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.
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.
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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*}