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I'm having trouble with the following integral

$$\int_0^{\infty} {x^2 \over (x^2+A)^n} \ dx \ \ \text{for} \ \ A \gt 0 $$


I had previously thought that the answer to the integral above was:

$$ {2\over(n-1)(n-2)(n-3)A^{n-3}}$$

But actually this is the answer to:

$$\int_0^{\infty} {x^5 \over (x^2+A)^n} \ dx \ \ \text{for} \ \ A \gt 0 $$

which is much easier to integrate (this is one book misprint I'll never forget). BD answer below made me see that (he tested it numerically).

Still, since I spent 4 days trying to solve the wrong integral, it is a relief to see a answer to that and to see how complicated it is. This is a sneaky integral right there, seems to be easy until you get down to business. Too bad I have to choose one answer, there are many good ones.

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Any idea where the $m$ came from if the integral only had variables $x$, $A$, and $n$? Or was $m$ a substitution for other variables in the original integral? – Envious Page Sep 8 '12 at 4:42
If you can compute the integralnwith A=1, then you can compute the others by a simple substitution. – Baby Dragon Sep 8 '12 at 5:01
I would suggest a trig substitution, $x=tan (t)$. – Baby Dragon Sep 8 '12 at 5:09
Sorry! It was really n in the answer! Edited to reflect that. – Forever_a_Newcomer Sep 8 '12 at 14:13
up vote 1 down vote accepted

This is intended to be a comment to André's answer (but I haven't got enough reputation!).

Speaking about $\displaystyle I_m = \int \frac1{(1+t^2)^m}dt$, I remember it was the hardest case to consider when integrating rational functions. On that occasion, I learned about a particular recursive formula, that I found useful: $$\displaystyle I_m = \frac1{2(m-1)}\left[(2m-3)I_{m-1} + \frac{t}{(1+t^2)^{m-1}}\right]$$

It goes under the name of Ostrogradsky and you can also find it in the Wiki page linked by André.

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Let $x=\sqrt{A}\,t$. Then $dx=\sqrt{A}\,dt$ and $x^2=At^2$. After we do the substitution, we end up wanting something like $$\int_0^\infty \frac{A^{3/2}}{A^n}\frac{t^2\,dt}{(1+t^2)^n}.$$ There is no sense in carrying the constant around. Note also that $$\frac{t^2}{(1+t^2)^n}=\frac{1+t^2}{(1+t^2)^n}-\frac{1}{(1+t^2)^{n}}=\frac{1}{(1+t^2)^{n-1}}-\frac{1}{(1+t^2)^n}.$$

So we will be through if we can handle $$\int_0^\infty \frac{dt}{(1+t^2)^m}.$$ There are various ways to do it. For example, let $t=\tan\theta$. Then $dt=\sec^2\theta\,d\theta$, and we arrive at $$\int_0^{\pi/2} \cos^{2m-2} \theta \,d\theta.$$ This can be done by using a Reduction Formula that we get by integrating by parts.
Indeed we could have made the trigonometric substitution already at the $\frac{t^2}{(1+t^2)^n}$ stage. We can also develop a reduction formula directly, without going through the trigonometric substitution.

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Make the change of variables $x=\sqrt{A \frac{u}{1-u}}$ for $0<u<1$. This reduces the integral to the Euler beta-integral: $$ \int_0^\infty \frac{x^2}{(x^2+A)^n}\mathrm{d}x = \frac{a^{3-2n}}{2}\int_0^1 u^{1/2} (1-u)^{n-5/2} \mathrm{d}u = \frac{a^{3-2n}}{2} \operatorname{B}\left(\frac{3}{2},n-\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4} a^{3-2n} \frac{\Gamma\left(n-\frac{3}{2}\right)}{\Gamma(n)} $$ where $\operatorname{B}(a,b)$ is the Beta-function.

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Residue Theory $$I=\int_0^{\infty} {x^2 \over (x^2+A)^n} \ dx \hspace{10mm} \text{suppose : } A=a^2 $$ $$ I= \frac{1}{2} \int_{-\infty}^{\infty} {x^2 \over (x^2+a^2)^n} \ dx=\frac{1}{2}\oint_{c^+} \frac{z^2}{(z^2+a^2)}dz =\frac{1}{2} \left[ 2 \pi i \sum_{i=1}^n R_i^+ \right]=\frac{1}{2}(2 \pi i. R_1)$$ $$ c^+ : \text{ Upper half plane of complex plane (uhp)} $$ $$ c^- : \text{ Lower half plane of complex plane (lhp)} $$ $$ R_i^+ \rightarrow \text{ Residue of function in singularity ponits ( that placed in uhp)}$$ $$ \text{Singularites : } z^2+a^2=0 \rightarrow \begin{cases} z_1=ia &\mbox{Acceptable (uhp)} \\ z_2=-ia & \mbox{Ineligible (lhp) } \end{cases} \text{poles of order of n}$$ $$ R_1=Residue(\frac{z^2}{(z^2+a^2)},z=ia,\text{pole of order of n}) $$ $$ R_1=\frac{1}{(n-1)!}\lim_{z \rightarrow ia } \left[ \frac{d^{n-1}}{dz^{n-1}}(\frac{(z-ia)^n z^2}{(z-ia)^n (z+ia)^n }) \right]$$ $$ R_1=\frac{1}{(n-1)!}\lim_{z \rightarrow ia } \left[ \frac{d^{n-1}}{dz^{n-1}}(\frac{ z^2}{ (z+ia)^n }) \right] $$ $$ \frac{d^{n}}{dz^{n}}(\frac{ z^2}{ (z+ia)^n })= \frac{ \left[ 2 (iaz)+(n-1) a^2 \right](-1)^{n+1}(n)!}{(z+ia)^{n+2}} $$

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