A Cool Problem

Challenging? Not really.

prove the identity:

\(\gamma_n = \displaystyle\int_1^\infty \frac{\{x\}}{x^2} (n-\log(x)) \log^{n-1} \, dx \)

We have,

\( \{x\}=x-i \, \text{for} \, i \leq x \leq i+1,\)

so,

\(\text{rhs}=\displaystyle\sum _{i=1}^{\infty } \displaystyle\int_i^{i+1} \frac{(x-i) (n-\log (x)) \log ^{n-1}(x)}{x^2} \, dx, \)

and now, since

\(\displaystyle\int \frac{(x-i) (n-x \log ) \left(x \log ^{n-1}\right)}{x^2} \, dx=\log ^n(x) \left(-\frac{i}{x}-\frac{\log (x)}{n+1}+1\right) \)

\(\displaystyle \int_i^{i+1} \frac{(x-i) (n-\log (x)) \log ^{n-1}(x)}{x^2} \, dx=\frac{(i+1) \log ^{n+1}(i)+(-(i+1) \log (i+1)+n+1) \log ^n(i+1)}{(i+1) (n+1)} \)

we get,

\(\text{rhs}=\underset{T\to \infty }{\text{lim}}\left(\frac{(n-(T+1) \log (T+1)+1) \log ^n(T+1)}{(n+1) (T+1)}+\sum _{i=2}^T \frac{\log ^n(i)}{i}\right) \)

and

\(\frac{\log ^n(T+1) (n-(T+1) \log (T+1)+1)}{(n+1) (T+1)}\overset{T}{\to}\frac{\log ^{n+1}(T)}{n+1} \)

Therefore:

\(\underset{T\to \infty }{\text{lim}}\left(\sum _{i=1}^T \frac{\log ^n i}{i}-\frac{\log ^{n+1}(T)}{n+1}\right)=\gamma _n \)