A cool problem

28 Mar, 2023

An integral representation of the Stieltjes constants:

We prove the identity

γ_{n} = \int_{1}^{\infty} \frac{{x}}{x^{2}} (n - \log x) \log^{n - 1} x d x .

We have

{x} = x - i for i \leq x < i + 1 .

So,

R H S & = \sum_{i = 1}^{\infty} \int_{i}^{i + 1} \frac{(x - i) (n - \log x) \log^{n - 1} x}{x^{2}} d x .

Now, since

\int \frac{(x - i) (n - \log x) \log^{n - 1} x}{x^{2}} d x = \log^{n} x (- \frac{i}{x} - \frac{\log x}{n + 1} + 1),

we obtain

\int_{i}^{i + 1} \frac{(x - i) (n - \log x) \log^{n - 1} x}{x^{2}} d x = \frac{(i + 1) \log^{n + 1} i + (- (i + 1) \log (i + 1) + n + 1) \log^{n} (i + 1)}{(i + 1) (n + 1)} .

Summing from $i = 1$ to $T$ , we get

R H S = {lim}_{T \to \infty} (\frac{(n - (T + 1) \log (T + 1) + 1) \log^{n} (T + 1)}{(n + 1) (T + 1)} + \sum_{i = 2}^{T} \frac{\log^{n} i}{i}) .

Moreover,

\frac{\log^{n} (T + 1) (n - (T + 1) \log (T + 1) + 1)}{(n + 1) (T + 1)} ~ - \frac{\log^{n + 1} T}{n + 1} .

Therefore,

R H S = {lim}_{T \to \infty} (\sum_{i = 1}^{T} \frac{\log^{n} i}{i} - \frac{\log^{n + 1} T}{n + 1}) .

By the defining limit for the $n$ -th Stieltjes constant,

γ_{n} = \int_{1}^{\infty} \frac{{x}}{x^{2}} (n - \log x) \log^{n - 1} x d x .

Fuck you nassim.