Calculus Challenge 4!

∫01(ln⁡ 1x)p−1 xq−1 dx1−axq=1aqpΓ(p) Lip(a)\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 Lia(b){\text{Li}}_{a}(b) is the polylogarithm function.

#Calculus

Note by Kartik Sharma
5 years, 9 months ago

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Comments

Consider the following Lemma.

Lemma: ∫0∞ts−1etz−1dt=Γ(s)⋅Li⁡s(z)\int_{0}^{\infty}\dfrac{t^{s-1}}{\frac{e^t}{z}-1} \mathrm{d}t = \Gamma(s) \cdot \operatorname{Li}_{s}(z)

Proof: ∫0∞ts−1etz−1dt=∫0∞ts−1ze−t1−ze−tdt\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

=∫0∞ts−1∑r=0∞(ze−t)(r+1)dt\displaystyle = \int_{0}^{\infty} t^{s-1} \sum_{r=0}^{\infty}(ze^{-t})^{(r+1)} \mathrm{d}t

=∑r=0∞zr+1∫0∞ts−1e−t(r+1)dt\displaystyle = \sum_{r=0}^{\infty} z^{r+1} \displaystyle \int_{0}^{\infty} t^{s-1}e^{-t(r+1)} \mathrm{d}t

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

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

Let,

I=∫01(ln⁡1x)p−1xq−11−axqdx \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=etx^{-q}=e^t,

  ⟹  I=1aqp∫0∞tp−1eta−1dt\displaystyle \implies \text{I} = \dfrac{1}{aq^p} \int_{0}^{\infty} \dfrac{t^{p-1}}{\frac{e^t}{a}-1} \mathrm{d}t

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

Q.E.D.

Ishan Singh - 5 years, 9 months ago

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

Kartik Sharma - 5 years, 9 months ago

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

Aditya Kumar - 5 years, 9 months ago

good one friend!

Aman Rajput - 5 years, 7 months ago
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