\[\displaystyle \int_{0}^{1}{{\left(\ln\ \frac{1}{x}\right)}^{p-1}\ \frac{{x}^{q-1} \ dx}{1-a{x}^{q}}} = \frac{1}{a{q}^{p}} \Gamma(p)\ {\text{Li}}_{p}(a)\]

Prove the identity above. where \({\text{Li}}_{a}(b)\) is the polylogarithm function.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestConsider the following Lemma.

Lemma:\[\int_{0}^{\infty}\dfrac{t^{s-1}}{\frac{e^t}{z}-1} \mathrm{d}t = \Gamma(s) \cdot \operatorname{Li}_{s}(z)\]Proof:\(\displaystyle \int_{0}^{\infty}\dfrac{t^{s-1}}{\frac{e^t}{z}-1} \mathrm{d}t = \int_{0}^{\infty}\dfrac{t^{s-1}ze^{-t}}{1-ze^{-t}}\mathrm{d}t\)\(\displaystyle = \int_{0}^{\infty} t^{s-1} \sum_{r=0}^{\infty}(ze^{-t})^{(r+1)} \mathrm{d}t\)

\(\displaystyle = \sum_{r=0}^{\infty} z^{r+1} \displaystyle \int_{0}^{\infty} t^{s-1}e^{-t(r+1)} \mathrm{d}t\)

\(\displaystyle \sum_{r=0}^{\infty} \dfrac{z^{r+1}}{(r+1)^s} \Gamma(s)\)

\(\displaystyle = \Gamma(s) \cdot \operatorname{Li}_{s}(z)\)

Let,\( \displaystyle \text{I}=\int_{0}^{1} \left(\ln \dfrac{1}{x}\right)^{p-1} \dfrac{x^{q-1}}{1-ax^q} \mathrm{d}x\)

Substitute \(x^{-q}=e^t\),

\(\displaystyle \implies \text{I} = \dfrac{1}{aq^p} \int_{0}^{\infty} \dfrac{t^{p-1}}{\frac{e^t}{a}-1} \mathrm{d}t\)

\( \displaystyle = \dfrac{1}{aq^p} \Gamma (p) \cdot \operatorname{Li}_{p}(a) \)

Q.E.D.Log in to reply

good one friend!

Log in to reply

Hats off! I'm getting to learn a lot!

Log in to reply

That's correct! You're leading the race by miles!

Log in to reply