1

How can I integrate:

$$\int \frac{1}{(t^2+k^2)^m}\, dt$$

without trygonometric substituition? where $t= (x+(p/2))$ and $k= (1-(p^2/4))$ coming from an equation with complex roots: $x^2 + px + q.$

Thank You.

Jack D'Aurizio
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Yassin Rany
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  • Are you really interested in the explicit primitive or the integral over $\mathbb{R}$/$\mathbb{R}^+$ is enough? – Jack D'Aurizio May 28 '16 at 14:37
  • $k$ is a pretty useless extra parameter, since you may remove it through the substitution $t=ku,,dt=k,du$. So it does not help very much to know where $k$ comes from. – Jack D'Aurizio May 28 '16 at 15:42
  • See also here: http://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/689932#689932 – Hans Lundmark May 28 '16 at 18:20

1 Answers1

4

I'll leave the details to you and go straight to the main integral.

Let $$I_n = \int \frac{du}{(1+u^2)^n}.\tag{1}$$ We clearly have $I_1=\arctan u$ and: $$\begin{eqnarray*} I_{n+1} = \int \frac{(1+u^2)-u^2}{(1+u^2)^{n+1}}\,du &=& I_n -\frac{1}{2}\int u\cdot\frac{2u}{(1+u^2)^{n+1}}\,du\\&\stackrel{IBP}{=}&I_n+\frac{u}{2n(1+u^2)^n}-\frac{1}{2n}\,I_n\tag{2}\end{eqnarray*}$$ so, by "unrolling" the recursion: $$ I_{n+1} = \frac{(2n-1)}{(2n)}\,I_n + \frac{u}{2n(1+u^2)^n}\tag{3} $$ we get:

$$ I_n = \frac{(2n-3)!!}{(2n-2)!!}\,\arctan u+\sum_{k=1}^{n-1}\frac{(2n-3)(2n-5)\cdot\ldots\cdot(2n+1-2k)}{(2n)\cdot(2n-2)\cdot\ldots\cdot(2n+2-2k)}\frac{u}{(1+u^2)^{n-k}} $$

or:

$$ I_n = \frac{1}{4^{n-1}}\binom{2n-2}{n-1}\arctan u+\sum_{k=1}^{n-1}\frac{\binom{2n-2}{n-1}}{2n\, 4^{k-1}\binom{2n-2k}{n-k}}\cdot\frac{u}{(1+u^2)^{n-k}}\tag{4} $$

Obviously, if you are just interested in the integral over $\mathbb{R}$ or $\mathbb{R}^+$, the second term in the RHS of $(4)$ is pretty useless.

Jack D'Aurizio
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