How compute $\int_0^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt ?$

Question

Put $$I_n= \int_0^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt \ (n\in \mathbb N).$$ We note that for each fixed $n\in \mathbb N, $ $I_n$ is convergent.

[ For $t\geq 1, \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} \leq \frac{1}{1+t^2}$, therefor $\int_1^{\infty} \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt $ is convergent by comparison test, and near near origin no singularity. So, $I_n$ is convergent for each fixed $n$. ]

Question: Can we compute $I_n$? Is $\{I_n\}_{n\in \mathbb N}$ a bounded sequence in $\mathbb R$?

Answer 1

Here is some place to start.

You can convert to partial fraction. After some work, you should get someting like $$ {\frac { \left( t+n \right) ^{2}{t}^{2}}{ \left( 1+ \left( t+n \right) ^{2} \right) ^{3}}}={\frac {-{n}^{2}-2-2\,tn}{ \left( 1+{t}^{ 2}+2\,tn+{n}^{2} \right) ^{2}}}+ \left( 1+{t}^{2}+2\,tn+{n}^{2} \right) ^{-1}+{\frac {1+{n}^{2}+2\,tn}{ \left( 1+{t}^{2}+2\,tn+{n}^{2 } \right) ^{3}}}. $$

Each of those can be integrated using standard techniques, like trig subs, u-subs, completing the square, etc.

This gives \begin{align*} \int _{0}^{\infty }\!{\frac { \left( t+n \right) ^{2}{t}^{2}}{ \left( 1+ \left( t+n \right) ^{2} \right) ^{3}}}{dt}&=-1/8\,\arctan \left( n \right) {n}^{2}-1/8\,n-3/8\,\arctan \left( n \right) +3/16\,\pi +1/16 \,\pi \,{n}^{2}\\ &= \left( -1/8\,\arctan \left( n \right) +1/16\,\pi \right) {n}^{2}-1/8 \,n-3/8\,\arctan \left( n \right) +3/16\,\pi \end{align*}

Now this appears to be decreasing, has limit $0$ and thus is bounded. I'm lacking time to complete.

Answer 2

$$ I_n = \int_0^\infty \frac{(t+n)^2t^2}{[1+(t+n)^2]^3} dt $$ $$ I_n = \int_0^\infty \frac{n^2+2nt+1}{(n^2+2nt+t^2+1)^3} + \frac{-n^2-2nt-2}{(n^2+2nt+t^2+1)^2} + \frac{1}{n^2+2nt+t^2+1} \, dt$$

using partial fraction expansion and from here you can integrate using $u$-substitutions to find

$$ I_n = \left[\frac{1}{8} \left( (n^2+3)\arctan(n+t) + \frac{n^3+n^2t+3n-5t}{n^2+2nt+t^2+1} - \frac{2(n^3 + n^2t + n -t) }{(n^2+2nt+t^2+1)^2} \right)\right]_{t=0}^{t=\infty}$$

take the limit as $t\to \infty$ and,

$$ I_n = \frac{\pi}{16}(n^2+3) - \frac{1}{8}(n^2+3)\arctan(n) - \frac{n}{8} $$

and $\underset{n\to\infty}{\lim} I_n = 0$ and let $f : \mathbb{R} \to \mathbb{R}, \,n \mapsto \frac{\pi}{16}(n^2+3) - \frac{1}{8}(n^2+3)\arctan(n) - \frac{n}{8}$,

$$ \frac{\partial f}{\partial n} = \frac{-(n^2+3)}{8(n^2+1)} + \frac{\pi n}{8} - \frac{n\arctan(n)}{4} - \frac{1}{8} < 0 \quad\forall n$$

So since $I_n = f \, \forall n\in \mathbb{N}$, $\frac{\partial f}{\partial n} < 0$, and $\underset{n\to\infty}{\lim} I_n = 0$ then $I_n$ has a lower bound of $y=0$.

In addition $ \underset{n\to -\infty}{\lim} I_n \to \infty $, so $I_n$ has no upper bound.