Recursive formula of $I_n=\int\frac{dx}{(x^2+a^2)^n}$

Question

I' m considering the following integral: $$I_n=\int\frac{dx}{(x^2+a^2)^n}$$ How to prove that: $$I_{n+1}=-\frac{1}{2na}\frac{d}{da}I_n$$

Answer 1

$I_n(a)=\displaystyle\int\frac{dx}{(x^2+a^2)^n}$

differentiate wrt $a$;

$I'_n(a)= \displaystyle\int\partial_a\frac{dx}{(x^2+a^2)^n}$

$I'_n(a) = \displaystyle\int\frac{-n\cdot2a}{(x^2+a^2)^{n+1}}\,dx$

$I'_n(a) = -2an\displaystyle\int\frac{1}{(x^2+a^2)^{(n+1)}}\,dx$

$\implies I'_n(a) = -2an \cdot I_{n+1}(a)$

$\implies I_{n+1}= \frac{-1}{2an}\cdot I'_n$

Answer 2

Indeed you can prove it, using Differentiation Under the Integral Sign a very useful "trick"

$$I_{n+1}=-\frac{1}{2na}\cdot\frac{d}{da}\left(\int{\frac{dx}{(x^2+a^2)^n}}\right)$$

$I_{n+1}=-\displaystyle\frac{1}{2na}\int{\frac{\partial}{\partial a}\frac{1}{(x^2+a^2)^n}}\; dx$

$I_{n+1}=-\displaystyle\frac{1}{2na}\int{\frac{-\frac{\partial}{\partial a}(x^2+a^2)^n}{(x^2+a^2)^{2n}}}\; dx$

$I_{n+1}=-\displaystyle\frac{1}{2na}\int{\frac{-n(x^2+a^2)^{n-1}\frac{\partial}{\partial a}(x^2+a^2)}{(x^2+a^2)^{2n}}}\; dx$

$I_{n+1}=-\displaystyle\frac{1}{2na}\int{\frac{-2an(x^2+a^2)^{n-1}}{(x^2+a^2)^{2n}}}\; dx$

$I_{n+1}=\displaystyle\int{\frac{(x^2+a^2)^{n-1}}{(x^2+a^2)^{2n}}}\; dx$

$I_{n+1}=\displaystyle\int{(x^2+a^2)^{(n-1)-2n}}\; dx$

$I_{n+1}=\displaystyle\int{(x^2+a^2)^{-(n+1)}}\; dx$

$$\int{\frac{dx}{(x^2+a^2)^{n+1}}}=\int{\frac{dx}{(x^2+a^2)^{n+1}}}$$ $$\mathbf{\star \ Proved\ \star}$$