<s>[INST] 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
 [/INST] Let $I_k=\int_{-\infty}^\infty\frac{dx}{(x^2+1)^{k+1}}$. An easy integration shows that $I_1=\frac{\pi}{2}$. 
Evaluate $I_k$ using integration by parts, letting $du=dx$ and $v=\frac{1}{(x^2+1)^{k+1}}$. We get $u=x$ and $dv=-(k+1)(2x)\frac{1}{(x^2+1)^{k+2}}$. Thus
$$I_k=\left.\frac{x}{(x^2+1)^{k+1}}\right|_{-\infty}^\infty+\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx=\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx.$$
Rewrite the top of the last integrand as $2(k+1)(x^2+1-1)$. Then we obtain
$$I_k=2(k+1)I_k -2(k+1)I_{k+1}.$$
We can rewrite this as
$$I_{k+1}=\frac{2k+1}{2(k+1)}I_k.$$
Now the only thing that remains is to show that if $A_k=\frac{(2k)!\pi}{2^{2k}(k!)^2}$ then $A_k$ satisfies the same initial condition as $I_k$ (easy) and the same recurrence relation (straightforward). Since $I_1=A_1$ and $I_k$ and $A_k$ satisfy the same recurrence, the two sequences must be the same.
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