Waste less time on Facebook — follow Brilliant.
×

Calculus Challenge 4!

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

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

Note by Kartik Sharma
2 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Consider 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.

Ishan Singh - 2 years, 4 months ago

Log in to reply

good one friend!

Aman Rajput - 2 years, 2 months ago

Log in to reply

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

Aditya Kumar - 2 years, 4 months ago

Log in to reply

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

Kartik Sharma - 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...