# Can we determinine the convergence of $\int_0^\infty \frac{x^{2n - 1}}{(x^2 + 1)^{n + 3}}\,dx$ without evaluating it?

Can we determine convergence without evaluating this improper integral? $$\int_0^\infty {\frac{x^{2n - 1}}{{\left( x^2 + 1 \right)}^{n + 3}}\,dx}\quad\quad n\geq 1\;,\; n\in\mathbb{Z}$$

When trying to bound the integrand, relating a known function, the integral does not converge.

But when I use the software package Maple I see that the integral converges to: $$\int_0^\infty {\frac{x^{2n - 1}}{{\left( x^2 + 1 \right)}^{n + 3}}\,dx}=\frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)n}}$$

Evaluating it is not difficult, but I want to know if you can determine convergence (delimiting the integrand) without evaluating such an improper integral.

• Comparison test is the first thing to pop into my mind. – Vim May 15 '15 at 10:04
• Show that $0\leq \frac{x^{2n-1}}{\left(x^2+1\right)^{n+3}}\leq \frac{1}{x^2+1}$ for $x>0$ and all $n\geq 1$ – Lozenges May 15 '15 at 11:44
• The integrand is asymptotically $x^{2n-1-2(n+3)}=x^{-7}$, so convergence is very fast. – Yves Daoust May 15 '15 at 14:10
• @Lozenges I find your suggestion very useful, but this inequality is difficult to prove. Anyway thanks. – mathsalomon May 15 '15 at 19:39
• quite easy to prove by induction on $n$ that $0\leq x^{2n-1}\leq \left(x^2+1\right)^{n+2}$ – Lozenges May 16 '15 at 8:57

By equivalents: $$\frac{x^{2\;n - 1}}{(x^2 + 1)^{n + 3}}\sim_{\infty}\frac{x^{2\;n - 1}}{x^{2(n + 3)}}=\frac1{x^7}$$ and the latter is convergent since the exponent of $x$ in the denominator is $>1$.

• And close to $x=0$, the integrand is $x^{2 n} \left(\frac{1}{x}-(n+3) x+O\left(x^3\right)\right)$ – Claude Leibovici May 15 '15 at 8:59
• The hypothesis is $n\ge1$, so there is no problem of convergence. – Bernard May 15 '15 at 9:06
• I noticed but I prefered to control ! Cheers :-) – Claude Leibovici May 15 '15 at 9:09
• @Bernard Thank you, just do not know how to justify to $x \in [0,1]$. Thanks to Ian comment now know how to justify it. – mathsalomon May 15 '15 at 19:45
• For $x\in [0,1]$, there's nothing to justify since the integrand is continuous on $[0,1]$, so it is an ordinary Riemann integral. – Bernard May 15 '15 at 20:02

The numerator is of degree $2n - 1$. The denominator looks like a polynomial of degree $2n + 6$, and is bounded away from $0$. As a general rule, integrals of the form $$\int_0^\infty \frac{1}{(1 + x)^n}dx$$ converge for $n > 1$ and diverge for $n \leq 1$. Your integral looks like this integral, but with $n = 7$. So it converges. You can make this argument more formal through limit comparison, if that's something you're familiar with.

The naive comparison would simply drop the $1$ in the denominator, but the result is not bounded, whereas your integrand is bounded. To fix this, keep the $1$ on, say, $[0,1]$ and then drop it on $[1,\infty)$. The effect is that on $[0,1]$ the integrand is bounded while on $[1,\infty)$ the integrand is bounded, and in particular bounded by $Cx^{-7}$ for some $C>0$.

• Thanks, that's just what he had considered to do, because of the divergence of $x^{-7}$ in $[0,1]$. – mathsalomon May 15 '15 at 19:31

Setting $x=\tan(t)$, we obtain $$\dfrac{x^{2n-1}}{\left(x^2+1\right)^{n+3}} = \dfrac{\tan^{2n-1}(t)}{\sec^{2n+6}(t)} = \sin^{2n-1}(t)\cos^7(t)$$ Hence, we have $$\int_0^{\infty}\dfrac{x^{2n-1}}{\left(x^2+1\right)^{n+3}} dx = \int_0^{\pi/2} \sin^{2n-1}(t)\cos^5(t)dt$$ which clearly exists, since the integrand is continuous and bounded on $[0,\pi/2]$.

Also, what better way to prove convergence than evaluating the integral. The integral can be easily evaluated using the identity here and the definition of the $\beta$ function. We have $$\beta(x,y) = 2 \int_0^{\pi/2}\sin^{2x-1}(t)\cos^{2y-1}(t)dt$$ which gives us that $$\int_0^{\pi/2} \sin^{2n-1}(t)\cos^5(t)dt = \dfrac{\beta(n,4)}2 = \dfrac{\Gamma(n)\Gamma(3)}{2\Gamma(n+3)} = \dfrac1{(n+2)(n+1)n}$$

• Did you manage the $dx$? (I think you should have another $\sec^2(t)$.) – Ian May 15 '15 at 14:23
• @Ian Thanks. corrected. – Leg May 15 '15 at 14:28