# How can I calculate $\displaystyle\int_0^\infty\frac{r^{n-1}}{(1+r^2)^\frac{n+1}{2}}\,dr$

How can I calculate: $$\displaystyle\int_0^\infty \frac{r^{n-1}}{(1+r^2)^\frac{n+1}{2}}\,dr$$ I encountered this integral in proving that a function is an approximation to the identity. But I don't know how to solve this integral. I would greatly appreciate any help.

• If $n$ is odd, we could let $u=1+r^2$ and apply binomial expansion. – Simply Beautiful Art Mar 3 '17 at 0:34
• $\, n \gt 0 \,$. – Hazem Orabi Mar 3 '17 at 0:49


Let $u=r^2$, so $du/u = 2dr/r$, and the integral becomes $$I = \frac{1}{2}\int_0^{\infty} \frac{u^{n/2-1}}{(1+u)^{(n+1)/2}} \, du.$$ Now, this is an integral that can be written in terms of the Beta-function: in particular the transformation $u=1/y-1$ turns it into $$\frac{1}{2} \int_0^1 (1-y)^{n/2-1} y^{-1/2} \, dy = \frac{1}{2}B(1/2,n/2) = \frac{\sqrt{\pi}\Gamma(n/2)}{2\Gamma((n+1)/2)}.$$

• All too easy. (+1) – Mark Viola Mar 3 '17 at 0:51

Let $x=r^2>0$ then $x^{\frac{n-2}{2}}=r^{n-2}$ and we have $dx=2rdr$ so that we have,

$$\frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{n-2}{2}}}{(1+x)^{\frac{n+1}{2}}} dx$$

Now use the Beta Function properties. See this.

As shown we have

$$I_n =\displaystyle\int_0^\infty \frac{r^{n-1}}{(1+r^2)^\frac{n+1}{2}}\,dr= \frac{\sqrt{\pi}\Gamma(n/2)}{2\Gamma((n+1)/2)}$$

Now use the formula

$$\Gamma \left(n+\frac{1}{2}\right) = \frac{(2n-1)!! \sqrt{\pi}}{2^n}$$

For even integers

$$I_{2k} = \frac{ (k-1)!}{(2k-1)!!}2^{k-1}$$

and for odd integers

$$I_{2k+1} =\frac{(2k-1)!!}{k!}\frac{\pi}{2^{k+1}}$$

where the double factorial

$$n!!= \begin{cases} n(n-2)\cdots 5\times 3 \times 1 & n \in \mathbb{2N}+1 \\ n(n-2)\cdots 6\times 4 \times 2 & n \in \mathbb{2N} \\ 1 & n \in \{-1,0\} \end{cases}$$