Two celestial objects of mass $m_{1}$ and $m_{2}$ are orbiting around each other on a very elongated ellipse($\epsilon\approx1$). The system is isolated, there is no other proximate celestial body. How would the parameters of the elliptical orbit change as a consequence of radiation of gravitational waves. How can the time of the process be estimated when the orbit changes from an ellipse to a circle?

Here is my work:

The power emitted by gravitational waves is given by:

$$P_{GW}= \frac{c^5}{G}\left(\frac{GM}{c^5l}\right)^5$$

Very compact binaries will lose energy rapidly by GW radiation.

If we assume that the two bodies making up the binary lie in the $x-y$ plane and their **orbits are circular** ($\epsilon=0$), then only non-vanishing components of quadrupole tensors are:

$$\boxed{Q_{xx}=-Q_{yy}=\frac{1}{2}(\mu)a^{2}\cos2\Omega t}$$ and

$$\boxed{Q_{xy}=Q_{ya}=\frac{1}{2}(\mu)a^{2}\sin2\Omega t}$$

Where $\Omega$ is the orbital velocity, $\mu=\dfrac{m_{1}m_{2}}{m}$ is the reduced mass and where $m=m_{1}+m_{2}$

The luminosity of the system can be deduced as:

$$L^{GM}=\frac{32}{5}\frac{G}{c^5}\mu^2 a^4\Omega^6=\frac{32}{5}\frac{G^4}{c^5}\frac{M^3 \mu^2}{a^5}$$

The latter part is obtained from Kepler third law: $\Omega^2=\frac{GM}{a^3}$

As the gravitating system loses energy by emitting radiation, the distance between the 2 bodies shrinks at a rate:

$$\boxed{\frac{da}{dt}=\frac{64}{5}\frac{G^{3}M\mu}{c^5 a^3}}$$

The binary would hence colase at a time :

$$\tau=\frac{5}{256}\frac{c^5}{G^3}\frac{a_{0}^4}{\mu M^4}$$

I am having problems in converting this concept to that mentioned in the question, I.e., for a system of binaries of highly elliptical orbit(in my calculations, I have assumed a circle) and how would the elliptical parameters change with time. I also require help in calculating the time of the process when the orbit changes from an ellipse to a circle.

I have obtained the following integral for finding the decay time of the orbits( the orbits would eventually decay into a circular one):

$$T(a_{0},e_{0})=\frac{12(c_{0}^4)}{19\gamma}\int_{0}^{e_0}{\frac{e^{29/19}[1+(121/304)e^2]^{1181/2299}}{(1-e^2)^{3/2}}}de\tag1$$

Where $$\gamma=\frac{64G^3}{5c^5}m_{1}m_{2}(m_{1}+m_{2})$$

For $e_{0}$ close to $1$ the equation becomes:

$$T(a_{0},e_{0})\approx\frac{768}{425}T_{f}a_{0}(1-e_{0}^2)^{7/2}\tag2$$

Where $$T_{f}=\frac{a_{0}^4}{4\gamma}$$

Here $T$ depends on both $a_{0}$ an de $e_{0}$ as the former determines the latter.

And $\gamma$ has the same value as define do above.

I am stuck with $(1)$ and $(2)$; any help would be appreciated.

**My approach:**

I have tried to use Appell hypergeometric integration and obtained the following result:(let $x=e$)

>$$\boxed{\int\frac{x^(\frac{3}{2}+y)[1+Ax^2]^{\frac{1}{2}+z}}{(1-x^{2})^\frac{3}{2}}dx=2x^{y+\frac{5}{2}}\frac{F_{1}(\frac{y}{2}+\frac{5}{4};\frac{3}{2},-z-\frac{y}{2}+\frac{9}{4};x^{2},-Ax^2)}{2y+5}}$$

Where $y=\frac{55}{38},z=\frac{63}{4598},A=\frac{121}{304}$

Here $F_{1}(a,\beta,\beta^{'},c;x,y)$ is appells function and I read that it can be expressed as the following integral:

>$$\boxed{F_{1}(a,\beta,\beta^{'},c;x,y)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_{0}^{1}u^{a-1}(1-u)^{b-a-1}(1-ux)^{-\beta}(1-uy)^{-\beta^{'}}du}$$

Is this right? How am I to solve this integral? And also how am I to solve for $e_{0}$ in equation (2)?