Convergence of series $\sum_{n=0}^{\infty} \dfrac{(-1)^n}{\alpha + \beta n}$ for $\alpha, \beta > 0$

Question

Shos that for $\alpha, \beta > 0$,

$$\int_{0}^{1} \dfrac{x^{\alpha -1}}{1+x^{\beta}} dx = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{\alpha + \beta n}$$

I used the geometric series identity and the Lebesgue Dominated Convergence Theorem and I arrived at

$$ \sum_{n=0}^{\infty} \dfrac{(-1)^n}{\alpha + \beta n} = \dfrac{1}{\alpha} \int_{0}^{1}\dfrac{1}{1+x^{\frac{\beta}{\alpha}}}$$

But from here, I don't know how to proceed or I don't even know if this is useful at all, so any hint would be very appreciated!

Thank you so much!

Answer 1

Hint set $u =x^{\frac1\alpha}$ that is $x=u^{\alpha}$ then you will recover the previous integral since $$dx = \alpha u^{\alpha-1}du $$