# Higher order reflected logarithms

• MHB
• alyafey22
In summary, the purpose of the conversation was to find a closed form for the integral $I(n,m) = \int^1_0 \log^n(x)\log^m(1-x)\,dx$ by using different approaches. The main goal was to tackle the question for the general case, but the computations were very complicated. The conversation discussed the special case $n=m$, where the dilogarithm reflection formula was used to compute the integral. It was also mentioned that the formula for $I(n,m)$ can be deduced by successive differentiation of the beta representation. The conversation then moved on to discussing another approach for computing the integral and showed how it can be generalized to the general case. It was also mentioned that
alyafey22
Gold Member
MHB
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.

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.

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:

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

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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 $z=-1$

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

## 1. What are higher order reflected logarithms?

Higher order reflected logarithms are a type of mathematical function that involves taking the logarithm of the logarithm of a number. They are used in complex mathematical calculations and have applications in fields such as physics, engineering, and finance.

## 2. How are higher order reflected logarithms different from regular logarithms?

The main difference between higher order reflected logarithms and regular logarithms is the number of logarithms involved. Regular logarithms only involve taking the logarithm of a number, while higher order reflected logarithms involve taking the logarithm of the logarithm of a number.

## 3. What is the formula for calculating higher order reflected logarithms?

The formula for calculating a higher order reflected logarithm is log(log(x)), where x is the number being evaluated. This can also be written as log²(x), where the ² represents the number of logarithms being taken.

## 4. What are some real-world applications of higher order reflected logarithms?

Higher order reflected logarithms have many applications in fields such as physics, engineering, and finance. They are often used in complex mathematical calculations involving exponential growth, such as population growth or financial investments.

## 5. Are higher order reflected logarithms difficult to work with?

Higher order reflected logarithms can be more challenging to work with compared to regular logarithms, as they involve multiple steps and require a good understanding of logarithmic properties. However, with practice and a solid understanding of the concept, they can be mastered like any other mathematical function.

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