MHB Higher order reflected logarithms

[alyafey22]

Let us define the following

$$I(n,m) = \int^1_0 \log^n(x)\log^m(1-x)\,dx$$

Our purpose is finding a closed form for the general case.

Note: for a given n and m the above formula can be deduced by succesive differentiation of the beta representation

$$B(p,q) = \int^1_0 x^{p-1} (1-x)^{1-q}\,dx$$

Yet , the computations are very complicated. The main goal is tackling the question using different approaches , possibly better.

This is NOT a tutorial , any suggestions or attempts are always welcomed.

 

[alyafey22]

We will start by the special case $n=m$.

The idea is using the dilogarithm reflection formula

$$\log(x)\log(1-x) =\zeta(2)-\mathrm{Li}_2(x)-\mathrm{Li}_2(1-x) $$

Let us look for the case $m=n=1$
$$
\begin{align}
I(1,1)=\int^1_0\log(x)\log(1-x) &=\zeta(2)-\int^1_0\mathrm{Li}_2(x)\,dx-\int^1_0\mathrm{Li}_2(1-x)\,dx\\ &=\zeta(2)-2\int^1_0\mathrm{Li}_2(x)\,dx\\
&=\zeta(2)-2\zeta(2)-2\int^1_0\log(1-x)\,dx\\
&=2-\zeta(2)
\end{align}
$$

We will see if we can generalize that to the general case.

 

[Amad27]

I seem to have found a more "interesting," way to compute the integral in the earlier post (not the general form though).

$$I = \int_{0}^{1} \log(x)\log(1-x) dx$$

We can use:

$$\log(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$$

$$\log(x)\log(1-x) = -\sum_{n=1}^{\infty} \frac{\log(x)x^n}{n}$$

$$\int_{0}^{1} \log(x)\log(1-x) = -\sum_{n=1}^{\infty} \frac{1}{n}\cdot \int_{0}^{1} \log(x) x^n dx$$

With differentiation under the integral sign (leibniz rule) trick for the integral in the RHS (with differentiation $d/dn x^n = \log(x)x^n$), it is easy to see that:

$$\int_{0}^{1} \log(x)\log(1-x) dx = \sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} = I$$

First consider a square contour $C$ below:

View attachment 3848

I am assuming, as $R \to \infty$ we have: $\displaystyle \oint_{C} f(z) dz \to 0$. I do not have a proof of this, any proof is welcome, but I will work with this assumption.Consider:

$$f(z) = \frac{\psi(-z)}{z(1+z)^2}$$

$$\mathrm{Res}_{z=n} f(z) = \mathrm{Res} \psi(-z) \cdot \frac{1}{n(n+1)^2} = \frac{1}{n(n+1)^2}$$

You can check that the residue of $\psi(-z) = 1$ by using series and finding the coefficient of $\frac{1}{z-n}$

Because $z=n$ for $n \ge 1$ we only have a simple pole (order 1).

$$\mathrm{Res}_{z=0} f(z) = -2 - \gamma$$ we show this by:

Digamma function - Wikipedia, the free encyclopedia

$$\psi(-z) = \frac{1}{z} - \gamma - \zeta(2)z + ...$$
$$\frac{1}{z(1+z)^2} = \frac{1}{z} - 2 + 3x + ...$$

Multiplying the two series we receive:

$$\frac{\psi(-z)}{z(1+z)^2} = \left( \frac{1}{z} - \gamma - \zeta(2)z + ... \right) \cdot \left( \frac{1}{z} - 2 + 3z + ... \right)$$

$$ = \left( \frac{1}{z^2} -\frac{2}{z} + 3 + ... \right) + \left( \frac{1}{z^2} -\frac{\gamma}{z}+ 2\zeta(2) + ... \right)$$

$$ = ... + \frac{1}{z}\cdot \left( -\gamma - 2 \right) + ...$$

Hence,

$$\mathrm{Res}_{z=0} f(z) = -\gamma - 2$$

Next we find:

$$\mathrm{Res}_{z=-1} f(z) = \gamma + \zeta(2)$$

This is a double pole we simply use the formula:

$$\mathrm{Res}_{z=-1} f(z) = \lim_{z = -1} - \frac{\psi(-z) + z\psi_{1}(-z)}{z^2} = -\psi(1) + \psi_{1}(1) = \gamma + \zeta(2)$$

The values for $\psi(1)$ and $\psi_{1}(1)$ I took from Wikipedia Tables, but I suppose ti s derivable by the series representations...

Anyway.

$$\frac{1}{2\pi i} \cdot \oint_{C} f(z) dz = \sum \text{Residues in contour} = 0$$

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} -(\gamma + 2) + \gamma + \zeta(2) = 0$$

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} = 2 - \zeta(2)$$

Finally, just to make sure of a clean ending:

$$I = \int_{0}^{1} \log(x)\log(1-x) dx = \sum_{n=1}^{\infty} \frac{1}{n(n+1)^2} = 2 - \zeta(2) \space \space \space \space \blacksquare$$

 

[alyafey22]

Suppose you have the function $f$ with a pole of order 2 around $b$ then you can write it as

$$f = \frac{a_{-2}}{(z-b)^2}+\frac{a_{-1}}{z-b}+a_0+a_1(z-b)+\cdots $$

Finding the residues means finding the value of $a_{-1}$.

To do that we can multiply by $(z-b)^2$

$$f(z)(z-b)^2 = a_{-2}+a_{-1}(z-b)+a_0(z-b)^2+a_1(z-b)^3+\cdots $$

Now differentiate with respect to $z$

$$(f(z)(z-b)^2)' = a_{-1}+2a_0(z-b)+3a_1(z-b)^2+\cdots $$

Now but $z=b$ to get

$$(f(z)(z-b)^2)'|_{z\to b} = a_{-1} $$

For example

$$f = \frac{\psi(-z)}{z(z+1)^2}$$

has a pole of order 2 at $z=-1$

$$\text{Res}_{z=-1}f = \frac{d}{dz} \frac{\psi(-z)}{z} |_{z=-1} = \frac{-\psi_1(-z)\cdot z-\psi(-z)}{z^2} |_{z=-1} = \gamma+\zeta(2)$$