Recurrence relation for $\int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$

Question

Let$$I_{n,a} = \int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$$ where $a>0$. Show that $$I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$

I have tried integrating by parts but it didn't work for me, and I don't know what else can I try. Can anyone please help?

Answer 1

$$I_n=\int_{0}^{\infty} \frac{1+x^2/a}{(1+x^2/a)^{n+1}}dx=\underbrace{\int_{0}^{\infty} \frac{1}{(1+x^2/a)^{n+1}}dx}_{=I_{n+1}}+\frac12\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx$$ Now since: $$\int \frac{2x/a}{(1+x^2/a)^{n+1}}dx=\int \frac{(1+x^2/a)'}{(1+x^2/a)^{n+1}}dx=-\frac{1}{n} \frac{1}{(1+x^2/a)^n}+C$$ $$\Rightarrow\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx=\int_0^\infty x\left(-\frac{1}{n}\frac{1}{(1+x^2/a)^{n}}\right)'dx$$ $$=-\underbrace{\frac{x}{n}\left(\frac{1}{\left(1+x^2/a\right)^n}\right)\bigg|_0^\infty}_{=0}+\frac{1}{n}\int_0^\infty \frac{1}{(1+x^2/a)^n}dx=\frac{1}{n} I_n$$ $$\Rightarrow I_n=I_{n+1}+\frac{1}{2n} I_n\Rightarrow I_{n+1}=\frac{2n-1}{2n}I_n$$

Answer 2

Hint:

Calculate the integral $I_{\color{red}n,a}$ by parts, setting \begin{align}&&u&=\frac 1{\biggl(1+\cfrac{x^2}{a}\biggr)^n},\qquad &\mathrm d v&=\mathrm dx,&\qquad&\\ &\text{whence}\qquad&\qquad\mathrm du&=-\frac n{\biggl(1+\cfrac{x^2}{a}\biggr)^{n+1}}\frac{2x}{a},&& v=x,\\ &\text{which yields } &I_{n,a}&=\begin{array}{c|}\dfrac x{\biggl(1+\tfrac{x^2}{a}\biggr)^n}\end{array}_0^\infty+2n\int_0^\infty\rlap{\dfrac{\tfrac{x^2}a}{\biggl(1+\tfrac{x^2}{a}\biggr)^{\!n}}\,\mathrm dx}. \end{align} Can you end the computation?

Answer 3

Just in case you're interested you can solve this for any $n \in \mathbb{R}^+$, $n \geq 1$ using the Beta and by extension the Gamma Function.

Here we will address your integral: \begin{equation} I(a,n) = \int_0^\infty \frac{1}{\left(1 + \frac{x^2}{a}\right)^n}\:dx \end{equation} We begin by making the substitution $x = \sqrt{a}t$: \begin{equation} I(a,n) = \int_0^\infty \frac{1}{\left(1 + \frac{at^2}{a}\right)^n} \cdot \sqrt{a}\:dt = \sqrt{a}\int_0^\infty \frac{1}{\left(1 + t^2\right)^n}\:dt \end{equation} We now let $t = \tan(\theta)$: \begin{align} I(a,n) &= \sqrt{a}\int_0^{\frac{\pi}{2}} \frac{1}{\left(1 + \tan^2(\theta)\right)^n} \cdot \sec^2(\theta)\:d\theta \nonumber \\ &= \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta \end{align} Now from here you could form your recurrence relationship by IBP twice, but alternatively you can call upon one of the common definitions of the Beta Function: \begin{equation} B(a,b) = 2\int_0^{\frac{\pi}{2}} \cos^{2a - 1}(x)\sin^{2b - 1}(x)\:dx \end{equation} For $I(a,n)$ we observe that (1) $2a - 1 = 2n - 2$ and so $a = \frac{2n - 1}{2}$ and (2) $2b - 1 = 0$ and so $b = \frac{1}{2}$.We also observe that we need to scale the integral by $\frac{1}{2}$: \begin{equation} I(a,n) = \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta = \frac{\sqrt{a}}{2}B\left( \frac{2n - 1}{2}, \frac{1}{2} \right) \end{equation} Using the relationship between the Beta and Gamma Function, this reduces to: \begin{equation} I(a,n) = \frac{\sqrt{a}}{2}\cdot \frac{\Gamma\left(\frac{2n - 1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{2n - 1}{2} + \frac{1}{2} \right)} = \frac{\sqrt{a}\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{2n - 1}{2}\right)}{\Gamma(n)} \end{equation}

We can also see that this satisfies the recurrence relationship you sought.