# Integration by parts, the Möbius function and the Apéry's contant

I believe that by integration by parts for $\int_0^1u\cdot dv$, see the dot here to define the formula (after we sum for a sufficently large integer) $$\sum_{k=3}^\infty\int_0^1\frac{d}{dx}x^{\mu(k)}\cdot\frac{x^{k-1}}{k^2}dx=\frac{1}{\zeta(3)}-(1-\frac{1}{2^3})-2\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{1}{k^3(k-2)},$$ where $\mu(k)$ is the Möbius function.

And in the other hand with a new integration by parts defined as $\int_0^1u\cdot dv$, see the dot here to define the formula $$\sum_{k=3}^\infty\int_0^1\frac{x^{k-1}}{k^2}\cdot\frac{d}{dx}x^{\mu(k)}dx= (\frac{\pi^2}{6}-1-\frac{1}{4})-\sum_{\substack{k\geq 3},\mu(k)=0}\frac{1}{k^2}$$

$$\qquad\qquad\qquad\qquad-\sum_{\substack{k\geq 3},\mu(k)=1}\frac{k+1}{k^3}-\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{k-1}{k^2(k-2)}.$$ Then notice that if there are no mistakes it is possible to get $\frac{1}{\zeta(3)}$ as function of $\sum_{\substack{k\geq 3},\mu(k)=0}\frac{1}{k^2}$, since the term $$-\left(\sum_{\substack{k\geq 3},\mu(k)=1}\frac{k+1}{k^3}-\sum_{\substack{k\geq 3},\mu(k)=-1}\frac{k+1}{k^3}\right)=$$ is equals to $$-\left(\frac{1}{\zeta(2)}-(1-\frac{1}{2^2})+\frac{1}{\zeta(3)}-(1-\frac{1}{2^3})\right).$$

Question. Please can you find where was the first of my mistakes? I say this because the computations were tedious. If it is possible/feasible, can you obtain an identity with mathematical meaning or the final identity that I've presumed (I say if you can get a simple identity involving the $\frac{1}{\zeta(3)}$ from this approach)? Thanks in advance.

I've tried different integrands in integrals as $$\int_0^1\frac{d}{dx}x^{\mu(k)}\cdot d\left(\frac{x^{k-1}}{k^2}\right),$$ defined also for $k$ sufficiently large (or leaving easy calculations for the corresponding tail), say us $k\geq 4$.

• Lemma. One has that for each $k\geq 1$ i) $\Big(\frac{d}{dx}x^{\mu(k)}\Big|_{x=1}=\mu(k)$, ii) $\frac{d}{dx}x^{\mu(k)} = \begin{cases} 0, & \text{if$\mu(k)=0$} \\ 1, & \text{if$\mu(k)=1$}\\ -1/x^{2}, & \text{if$\mu(k)=-1$} \end{cases},$ and iii) $\frac{d}{dx} \left( \frac{d}{dx}x^{\mu(k)} \right) = \begin{cases} 0, & \text{if$\mu(k)=0$} \\ 0, & \text{if$\mu(k)=1$}\\ 2/x^3, & \text{if$\mu(k)=-1$} \end{cases}.$ – user243301 Jul 5 '16 at 23:02
• I noticed that you ask a lot of questions about Apéry's constant. If you don't mind me asking, is your goal to find the exact sum ("closed form") of $\sum_{k=1}^\infty \frac{1}{k^3}$? I ask because that is one of my major goals in mathematics. – zerosofthezeta Jul 7 '16 at 6:01
• I do not care to answer it, my situation in my real life is not the best , and only with very motivating problems can encourage to me to study more mathematics. The best answer should be what I follow the advice of one of the users here, when they said to me to take a book, for example of complex analysis and read it. I lose a lot of the time, and additionaly if in next years I have a collection of failed attempts, and little knowledge in mathematics, perhaps I have not made ​​the best decisions to improve and move forward. Thanks @zerosofthezeta – user243301 Jul 7 '16 at 6:32