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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
2 years, 4 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.

- 2 years, 4 months ago

good one friend!

- 2 years, 2 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago