Find $a$ and $b$ such that the improper integral converges

Question

Find $a$ and $b$ so that the integral $$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$ converges.

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I know that $\int_1^\infty \frac{1}{x^p}$ converges if $p>1$ and I've tried integration by parts but neither has helped. Can someone give me a hint as to how I should approach this?

Answer 1

We have that

$$\int_1^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx+\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}\, dx$$

and

$$\int_1^2 \frac{x}{(x-1)^a(1+x^2)^b}\, dx=\int_0^1 \frac{y+1}{y^a(y^2+2y+2)^b}\, dy$$

which converges for $a<1$ since as $y \to 0^+$

$$ \frac{y+1}{y^a(y^2+2y+2)^b} \sim\frac1{y^{a}}$$

and

$$\int_2^\infty \frac{x}{(x-1)^a(1+x^2)^b}dx, \quad \frac{x}{(x-1)^a(1+x^2)^b}\sim\frac1{x^{a+2b-1}} \quad x\to \infty$$

converges for $a+2b-1>1 \iff b>1-\frac a 2$.

Therefore the given integral converges for $a<1$ and $b>1-\frac a 2$.