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Integrate using differentiation with respect to parameter only: $$ I=\int_0^\infty \frac{1}{(x^2+p)^{n+1}}dx, \ n\geq 0, \ p\geq1 $$ No complex methods allowed. This is a rather useful integral to know...By differentiation w.r.t parameter I mean writing something like $$ I(\alpha)=\int_0^\infty f(\alpha,x) dx,\ \to I'(\alpha)=\int_0^\infty \partial_\alpha f(\alpha,x) dx=\frac{d}{d\alpha}\int_0^\infty f(\alpha,x) dx. $$ These integrals are mostly taken from bulgarian math team questions. Thanks again for the assistance and your time and help!

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  • $\begingroup$ Differentiate with respect to $p$ and you will end up with a nice integral. $\endgroup$ – Mhenni Benghorbal Feb 27 '14 at 3:29
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Hint: Let $I(p)=\displaystyle\int_0^\infty\frac{dx}{x^2+p}$ , and then try to express $\displaystyle\int_0^\infty\frac{dx}{(x^2+p)^{n+1}}$ in terms of $I^{(n)}(p)=$ $=(-1)^n\dfrac{(2n-1)!!}{2^{n+1}p^n\sqrt p}\cdot\pi$.

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