I'm attempting to calculate an approximate "closed form" of the integral $$\int \frac{dp}{1 + a p^4 + b p^6}$$

as a function of $a$ and $b$, two small parameters (of the order of $10^{-2}$). I'm really not sure how to go about doing this.

First attempt at a solution:

I naively considered performing a series expansion with respect to $a$ and $b$, and then performing the integral, but when one of the limits is infinity the integral diverges.

I'm not really used to approximating integrals so I'm completely at a loss!


The title earlier showed a definite integral instead of an indefinite one.

I also forgot to mention that $a>0$ and $b>0$.

  • $\begingroup$ Yeah, I believe this will not have a usable closed form, if one exists. If you're feeling bored, try $a = b = 1$ and let WolframAlpha show you the mess (scroll down to 'alternate form of the integral'). $\endgroup$
    – orlp
    May 24 '17 at 0:20
  • $\begingroup$ Yeah, that's sort of why I'm here. :( I tried to get Mathematica to do a series approximation of the integral in $a$ and $b$, but it's taking far too long! Was hoping that I could use the fact that $a$ and $b$ were small to my advantage to simplify the job... $\endgroup$
    – Philip
    May 24 '17 at 1:43

For an exact solution, what I would try is first to change variable $$B p^6=x^3\implies p=\frac{\sqrt{x}}{\sqrt[6]{B}}\implies dp=\frac{dx}{2 \sqrt[6]{B} \sqrt{x}}$$ which gives $$I=\int \frac{dp}{1 + A p^4 + B p^6}=\frac{1}{2 \sqrt[6]{B}}\int\frac{dx}{ \sqrt{x} \left(x^3+{A }{B^{-2/3}}x^2+1\right)}$$ Let us call $a,b,c$ the roots of the cubic equation ( two of them will be non-real complex conjugate since $\Delta=-{4 A^3}{B^{-2}}-27<0$) and use partial fraction decomposition to get $$\frac{1}{(x-a) (x-b) (x-c)}=\frac{1}{(a-b) (a-c) (x-a)}+\frac{1}{(b-a) (b-c) (x-b)}+\frac{1}{(c-a) (c-b) (x-c)}$$ So, we are left with the problem of $$J(d)=\int \frac{dx}{\sqrt{x} (x-d)}=-\frac{2}{\sqrt{d}}\tanh ^{-1}\left(\frac{\sqrt{x}}{\sqrt{d}}\right)$$ which seems to be manageable.

  • $\begingroup$ Thanks so much! I can't believe I didn't think of partial fractions. $\endgroup$
    – Philip
    May 27 '17 at 12:30
  • 1
    $\begingroup$ @PhilipCherian. You are welcome ! It looks simple but this is the same as in Robert Israel's answer. $\endgroup$ May 27 '17 at 12:32
  • $\begingroup$ That's true, but this method was easier for me to handle with Mathematica, since the series of the sum of the logs was taking too long on my computer, for some reason. (I'm no expert, I might not have been doing it optimally!) That's why I chose this as the answer :) $\endgroup$
    – Philip
    May 27 '17 at 12:36

I'll assume $a > 0$ and $b > 0$.

By symmetry, the integral from $0$ to $\infty$ is half the integral from $-\infty$ to $\infty$. You can use the Residue Theorem with a semicircular contour. The result is that your integral is $\pi i$ times the sum of the residues of $1/(1 + a p^4 + b p^6)$ at the roots of $1+a p^4 + b p^6$ in the upper half plane.

The residue of $1/f(p)$ at a simple root $p=r$ is $1/f'(r)$.

EDIT: The indefinite integral can be written as

$$ \sum_r \dfrac{\ln(p-r)}{6 b r^4 + 4 a r^3}$$

where the sum is over the roots of $f(p) = 1 + a p^4 + b p^6$, assuming these are distinct: that is in fact the case when $a > 0$ and $b > 0$, since the discriminant of that polynomial is $-64 b (4 a^3 + 27 b^2)^2$.

A series expansion in powers of $a$ and $b$, valid for fixed $p$, is easy: just write

$$ \dfrac{1}{f(p)} = \sum_{k=0}^\infty (-a p^4 - b p^6)^k $$

and integrate term by term. Thus the first few terms are

$$ \dfrac{1}{f(p)} = 1 - a p^4 - b p^6 + a^2 p^8 + 2 a b p^{10} + b^2 p^{12} + \ldots $$ $$ \int \dfrac{dp}{f(p)} = C + p - \frac{a}{5} p^5 - \frac{b}{7} p^7 + \frac{a^2}{9} p^9 + \frac{2ab}{11} p^{11} + \frac{b^2}{13} p^{13} + \ldots$$

  • $\begingroup$ I'm so sorry, I forgot the change the limits in the title (I changed it in the body), but I was hoping for the improper integral, since this itself is an approximation of a more complicated integral that I'm having no luck with. I was hoping that with a "closed" form, I could not only have the behavior with $a$ and $b$, but I could also differentiate it with respect to them and approximate other integrals I need to calculate (like $\int \frac{p^6}{(1 + a p^4 + b p^6)^2}$. The simple root residue will definitely come in handy, thanks! $\endgroup$
    – Philip
    May 24 '17 at 1:34
  • $\begingroup$ Thanks! Your answer was very helpful, but @Claude's answer actually helped me approximate the integral more efficiently, since once I found the exact solution I could then expand the solution as a series in Mathematica! $\endgroup$
    – Philip
    May 27 '17 at 12:29

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