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Through some work, I've managed to solve the following sum in the form of integrals:

$$\sum_{n=1}^x\frac{r^n}{n^k}=\int_0^r\frac1{a_{k-1}}\int_0^{a_{k-1}}\frac1{a_{k-2}}\int_0^{a_{k-2}}\dots\int_0^{a_1}\frac{t^x-1}{t-1}dtda_1da_2da_3\dots da_{k-1}$$

And as such, I can calculate the sum for arbitrary values of $x$.

More interestingly, if we have $x\to\infty$ and $r=\pm1$, we result with either the Riemann Zeta function or the Dirichlet Eta function for positive whole values of $k$.

I start by assuming $t^x\to0$ as $x\to\infty$, then proceed to solve.

$$\zeta(1)=\int_0^1\frac1{1-t}dt\to\infty$$

$$\eta(1)=\int_0^{-1}\frac1{1-t}dt=\ln(2)$$

Well known results to start with, and it appears to pass for other values of $k$.

However, I have to wonder if it is possible to make this work for $k\notin\mathbb N$. Naturally, I doubt it would work for $k<1$, since the limit shouldn't converge... but for $k>1$, it doesn't seem to make much sense to do a fractional amount of integrals.

On top of that, I can't tackle this problem like a normal fractional calculus problem and simply use Cauchy's repeated integral formula, since the integral is in the form

$$\int_0^r\int_0^{a_{k-1}}\int_0^{a_{k-2}}\dots \int_0^{a_1}f(t,a_1,a_2,a_3,\dots a_{k-1})dt\dots d_{k-1}$$

and each variable is dependent on the previous.

Perhaps there is a multivariable version of Cauchy's repeated integral I do not know of?

What can we do to generalize this for arbitrary $k$?

Simply Beautiful Art
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1 Answers1

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You multiple-integrals approach works, but there is a more compact one, through:

$$ \frac{1}{n^k}=\frac{1}{(k-1)!}\int_{0}^{1}t^{n-1}\left(-\log t\right)^{k-1}\,dt \tag{1}$$ that leads to: $$ (k-1)!\sum_{n=1}^{N}\frac{r^n}{n^k} = \int_{0}^{1}\sum_{n=1}^{N} r^n t^{n-1}\left(-\log t\right)^{k-1}\,dt=\color{red}{r\int_{0}^{1}\frac{1-(rt)^N}{1-rt}(-\log t)^{k-1}\,dt}\tag{2} $$ and the RHS of $(2)$ boils down to a combination of dilogarithms by integration by parts.

Jack D'Aurizio
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  • Wait, where did $(1)$ come from? And I'm not going to accept this answer because I want to see if there is anything doable with my integrals. Thanks for the answer. – Simply Beautiful Art Jun 26 '16 at 15:00
  • @SimpleArt: $(1)$ is well-known and straightforward to prove. It is just the integral definition of the $\Gamma$ function in disguise: set $t=e^{-u}$. – Jack D'Aurizio Jun 26 '16 at 15:07
  • Anyway, $(2)$ leads to dilogarithms in a straightforward way. Multiple integrals have to give the same outcome, and we are just skipping repeated integration. – Jack D'Aurizio Jun 26 '16 at 15:09
  • Ah, much complicated way of writing it. I haven't practiced my maths in about a week, so I'm slower than I should be. And yes, wolfram alpha gave me the same logarithm stuff. – Simply Beautiful Art Jun 26 '16 at 15:09
  • Oh, I guess my formula is the partial sum of the definition of the polylogarithm according to the Wikipedia. I'll read up on it. – Simply Beautiful Art Jun 26 '16 at 15:12