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Could someone please show me how to do the indefinite integral of

$$\frac{1}{(ax^2+bx+c)^n}$$

a) using real analysis (hard)

b) using complex analysis (nice factoring)

and show they give the same answer, without using any simplifiers into $1 + t^2$ or other stuff making it easier (but assumes you remember the answer or what to do).

I keep getting incorrect answers and forgetting what to do :(

bobby
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    Integrating over what? The substitution $t=\tan\theta$ brings $\int_{0}^{+\infty}\frac{dt}{(1+t^2)^n}$ into $\int_{0}^{\pi/2}\cos^{2n-2}(\theta),d\theta$, that is a well-known integral, manageable through real or complex analytic techniques (it leads to Wallis product, for instance). – Jack D'Aurizio May 26 '14 at 12:31
  • Indefinite integrals, but without tricks to turn it into anything involving $1 + t^2$ because you'd have to remember that which is tantamount to just memorizing the answer – bobby May 26 '14 at 12:36
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    I cannot understand - turning $ax^2+bx+c$ into $x^2+1$ by the proper affine map is straightforward, do you prefer to have a baked expression depending on $a,b,c$ (and the domain of integration) and memorize it, but spending two nanoseconds in the determination of the right affine map? – Jack D'Aurizio May 26 '14 at 12:43
  • You can nicely integrate $\frac{1}{\sqrt{ax^2+bx+c}}$ without any of that affine map stuff, you can simply work it out in the form it's given without any little memory tricks - can the same be done for the integral I've given above? – bobby May 26 '14 at 12:47
  • These are not memory tricks, they are basic integration methods. If you don't expect to know the basics, no need to dig deeper. – Jean-Claude Arbaut May 26 '14 at 16:30
  • Seems like a memory trick to me when it can be done without any of those intermediate memory steps – bobby May 27 '14 at 01:05

1 Answers1

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$$\int_{0}^{+\infty}\frac{dt}{(1+t^2)^n}=\int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{2\pi}{4^{n}}\cdot\binom{2n-2}{n-1},\tag{1}$$ where the first identity depends on the substitution $t=\tan\theta$ and the second one follows by integrating by parts. In the parametric case: $$\begin{eqnarray*}\int_{-\infty}^{+\infty}\frac{dx}{(ax^2+bx+c)^n}&=&(4a)^n\int_{-\infty}^{+\infty}\frac{dy}{((2ay+b)^2+(4ac-b^2))^n}\\ &=&(4a)^n\int_{-\infty}^{+\infty}\frac{dz}{((2az)^2+(4ac-b^2))^n}\\ &=& 2^{2n-1}a^{n-1}\int_{-\infty}^{+\infty}\frac{dz}{(z^2+(4ac-b^2))^n}\\ &=& 2^{2n-1}a^{n-1}(4ac-b^2)^{1/2-n}\int_{-\infty}^{+\infty}\frac{dw}{(w^2+1)^n}\\ &=& 2\pi\cdot a^{n-1}(4ac-b^2)^{1/2-n}\cdot\binom{2n-2}{n-1}.\\ \end{eqnarray*}$$ Indefinite integration can be carried over through the following observation: it is easy to integrate $\cos^{2n}(\theta)$ because from the De Moivre identity you know its Fourier series: $$\cos^{2n}(\theta)=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-2k)\theta}=\frac{2}{4^n}\left(\binom{2n}{n}+2\sum_{k=0}^{n-1}\binom{2n}{k}\cos((2n-2k)\theta)\right),$$ $$\int\cos^{2n}(\theta)d\theta = \frac{2}{4^n}\left(\binom{2n}{n}\theta+2\sum_{k=0}^{n-1}\binom{2n}{k}\frac{\sin((2n-2k)\theta)}{2n-2k}\right).$$ To recover the indefinite integral as a function of $t$, it suffices to expand the Chebyshev polynomials: $$\sin((2n-k)\theta) = \sin(\theta)\cdot U_{2n-k-1}(\cos\theta) = \frac{t}{\sqrt{1+t^2}}U_{2n-k-1}\left(\frac{1}{\sqrt{1+t^2}}\right).$$ As an alternative, consider that integration by parts gives: $$ I_n=\int \frac{dt}{(1+t^2)^n} = I_{n-1}-\int t\cdot\frac{t}{(1+t^2)^n} = I_{n-1}+\frac{t}{(2n-2)(1+t^2)^{n-1}}-\frac{I_n}{2n-2},$$ $$I_{n}=\frac{2n-3}{2n-2}I_{n-1}+\frac{1}{2n-2}\cdot\frac{t}{(1+t^2)^{n-1}},$$ hence:

$$ I_n = \frac{(2n-3)!!}{(2n-2)!!}\left(\arctan t + \sum_{k=1}^{n-1}\frac{(2k-2)!!}{(2k-1)!!}\cdot\frac{t}{(1+t^2)^k}\right).$$

Jack D'Aurizio
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  • a) You ignored the fact I was talking about indefinite integrals b) You ignored me twice saying I didn't want any of those simplifications you gave :( – bobby May 26 '14 at 13:04
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    a) For indefinite integrals, just integrate $\cos^{2n-2}(\theta)$ by parts. b) without "those simplifications", there is no way to pull out the $(4ac-b^2)^{1/2-n}$ factor in this specific case, and I'm unable to do mathematics if I cannot use mathematics. It is like asking someone to hug you without using his own arms. – Jack D'Aurizio May 26 '14 at 13:14
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    Always about a), it is easy to integrate $\cos^{2n}(\theta)$ because from the De Moivre identity you know its Fourier series: $$\cos^{2n}(\theta)=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}e^{i(n-2k)\theta}=\frac{2}{4^n}\cdot\left(\binom{2n}{n}+2\sum_{k=0}^{n-1}\cos((n-2k)\theta)\right).$$ – Jack D'Aurizio May 26 '14 at 13:22