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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
1 year, 9 months ago

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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 · 1 year, 8 months ago

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@Ishan Singh good one friend! Aman Rajput · 1 year, 7 months ago

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@Ishan Singh Hats off! I'm getting to learn a lot! Aditya Kumar · 1 year, 8 months ago

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@Ishan Singh That's correct! You're leading the race by miles! Kartik Sharma · 1 year, 8 months ago

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