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For any positive integer $n$, consider $$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr.$$ I would like to show that it is positive. I try to write it as $$\int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr,$$ but I am not sure that it helps.

EDIT: According to sjasonw, the integral may not be positive as I think. I would be happy to see a proof showing that it's not positive.

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Let $u = r^2 + 1$, so that $r^2 = u - 1$ and $dr = 2rdu$. Suppose $n$ is even. Then the integral is $$ \frac{1}{2} \int_1^\infty \frac{(u-2)(u-1)^{n/2}}{u^{n+3}}du. $$ Let's say $n = 2$. We get $$ \frac{1}{2} \int_1^\infty \frac{u^2 - 3u + 2}{u^{5}}du = \frac{1}{2}\left(\frac{1}{2} - 3\cdot\frac{1}{3} + 2\cdot\frac{1}{4} \right) = 0. $$ – user29743 Dec 22 '12 at 4:15

3 Answers 3

up vote 3 down vote accepted

Let $I$ denote the integral, then we have that $$I = \int_{0}^\infty\frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr = \underbrace{\int_{0}^\infty\frac{r^{n+3}}{(r^2+1)^{n+3}}dr}_J - \underbrace{\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr}_K$$ Consider $J$. Let $r=1/t$, we then get that $$J = \int_{\infty}^0 \dfrac{1/t^{n+3}}{(1/t^2+1)^{n+3}} \left(- \dfrac{dt}{t^2} \right) = \int_{0}^\infty\frac{t^{n+1}}{(t^2+1)^{n+3}}dt = K$$ Hence, $$I=0$$

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That's exactly what I need. Thanks a lot! – Paul Dec 22 '12 at 4:20

If $$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=I$$

$$\int \frac{(r^2-1)r^{n+1}}{(r^2+1)^{n+3}}dr=\int \frac{r^{n+1}}{(r^2+1)^{n+2}}dr-2\int_{0}^\infty\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$

Now $$\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr=\frac1{(r^2+1)^{n+2}}\int r^{n+1} dr-\int\left(\frac{d\frac1{(r^2+1)^{n+2}}}{dr}\int r^{n+1} dr\right)dr$$

$$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}-\int\left( \frac{-(n+2)2r}{(r^2+1)^{n+3}}\cdot\frac{r^{n+2}}{n+2} \right)dr$$ $$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr$$ (assuming $n+2\ne 0$)

So, $$I=\int\frac{r^{n+1}}{(r^2+1)^{n+2}}dr$$ $$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2\int\frac {r^{n+3}}{(r^2+1)^{n+3}}dr-2\int\frac{r^{n+1}}{(r^2+1)^{n+3}}dr$$ $$=\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}+2I$$

So, $$I=-\frac{r^{n+2}}{(n+2)(r^2+1)^{n+2}}$$

Now apply the limit.

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Mathematica gives

$$\int\frac{\left(r^2-1\right)r^{n+1}}{\left(r^2+1\right)^{n+3}}dr=-\frac{r^{2+n} \left(1+r^2\right)^{-2-n}}{2+n}$$

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