Let's do some calculus! (39)

Calculus Level 5

∫ 0 2 Li 3 ( 4 x ) d x \large \displaystyle \int_0^2 \text{Li}_3 (4^x) \, dx

Given the definite integral above, if its value can be represented in the closed form as

a b log ⁡ b ( Li c ( d ) − ζ ( c ) ) \dfrac{a}{b \log b} \bigg( \text{Li}_c (d) - \zeta (c) \bigg)

Find a + b + c + d a+b+c+d .

Notations:

  • Li n ( a ) \text{Li}_n (a) denotes the polylogarithm function , that is Li n ( a ) = ∑ k = 1 ∞ a k k n \text{Li}_n (a) = \displaystyle \sum_{k=1}^{\infty} \dfrac{a^k}{k^n} .
  • ζ ( s ) \zeta (s) represents the Riemann zeta function , that is ζ ( s ) = ∑ n = 1 ∞ 1 n s \zeta (s) = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^s} .
  • log ⁡ ( ⋅ ) \log (\cdot) denotes the natural logarithm function , that is log ⁡ e ( ⋅ ) \log_{e}{(\cdot)} or ln ⁡ ( ⋅ ) \ln(\cdot) .

Inspired by Hummus A .

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The answer is 23.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Tapas Mazumdar
Oct 14, 2016

∫ 0 2 Li 3 ( 4 x ) d x = ∫ 0 2 ∑ k = 1 ∞ 4 k x k 3 d x = ∑ k = 1 ∞ 1 k 3 ∫ 0 2 4 k x d x = ∑ k = 1 ∞ 1 k 3 [ 4 k x k log ⁡ 4 ] 0 2 = ∑ k = 1 ∞ 1 k 4 ⋅ 4 2 k − 1 log ⁡ 4 = 1 log ⁡ 4 ⋅ ∑ k = 1 ∞ 4 2 k − 1 k 4 = 1 2 log ⁡ 2 ⋅ ∑ k = 1 ∞ ( 1 6 k k 4 − 1 k 4 ) = 1 2 log ⁡ 2 ( ∑ k = 1 ∞ 1 6 k k 4 − ∑ k = 1 ∞ 1 k 4 ) = 1 2 log ⁡ 2 ( Li 4 ( 16 ) − ζ ( 4 ) ) \begin{aligned} \displaystyle \int_0^2 \text{Li}_3 (4^x) \, dx &= \displaystyle \int_0^2 \sum_{k=1}^{\infty} \dfrac{4^{kx}}{k^3} \, dx \\ &= \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^3} \int_0^2 4^{kx} \, dx \\ &= \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^3} \left[ \dfrac{4^{kx}}{k \log 4} \right]^2_0 \\ &= \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^4} \cdot \dfrac{4^{2k}-1}{\log 4} \\ &= \dfrac{1}{\log 4} \cdot \displaystyle \sum_{k=1}^{\infty} \dfrac{4^{2k}-1}{k^4} \\ &= \dfrac{1}{2 \log 2} \cdot \displaystyle \sum_{k=1}^{\infty} \bigg( \dfrac{16^k}{k^4} - \dfrac{1}{k^4} \bigg) \\ &= \dfrac{1}{2 \log 2} \bigg( \displaystyle \sum_{k=1}^{\infty} \dfrac{16^k}{k^4} - \displaystyle \sum_{k=1}^{\infty} \dfrac{1}{k^4} \bigg) \\ &= \dfrac{1}{2 \log 2} \bigg( \text{Li}_4 (16) - \zeta (4) \bigg) \end{aligned}

Thus

( a , b , c , d ) = ( 1 , 2 , 4 , 16 ) (a,b,c,d) = (1,2,4,16)

Which gives

a + b + c + d = 23 a+b+c+d = \boxed{23}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

This has a right approach. However, there is some gap that needs addressing:

The part which needs justifying (even though it's fairly obvious) is why we can interchange the order of the sum/integral in the second step in your working?

Pi Han Goh - 4 years, 8 months ago

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