Can be justified this $\zeta(3)=\int_0^1\frac{1}{x}\sum_{n=0}^\infty\frac{x^{(n+1)^{3/2}}}{(n+1)^{3/2}}dx$?

Question

It is well known the unsolved problem concerning to Àpery constant $\zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}$ (see this site or [1] for example). The following computations will be viewed with the purpose to find what steps could be wrongs, or what improvements we can made (if you, optionally, answer b) of my question), well with the goal to refresh my mathematics with interesting topics and easy facts. Thus could have mistakes but too I have the opportunity to learn.

We have the following

Fact. For every integer $n\geq 0$, $(n+1)^3=(n+1)^{3/2}\cdot((n+1)^{3/2}-1+1)$.

Thus by Tonelli (if it is justified) $$\int_0^1\sum_{n=0}^\infty\frac{x^{(n+1)^{3/2}-1}}{(n+1)^{3/2}}dx=\sum_{n=0}^\infty\frac{1}{(n+1)^{3/2}}\cdot \Big(\frac{x^{(n+1)^{3/2}}}{(n+1)^{3/2}}\Big|_0^1=\zeta(3),$$ and there is convergence of this function series to claim (if it isn't wrong) that $$LHS=\int_0^1\frac{1}{x}\sum_{n=0}^{\infty}\frac{x^{(n+1)^{3/2}}}{(n+1)^{3/2}}dx.$$

Since these computations are very easy to write, I suspect that or there isn't convergence, or aren't useful because it is difficult to compute $f(x)=\sum_{n=0}^\infty\frac{x^{(n+1)^{3/2}}}{(n+1)^{3/2}}$, for $0\leq x\leq 1$.

My question is ( I choose the answer more useful to this community for a) of this)

Question. a) A proof verification of previous computations and claim is required and appreciated (is valid the use of Tonelli's theorem?, is there convergence for $f(x)$?). b) Optionally are welcome suggestions, references if this exercise is easy and known, or if you can claim properties of $f(x)$. Then I learn from your computations. Thanks in advance.

**I am looking an useful answer for a) of previous questions. If it is possible compute LHS (if you want leave some useful comment to b), we learn from you). I would like an answer with details of the use of Tonelli's theorem, what hypothesis you need to prove and if it is possible statement's of Tonelli that you are using. Too details about the convergence, to claim $LHS$ as I've written and $f(x)$, if it is rigth.

References:

[1] Wikipedia, Àpery constant https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant

Answer 1

All we need to successfully apply Tonelli's theorem is that $f_n(x) = \frac{x^{n^{3/2}-1}}{n^{3/2}} \geq 0$ for all $x\in[0,1]$ plus we need $f_n(x)$ to be integrable. This is indeed the case here so your calculations are justified and you are allowed to write

$$\int_0^1\sum_{n=1}^\infty f_n(x){\rm d}x = \sum_{n=1}^\infty\int_0^1 f_n(x){\rm d}x = \zeta(3)$$

For a good discussion on when you can interchange sums and integrals see this MSE answer.

The uniform convergence of the series $f(x) = \sum_{n=1}^\infty \frac{x^{3/2}}{n^{3/2}}$ on $[0,1]$ follows by Weierstass M-test since $f_n(x) \leq \frac{1}{n^{3/2}}$ for all $x\in[0,1]$ and the series $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$ converges (by for example the integral test)

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I don't know how to answer your other question, but it's hard to see a reason for why studying $f(x) = \sum_{n=1}^\infty \frac{x^{n^{3/2}-1}}{n^{3/2}}$ should help you understand any properties (e.g. irrationality; trancendental nature; or just estimating the sum) of $\zeta(3)$ better than simply working with some of the other well known representations like $\zeta(3) = \sum_{n=1}^\infty\frac{1}{n^3}$ or $\zeta(3) = \frac{5}{2}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3{2n\choose n}}$ as is used in Apéry's proof.