0}^{1}\frac{\operatorname{Li}_2(x)\log x}{1-x}\,dx,$$ [This part is merely optional] so the identity $I=-\frac{\zeta(4)}{2}$ can probably be proved through dilogarithm identities. By using the Taylor series of $\operatorname{Li}_2(x)$ and $\frac{\log(1-x)}{x}$ and the well-known identity $\int_{0}^{1}(1-x)^m x^n dx=\frac{m! n!}{(m+n+1)!}$ it is quite easy to notice that:

$$\int_{0}^{1}\frac{\operatorname{Li}_2(1-x)\log(1-x)}{x}\,dx = -\sum_{m,n=1}^{+\infty}\frac{m! n!}{m^2 n^2 (m+n)!},$$ and I would not be surprised if someone manage to put the RHS in $c\cdot\zeta(2)^2$ form though "reverse" creative telescoping or simple fractions decomposition. [End of the optional part]

For istance: $$\int_{0}^{1}\frac{\operatorname{Li}_2(x)\log x}{1-x}dx = -\sum_{n=1}^{+\infty}\frac{1}{(n+1)^2}\sum_{m=1}^{n}\frac{1}{m^2}.\tag{1}$$ Finally, here is the trick: $$\frac{\pi^4}{120}=\frac{\zeta(2)^2-\zeta(4)}{2} = \sum_{m>n}\frac{1}{m^2 n^2}$$ is exactly the opposite of the RHS in $(1)$. By collecting pieces we get: $$ I = -\frac{\zeta(2)^2}{2}+\frac{\zeta(2)^2-\zeta(4)}{2} = -\frac{\zeta(4)}{2}$$ QED.

Jack D'Aurizio
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  • Thanks very much for answering thoroughly. So is the crux of the existence of the limit that although the limit of a function may exist, the limit of its derivative may exist? – Meow Dec 20 '13 at 20:11
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    Exactly. It is not an unusual behaviour for a differentiable function over an open set $\Omega$ that the limit of (partial or directional) derivatives exists but does not approach the derivative of the function in a point $z\in\partial\Omega$, or even $\lim_{w\to z}f(w)$ does not exist. A classical example is given by the function $g(x,y)=\frac{x^2-y^2}{x^2+y^2}$. – Jack D'Aurizio Dec 21 '13 at 00:55
  • Thank you. In this case, is it possible to easily calculate the integral directly from the beta limit, or is the above method much easier? – Meow Dec 21 '13 at 12:15
  • That is just a matter of taste between the polylogarithm and the polygamma function :) – Jack D'Aurizio Dec 25 '13 at 13:42