Show that this integral is convergent for n in natural numbers:

Question

Is $\int_{0}^{∞} \frac{x^{2n−1}}{(1 + x^2)^{n+3}}dx$ convergent for n in natural numbers? I've already found the indefinite integral which is that but I don't know how to continue.

Answer 1

Hint: observe that for $x \to +\infty$, we have $$\frac{x^{2n-1}}{(1+x^2)^{n+3}} \sim \frac{1}{x^{2n+6 - 2n +1}}=\frac{1}{x^7}$$

Answer 2

$\int_1^{\infty} \frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx \leq \int_1^{\infty} \frac {x^{2n-1}} {(x^{2})^{n+3}}dx=\int_1^{\infty} x^{-7}dx <\infty$. Can you show that $\int_0^{1} \frac {x^{2n-1}} {(1+x^{2})^{n+3}}dx<\infty$?