Find the value of the following integral:
$$ \int_{0}^{\infty} \frac{\ln(1+x)}{x^{1+a}} \, dx, $$
where $0 < a < 1$ is a real number.
I can bound the integral using $\ln(1+x) < x$ for $0 < x < M$ and $\ln(1+x) < x^{a/2}$ for $x>M$, which means that the integral exists and may be calculated by hands.
I think I must interpret the integral in the viewpoint of complex analysis, but I could not handle the log function and $x^{1+a}$, because I am not good at stuffs like branch cuts.
I would be happy if you let me know how to tackle this problem.
The integral in its current form is not easy to deal with as it has two logarithms, one $\log (1+x)$ and the other $x^{1+a} = e^{(1+a)\log x}$. So removing one of them does help. Taking integration by parts,
$$ \int_{0}^{\infty} \frac{\log(1+x)}{x^{a+1}} \, dx = \underbrace{ \left[ -\frac{1}{a}\frac{\log(1+x)}{x^a} \right]_{0}^{\infty} }_{=0} + \frac{1}{a} \int_{0}^{\infty} \frac{dx}{x^a(1+x)}. \tag{*} $$
Now this integral can be attacked by some standard techniques. For instance, you can utilize the keyhole contour. For this technique, it would be good to read this to see how we choose the branch cut of $\log$ and the contour to compute the integral.
So let me leave the rest of computation to you, but I will write a more detailed answer in case you find it hard to apply it to $\text{(*)}$.
(There are solutions that relies on real-analytic methods if you are familiar with some special functions such as the gamma function. But I will skip them too.)