Proving that $\int_{0}^{1}\left(\zeta(t)+\frac{1}{1-t}\right)dt=\sum_{n=0}^{\infty}\frac{\gamma_{n}}{(n+1)!}$ [closed]

Question

It seems that the above identity is true. Can this be proven? Or are there references treating sums like the right hand side?

The above constants, $\gamma_{n}$, are the Stieltjes constants.

Thanks.

Answer 1

With a quick search in wikipedia: $$ \zeta \left( 1+z \right) =\frac{1}{z}+\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{n!}\gamma _nz^n} $$ So \begin{align*} \int_0^1{\left( \zeta \left( t \right) +\frac{1}{1-t} \right) \mathrm{d}t}=&\int_0^1{\left( \zeta \left( 1+\left( t-1 \right) \right) -\frac{1}{t-1} \right) \mathrm{d}t} \\ =&\int_0^1{\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n}\mathrm{d}t} \\ =&\sum_{n=0}^{\infty}{\frac{\gamma _n}{n!}}\int_0^1{\left( 1-t \right) ^n\mathrm{d}t}=\sum_{n=0}^{\infty}{\frac{\gamma _n}{\left( n+1 \right) !}} \end{align*} The only question now is why we can interchange the sum with the integral. From the same article, we have $$ \left| \gamma _n \right|\leqslant \frac{n!}{2^{n+1}} $$ Therefore when $t\in [0,1]$ \begin{align*} \left| \sum_{n=0}^N{\frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n} \right|&\leqslant \sum_{n=0}^N{\left| \frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n \right|}\leqslant \sum_{n=0}^N{\frac{\left| \gamma _n \right|}{n!}}\left( 1-t \right) ^n \\ &\leqslant \frac{1}{2}\sum_{n=0}^N{\left( \frac{1-t}{2} \right) ^n}\leqslant \frac{1}{1+t}\leqslant 1 \end{align*} By dominated convergence theorem: \begin{align*} \int_0^1{\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n}\mathrm{d}t}=&\int_0^1{\lim_{N\rightarrow \infty} \sum_{n=0}^N{\frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n}\mathrm{d}t} \\ =&\lim_{N\rightarrow \infty} \int_0^1{\sum_{n=0}^N{\frac{\left( -1 \right) ^n}{n!}\gamma _n\left( t-1 \right) ^n}\mathrm{d}t} \\ =&\lim_{N\rightarrow \infty} \sum_{n=0}^N{\frac{1}{n!}\gamma _n\int_0^1{\left( 1-t \right) ^n}}\mathrm{d}t \\ =&\sum_{n=0}^{\infty}{\frac{\gamma _n}{n!}}\int_0^1{\left( 1-t \right) ^n\mathrm{d}t} \end{align*}