Integrating factor lemma

Statement

Let $D\subset\mathbb R^2$ be an open domain and let $M,N\in C^1(D)$. Consider the $1$-form

A nonvanishing function $\mu$ on $D$ is an integrating factor for $\omega$ if the rescaled form $\mu\,\omega$ is exact (equivalently, $\mu\,\omega=dF$ for some potential $F$).

A common sufficient criterion (two-variable, one-function dependence)

Assume $N\neq 0$ on $D$ and define

If $\Phi(x,y)$ depends only on $x$, i.e. $\Phi(x,y)=f(x)$, then

is an integrating factor on $D$.

Similarly, if $M\neq 0$ on $D$ and

depends only on $y$, i.e. $\Psi(x,y)=g(y)$, then

is an integrating factor.

Key hypotheses

• $M,N$ are $C^1$ on $D$.
• $\mu$ is nonzero on $D$ (so multiplication does not change the set of admissible directions).
• For the explicit formulas above, the indicated dependence restriction holds (e.g. $\Phi(x,y)=f(x)$).

Key conclusions

• Under the criterion, $\mu\,\omega$ satisfies the exact differential criterion , hence $\mu\,\omega=dF$ for some state function $F$.
• The integrating factor is determined (up to an overall multiplicative constant) by a one-dimensional integral.

Proof idea / significance

Requiring exactness of $\mu\,\omega$ means

Expanding gives a first-order linear PDE for $\mu$. Under the stated dependence assumption (e.g. $\Phi(x,y)=f(x)$), the PDE reduces to an ODE in one variable, yielding the exponential formula.

Thermodynamic significance: in equilibrium thermodynamics, the reversible heat form $\delta Q_{\mathrm{rev}}$ typically fails to be exact, but the second law asserts the existence of an integrating factor $1/T$ (where $T$ is temperature ) such that

defining entropy (cf. Clausius theorem and Clausius inequality ).