integration - show that $int_{-infty}^{+infty} frac{dx}{(x^2+1)^{n+1}}=frac {(2n)!pi}{2^{2n}(n!)^2}$

show that:

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

where $n=0,1,2,3,\ldots$.

is there any help?

thanks for all

Answer

Write $${\vartheta _n} = \int_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {1 + {x^2}} \right)}^n}}}} \frac{{dx}}{{1 + {x^2}}}$$

Put $x=\tan\vartheta$. Then $${\vartheta _n} = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\cos }^{2n}}\vartheta } d\vartheta$$

so $${\vartheta _n} = 2\int_0^{\frac{\pi }{2}} {{{\cos }^{2n}}\vartheta } d\vartheta$$

We can come up with a recursion for $\vartheta_n$ using integration by parts, namely $${\vartheta _n} = \frac{{2n - 1}}{{2n}}{\vartheta _{n - 1}}$$

This means that $$\prod\limits_{k = 1}^n {\frac{{{\vartheta _k}}}{{{\vartheta _{k - 1}}}}} = \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}}$$

so by telescopy $$\frac{{{\vartheta _n}}}{{{\vartheta _0}}} = \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}}$$ but ${\vartheta _0} = \pi$ so $$\begin{align} {\vartheta _n} &= \pi \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} \cr &= \pi \prod\limits_{k = 1}^n {\frac{{2k - 1}}{{2k}}} \frac{{2k}}{{2k}} \cr &= \pi \frac{{\left( {2n} \right)!}}{{{2^{2n}}n{!^2}}}=\frac{\pi}{4^n}\binom{2n}{n} \end{align}$$

as desired.