Integration of rational functions. Integration techniques. Hermite-Ostragradski method.

Question

By the fundamental theorem of algebra a real nonconstant polynomial $Q$ has factorisation into real prime factors $$Q=g_1^{k_1}g_2^{k_2}\cdots g_l^{k_l}$$ The prime factors $g_j$, all distinct, are first degree polynomials or second degree polynomials without real roots. The exponents are positive integers. The polynomial $Q$ is called square-free if all the exponents $k_j$ equal 1.

Exercise:

Let $\psi$ be square-free polynomial of degree $d$ and $P$ have degree less than $2d$ and no prime factor in common with $\psi$. Show that the integral $\int\frac{P}{\psi^2}$ is rational if and only if $\psi$ divides $P'\psi'-P\psi''.$

My sratch work:

I have noticed that $$(\frac{P}{\psi'})'=\frac{P'\psi'-P\psi''}{(\psi')^2}$$ So I decided use this pattern in the given integral bu using integration by parts :$$\int\frac{P}{\psi^2}=\int(\frac{P}{\psi'})(\frac{1}{\psi^2}\psi')=\frac{P}{\psi'}(\frac{-1}{\psi})+\int\frac{1}{\psi}\frac{P'\psi'-P\psi''}{(\psi')^2}$$ Meanwhile the first sum in the right is rational we should care about the rightside integral. Lets consider $$\int\frac{1}{\psi}\frac{P'\psi'-P\psi''}{(\psi')^2}=\frac{A}{B}$$ Where A and B are polynomials. So we get $$\frac{1}{\psi}\frac{P'\psi'-P\psi''}{(\psi')^2}=\frac{A'B-B'A}{B^2} $$but how judge after I couldn't proceed.

Maybe, another approach is using Hermite-Ostragradski method of integration. In this case the integral becomes $$\int \frac{P}{\psi^2}=\frac{P_1}{\psi}+\int \frac{P_2}{\psi}$$ The integral on the right hand side is transcendental function since the denominator $\psi$ is square free and the numerator has lower degree than the denominator. In order integral to be rational the rightside integral should be equal $0$. So we get an equality $$\int \frac{P}{\psi^2}=\frac{P_1}{\psi}$$ and diferrentiate both side :$$\frac{P}{\psi^2}=\frac{P_1'\psi-P_1\psi'}{\psi^2}$$ However I am still unable to proceed further. Could anybody help me please.

Answer 1

You just need to combine the two approaches in your answer!

Let $\psi Q=P'\psi'-P\psi''$. Your first approach gives $$\int\frac P{\psi^2}=-\frac P{\psi\psi'}+\int\frac Q{\psi'^2}$$ and your second approach gives $$\int\frac P{\psi^2}=\frac{P_1}\psi$$ where upon differentiation, $P_1$ satisfies $P_1'\psi=P_1\psi'+P$.

Therefore, $$\frac{P_1}\psi=-\frac P{\psi\psi'}+\int\frac Q{\psi'^2}\iff\int\frac Q{\psi'^2}=\frac{P_1\psi'+P}{\psi\psi'}=\frac{P_1'}{\psi'}$$ and differentiating both sides yields $Q=P_1''\psi'-P_1'\psi''$ which is a polynomial.

Answer 2

The OP almost have the solution using Ostogradski's method. One can proceed as follows:

• First notice that from $$\frac{P}{\psi^2}=\frac{P_1'\psi-P_1\psi'}{\psi^2}$$ one gets that $P=P_1'\psi-P_1\psi'$ and so, $$P'=\psi P''_1+P'_1\psi'-P'_1\psi'-P_1\psi''_1=\psi P''_1-P_1\psi''$$
• Then \begin{align} P'\psi'-P\psi'' &=(\psi P''_1-P_1\psi'')\psi'-(\psi P'_1-P_1\psi')\psi''\\ &= \psi P''_1\psi' -\psi P'_1\psi''=\psi(P''_1\psi'-P'_1\psi'') \end{align}

This is the desired result

---

I am just writing the Ostograski decomposition to refresh my memory a little:

Since $\psi$ has no squares, the maximum common devisor of $Q=\psi^2$ and $Q'=2\psi\psi'$ is $\psi$.

Assuming $P/\psi^2$ is a proper fraction ($\operatorname{deg}(P)\leq \operatorname{deg}(Q)-1$), Ostogradski's method applied to the integral $\int\frac{P}{\psi^2}$ yields $$\int\frac{P}{\psi^2}=\frac{P_1}{\psi}+\int\frac{P_2}{\psi}$$ where $P_2/Q$ is a sum of terms of the form $\frac{c_j}{g_j}$, $c_j\in\mathbb{R}$, when $g_j$ is linear and $\frac{a_jx+b_j}{g_j}$, $a_j,b_j\in\mathbb{R}$ when $g_j$ is quadratic with complex roots.

If the integral (or rather the primitive of $P/Q$) is rational, then the transcendental part $\int\frac{P_2}{\psi}=0$ modulo a constant., that is, $P_2=0$. From $$\frac{P}{Q}=\Big(\frac{P_1}{\psi}\Big)'+\frac{P_2}{\psi}=\frac{\psi P'_1-P_1\psi'}{\psi^2}$$ We get that $$P=\psi P'_1 -P_1\psi'$$