Condition for $\int_{1}^{+\infty} x^a(x-1)^b dx$ be convergent?

Question

I did some full tests $\int_{1}^{+\infty} x^a(x-1)^b dx$ and apparently it is always convergent if $a<-1$, $b>-1$ and $a+b<-1$. Intuitively I think that b must be greater than $-1$ because close to $1$, integrating only the term $(x-1)^b$ and applying it to $1$, we get $0$. As well as $a<-1$, for the term $x^a$, when integrated and applied at infinity results in $0$. The sum $a+b<-1$ so that the product $x^a(x-1)^b$ applied at infinity results in zero. Is this conjecture correct?

Answer 1

There are two aspects that would make the integral diverge:

• potential singularity at $x=1$. This is integrable iff $b > -1$.
• the integrand decaying more slowly as $1/x$ as $x \to \infty$, i.e. like $x^{\alpha}$ with $\alpha \ge 1$. We have (in the limit $x\to\infty$) that $$x^a(x-1)^b\sim x^{a+b},$$ which will be integrable on $[1,\infty)$ iff $a+b < -1$.

So your conditions for convergence are: $b > -1$ and $a+b < -1$. In particular, this implies $a < 0$.