How can I start this sum

Question

anyone have an idea how to find this sum $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 \binom{x}{n+1} \, dx$$ My trial is to transfer the integral into gamma form then find the the integral but it's very hard also I think that may be induction could be a good idea for this

Answer 1

Using the dominated convergence theorem and the Newton series of the Digamma function we have $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}\int_{0}^{1}\dbinom{x}{n+1}dx=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}\dbinom{x}{n+1}dx$$ $$=\int_{0}^{1}x\left(\psi\left(x+1\right)+\gamma-1\right)dx$$ and the integral can be calculated using integration by parts and the Raabe's formula$$\int_{0}^{1}x\psi\left(x+1\right)dx=-\int_{0}^{1}\log\left(\Gamma\left(x+1\right)\right)dx=1-\frac{1}{2}\log\left(2\pi\right)$$ hence $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n}\int_{0}^{1}\dbinom{x}{n+1}dx=\color{red}{\frac{\gamma+1-\log\left(2\pi\right)}{2}}.$$