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$$\int_{0}^{1} Li_k(x) dx$$

$$Li_k(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^k}$$

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \int_{0}^{1} \frac{x^n}{n^k} \,dx$$

From Fubini's theorem, I suppose we were allowed to interchange.

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} [\frac{x^{n+1}}{(n+1)n^k}]_{0}^{1}$$

$$\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}$$

Let $$S = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}$$

I am having the big trouble in evaluating the sum $S$.

Amad27
  • 10,465

1 Answers1

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Set $$S_k := \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k},\quad k=1,2,3,\ldots.$$ Observe that, for $k=1,2,3,\ldots$, we have $$ \begin{align} S_{k+1} &= \sum_{n=1}^{\infty} \frac{1}{(n+1)n^{k+1}}\\ &= \sum_{n=1}^{\infty} \frac{(n+1)-n}{(n+1)n^{k+1}}\\ &= \sum_{n=1}^{\infty} \frac{1}{n^{k+1}}-\sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}\\ &= \zeta(k+1)-S_{k+1} \tag1 \end{align} $$ where $\zeta$ denotes the Riemann zeta function.

Rewrite $(1)$ as $$ (-1)^{k+1}S_{k+1}-(-1)^{k}S_{k}=(-1)^{k+1}\zeta(k+1) \tag2 $$ sum $(2)$ from $k=1$ to $k=p-1$, $p=2,3,4,\ldots$, then, by telescoping terms, you get

$$ \sum_{n=1}^{\infty} \frac{1}{(n+1)n^p}=(-1)^{p-1}+\sum_{k=2}^{p}(-1)^{p-k}\zeta(k), \quad p=1,2,3,\ldots. $$

where, by a direct telescoping, using $\displaystyle \frac{1}{(n+1)n}=\frac{1}{n}-\frac{1}{(n+1)}$, we have obtained $\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+1)n}=1$ .

Olivier Oloa
  • 120,989