1

Problem. Define $I_n=\int_0^\infty\frac{\operatorname du}{(1+u^2)^n}$. Prove that $(2n-1)I_n=2nI_{n+1}$.

I tried to integrate by parts two times (the second time in the "$\cdots$") \begin{align*} I_n&=\int_0^\infty\frac1{(1+u^2)^n}\cdot1\operatorname du=\left[\frac1{(1+u^2)^n}\cdot u\right]_0^\infty-\int_0^\infty\frac{-2nu}{(1+u^2)^{n+1}}\cdot u\operatorname du\\ &=0+\int_0^\infty\frac{2nu^2}{(1+u^2)^{n+1}}\operatorname du=n\int_0^\infty\frac{2u}{(1+u^2)^2}\cdot\frac u{(1+u^2)^{n-1}}\operatorname du\\ &=\cdots=nI_n-n(n-1)\int_0^\infty\frac{2u^2}{(1+u^2)^{n+1}}\operatorname du. \end{align*} The second time wasn't really meaningful as it just brought be to where I had been.

Du Doudou
  • 181

1 Answers1

3

Integration by parts: $$\begin{aligned}I_n&=\int_0^\infty\frac{\mathrm du}{(1+u^2)^n}\\&=\left.\frac{u}{(1+u^2)^n}\right|_{u=0}^\infty+2n\int_0^\infty\frac{u^2}{(1+u^2)^{n+1}}\,\mathrm du\\ &=2n\int_0^\infty\frac{\mathrm du}{(1+u^2)^n}-2n\int_0^\infty\frac{\mathrm du}{(1+u^2)^{n+1}}\\I_n&=2nI_n-2n I_{n+1}\\\color{red}{\implies(2n-1)I_n}&\color{red}{=2nI_{n+1}\qquad\text{proved.}}\end{aligned} $$

Note: The integral $I_n$ converges only for $2n>1$ or $n>\frac12$, i.e., $\frac1n<2$, and thus $\displaystyle\lim_{u\to\infty}\frac{u}{(1+u^2)^n}=\lim_{u\to\infty}\left(\frac{u^{\frac1n-2}}{1+u^{-2}}\right)^n=0$. This limit is $\infty$ for $n<\frac12$, and $1$ for $n=\frac12$. The integral does not converge for $n\leq\frac12$.

Pustam Raut
  • 2,490