Who's up to the challenge? 49
If the integral above can be expressed as
where and are positive integers, with coprime, find .
Notation
:
denotes the
polylogarithm
function,
The answer is 31.
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1 solution
\begin{aligned} I & = \int_0^1 x\ce{Li}_2 (x^2-1) \, dx \\ & = \int_0^1 x \sum_{n=1}^\infty \frac{( \color{#3D99F6}{x^2-1})^n}{n^2} \, dx \quad \quad \small \color{#3D99F6}{\text{Let } u = x^2 - 1 \, \implies du = 2x dx} \\ & = \frac{1}{\color{#3D99F6}{2}} \int_\color{#3D99F6}{-1}^\color{#3D99F6}{0} \sum_{n=1}^\infty \frac{\color{#3D99F6}{u}^n}{n^2} \, d\color{#3D99F6}{u} \\ & = \frac{1}{2} \sum_{n=1}^\infty \frac{u^{n+1}}{n^2(n+1)} \bigg|_{-1}^0 \\ & = \frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^n}{n^2(n+1)} \quad \quad \small \color{#3D99F6}{\text{By partial fractions}} \\ & = \frac{1}{2} \sum_{n=1}^\infty \left(\frac{(-1)^n}{n^2} - \frac{(-1)^n}{n} + \frac{(-1)^n}{n+1} \right) \\ & = \frac{1}{2} \left(\sum_{n=1}^\infty \frac{(-1)^n}{n^2} - \sum_{n=1}^\infty \frac{(-1)^n}{n} - \sum_{n=2}^\infty \frac{(-1)^n}{n} \right) \\ & = \frac{1}{2} \left(\sum_{n=1}^\infty \frac{(-1)^n}{n^2} - 2 \sum_{n=1}^\infty \frac{(-1)^n}{n} - 1 \right) \\ & = \frac{1}{2} \sum_{n=1}^\infty \frac{(-1)^n}{n^2} - \sum_{n=1}^\infty \frac{(-1)^n}{n} - \frac{1}{2} \\ & = -\frac{1}{2} \left( \sum_{n=1}^\infty \frac{1}{n^2} - \frac{2}{4} \sum_{n=1}^\infty \frac{1}{n^2} \right) + \ln 2 - \frac{1}{2} \\ & = -\frac{1}{4} \zeta(2) + \ln 2 - \frac{1}{2} \\ & = - \left[\frac{1}{2} + \frac{\pi^2}{24}\right] + \ln 2 \end{aligned}
⟹ a + b + c + d + f = 1 + 2 + 2 + 2 4 + 2 = 3 1