For $\alpha<0,$ $0\le \int_1^\infty \frac{x^\alpha}{1+x}dx\le \int_1^\infty x^{\alpha-1}dx$, and $\int_1^\infty x^{\alpha-1}dx$ converges. Thus, by comparison test, $\int_1^\infty \frac{x^\alpha}{1+x}dx$ converges.
I'd like to know if it is possible to calculate the value of $\int_1^\infty \frac{x^\alpha}{1+x}dx$ for $\alpha <0$.
Please let me know if you have any idea or comment for it.
Thanks in advance.
Expanding as a power series in $1/x$, $$ \frac{x^{\alpha}}{1+x} = \sum_{n=0}^{\infty} (-1)^{n} x^{a-n-1}, $$ and interchanging the order of integration and summation then gives $$ \int_1^{\infty} \frac{x^{\alpha}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \int_1^{\infty} x^{a-n-1} \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{n-a}. $$ This sum can be written in terms of the digamma function by splitting it into odd and even terms, $$ \sum_{n=0}^{\infty} \frac{(-1)^n}{n-a} = \sum_{n=0}^{\infty} \frac{1}{2n-a} - \frac{1}{2n+1-a} = \frac{1}{2} \left(\psi\left(\frac{1-a}{2}\right)+\psi\left(-\frac{a}{2}\right)\right). $$ This in general will not simplify any further, although some special values are probably known.