A Dilogarithmic integration by parts

[DreamWeaver]

From the logarithmic integral representation of the Dilogarithm, $$\text{Li}_2(x)$$, $$|x| \le 1$$, prove the reflection formula for the Dilogarithm. Dilogarithm definition:$$\text{Li}_2(x) = -\int_0^1\frac{\log(1-xt)}{t}\, dt = \sum_{k=1}^{\infty}\frac{x^k}{k^2}$$Dilogarithm reflection formula:$$\text{Li}_2(x) + \text{Li}_2(1-x) = \frac{\pi^2}{6}-\log x\log (1-x)$$Where$$\text{Li}_2(1) = \sum_{k=1}^{\infty}\frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6}$$Hint:

Spoiler

The clue is in the thread title...

 

[polygamma]

Spoiler

$$\text{Li}_{2}(x) = -\int_{0}^{x} \frac{\ln(1-t)}{t} \ dt$$

Let $u = 1-t$.

Then

$$ \text{Li}_{2}(x) = -\int_{1-x}^{1} \frac{\ln w}{1-w} \ dw$$

Now integrate by parts.

$$ \text{Li}_{2}(x) = \ln(1-w) \ln(w) \Big|^{1}_{1-x} - \int_{1-x}^{1} \frac{\ln(1-w)}{w} \ dw$$

$$ = - \ln(x) \ln(1-x) - \int_{0}^{1} \frac{\ln(1-w)}{w} \ dw + \int_{0}^{1-x} \frac{\ln(1-w)}{w} \ dw$$

$$ =- \ln(x) \ln(1-x) + \frac{\pi^{2}}{6} - \text{Li}_{2}(1-x)$$

 

[DreamWeaver]

Spoiler

$$\text{Li}_{2}(x) = -\int_{0}^{x} \frac{\ln(1-t)}{t} \ dt$$

Let $u = 1-t$.

Then

$$ \text{Li}_{2}(x) = -\int_{1-x}^{1} \frac{\ln w}{1-w} \ dw$$

Now integrate by parts.

$$ \text{Li}_{2}(x) = \ln(1-w) \ln(w) \Big|^{1}_{1-x} - \int_{1-x}^{1} \frac{\ln(1-w)}{w} \ dw$$

$$ = - \ln(x) \ln(1-x) - \int_{0}^{1} \frac{\ln(1-w)}{w} \ dw + \int_{0}^{1-x} \frac{\ln(1-w)}{w} \ dw$$

$$ =- \ln(x) \ln(1-x) + \frac{\pi^{2}}{6} - \text{Li}_{2}(1-x)$$

Nicely done, RV! (Poolparty)

Too bad there's no such easy relation for the Trilogarithm, eh? (Headbang)

 

[polygamma]

There is a quasi-reflection formula for the trilogarithm, but it also involves $\text{Li}_{3} \left( 1-\frac{1}{x} \right)$. But I'm sure you already knew that.

 

[DreamWeaver]

There is a quasi-reflection formula for the trilogarithm, but it also involves $\text{Li}_{3} \left( 1-\frac{1}{x} \right)$. But I'm sure you already knew that.

That sounds vaguely familiar. Although, to be fair, it's been a very long time since I've worked on Trilogarithmic identities... (Bad mammal! (Headbang) )

 

[alyafey22]

Spoiler

Let

$$f(x)=\text{Li}_2(x)+\text{Li}_2(1-x)$$

By differentiation

$$f'(x)=-\frac{\log(1-x)}{x}+\frac{\log(x)}{1-x}$$

Now integrate to obtain

$$f(x)=-\log(x)\log(1-x)+C$$

Take $x\to 1$ we get $C=\text{Li}_2(1)$

$$f(x)=\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$$

 

[DreamWeaver]

Spoiler

Let

$$f(x)=\text{Li}_2(x)+\text{Li}_2(1-x)$$

By differentiation

$$f'(x)=-\frac{\log(1-x)}{x}+\frac{\log(x)}{1-x}$$

Now integrate to obtain

$$f(x)=-\log(x)\log(1-x)+C$$

Take $x\to 1$ we get $C=\text{Li}_2(1)$

$$f(x)=\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$$

That's a beautiful proof, Zaid! Very nicely done... (Rock)