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There is a solution for the integral $\int_0^\infty \frac{1}{(1+x^2)^n}$ for $n<1$?

for case $n> 1$, we need to integrate by parts and use the recurrence relation. In case of $n <1$, I believe that we also need to integrate by parts and deduce a recurrence relationship, however, what observations should be made?

Gary
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megazord
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    The integral converges only for $n> 1/2$. It can be expressed in terms of the beta function, and that representation will hold for all $n>1/2$. – Gary Apr 27 '20 at 16:45
  • @Gary and considering $n> 1/2$ we can use the same equation for the case of $ n> 1$? – megazord Apr 27 '20 at 16:48
  • $n$ usually denotes a natural. Do you really need fractional values ? –  Apr 27 '20 at 17:00

1 Answers1

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There is an explicit form in terms of the $\Gamma$ function for any $n>\frac12$.
By letting $\frac{1}{1+x^2}=u$ we have $x=\frac{\sqrt{1-u}}{\sqrt{u}}$ and $dx=\frac{-1}{2u\sqrt{u(1-u)}}$, so, by Euler's Beta function

$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{(1+x^2)^n} &=& \frac{1}{2}\int_{0}^{1} u^{n-3/2}(1-u)^{-1/2}\,du = \frac{\Gamma(n-1/2)\Gamma(1/2)}{2\Gamma(n)}\\&=&\frac{\sqrt{\pi}\,\Gamma(n-1/2)}{2\,\Gamma(n)}=\frac{2\pi}{4^n}\binom{2n-2}{n-1}.\end{eqnarray*}$$

(Curiously enough, this is exactly $1$ for $n=\frac{3}{2}$)
The integral behaves like $2\sqrt{\frac{\pi}{n}}$ for large values of $n$.

Jack D'Aurizio
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