# Integral involving ratios of polynomial and square roots

Is there a simple way to compute the following integral

$$I(y,\beta,\gamma)=\int_1^y \frac{x^2}{\sqrt{\left(x^2-1\right)\left(x^2-\gamma ^2\right) \left(x^2+\beta ^2\right) }}\mathrm{d}x$$

with $$y>1$$, $$0<\gamma<1$$ and $$\beta>0$$?

• Do you have a reason to think that it converges? Feb 6, 2021 at 21:53
• Well, I know that $\int_1^y 1/\sqrt{x-1} dx$ does converge if $y>1$, and since $\gamma<1$ I think we are in good shape so long as we do not take the limit $y\to+\infty$. Feb 6, 2021 at 22:11
• @user12588 Is it safe to assume $\gamma$ is positive? Feb 7, 2021 at 1:28
• Sorry, I fixed. Yes, we want $\gamma$ to be positive as well. Feb 7, 2021 at 7:44

Define the function $$\mathcal{I}:\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)\rightarrow\mathbb{R}$$ via the improper integral

$$\mathcal{I}{\left(p,q,y\right)}:=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}.$$

Suppose $$\left(p,q,y\right)\in\left(0,1\right)\times\mathbb{R}_{>0}\times\left(1,\infty\right)$$, and set $$z:=y^{2}\land a:=1\land b:=p^{2}\land c:=0\land d:=-q^{2}$$.

Note that $$z>a>b>c>d$$, and then also

$$0<\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}<1\land0<\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}<1.$$

The integral $$\mathcal{I}$$ can be transformed into an elliptic integral by means of a simple power substitution:

\begin{align} \mathcal{I}{\left(p,q,y\right)} &=\int_{1}^{y}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{\left(x^{2}-1\right)\left(x^{2}-p^{2}\right)\left(x^{2}+q^{2}\right)}}\\ &=\int_{1}^{y^{2}}\mathrm{d}t\,\frac{t}{\sqrt{t}\sqrt{\left(t-1\right)\left(t-p^{2}\right)\left(t+q^{2}\right)}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}.\\ \end{align}

Next, consider the linear fractional transformation given implicitly by the relation

$$\left(t-b\right)\left(1-u\right)=\left(a-b\right).$$

Then,

$$t=\frac{a-bu}{1-u}\implies dt=du\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}},$$

and

$$\frac{t-a}{t-b}=u\land\frac{t-c}{t-b}=\frac{a-c}{a-b}-\frac{b-c}{a-b}u\land\frac{t-d}{t-b}=\frac{a-d}{a-b}-\frac{b-d}{a-b}u.$$

Applying this LFT to $$\mathcal{I}$$, we obtain

\begin{align} \mathcal{I}{\left(p,q,y\right)} &=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\sqrt{\left(t-a\right)\left(t-b\right)\left(t-c\right)\left(t-d\right)}}\\ &=\int_{a}^{z}\mathrm{d}t\,\frac{t}{\left(t-b\right)^{2}\sqrt{\left(\frac{t-a}{t-b}\right)\left(\frac{t-c}{t-b}\right)\left(\frac{t-d}{t-b}\right)}}\\ &=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{\left(a-b\right)}{\left(1-u\right)^{2}}\cdot\frac{a-bu}{1-u}\cdot\frac{\left(1-u\right)^{2}}{\left(a-b\right)^{2}}\\ &~~~~~\times\frac{1}{\sqrt{u\left(\frac{a-c}{a-b}-\frac{b-c}{a-b}u\right)\left(\frac{a-d}{a-b}-\frac{b-d}{a-b}u\right)}};~~~\small{\left[t=\frac{a-bu}{1-u}\right]}\\ &=\int_{0}^{\frac{z-a}{z-b}}\mathrm{d}u\,\frac{a-bu}{1-u}\cdot\frac{1}{\sqrt{u\left[\left(a-d\right)-\left(b-d\right)u\right]\left[\left(a-c\right)-\left(b-c\right)u\right]}}\\ &=\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\frac{a-b\left(\frac{a-d}{b-d}\right)v}{1-\left(\frac{a-d}{b-d}\right)v}\\ &~~~~~\times\frac{1}{\sqrt{v\left(1-v\right)\left[\left(a-c\right)\left(b-d\right)-\left(b-c\right)\left(a-d\right)v\right]}};~~~\small{\left[u=\left(\frac{a-d}{b-d}\right)v\right]}\\ &=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\left(\frac{a-d}{b-d}\right)v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left[1-\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}v\right]}}\\ &=\frac{1}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin^{2}{\left(\varphi\right)}}\mathrm{d}v\,\left[\frac{a-b}{1-\eta\,v}+b\right]\frac{1}{\sqrt{v\left(1-v\right)\left(1-\kappa^{2}v\right)}}\\ &=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\int_{0}^{\sin{\left(\varphi\right)}}\mathrm{d}x\,\left[\frac{a-b}{1-\eta\,x^{2}}+b\right]\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-\kappa^{2}x^{2}\right)}};~~~\small{\left[v=x^{2}\right]}\\ &=\frac{2}{\sqrt{\left(a-c\right)\left(b-d\right)}}\left[\left(a-b\right)\Pi{\left(\varphi,\eta,\kappa\right)}+bF{\left(\varphi,\kappa\right)}\right],\\ \end{align}

where we've introduced the auxiliary parameters

$$\kappa:=\sqrt{\frac{\left(a-d\right)\left(b-c\right)}{\left(a-c\right)\left(b-d\right)}}\in\left(0,1\right),$$

$$\varphi:=\arcsin{\left(\sqrt{\frac{\left(b-d\right)\left(z-a\right)}{\left(a-d\right)\left(z-b\right)}}\right)}\in\left(0,\frac{\pi}{2}\right),$$

$$\eta:=\frac{a-d}{b-d}\in\left(1,\infty\right).$$

Note: even though $$\eta>1$$, we can still define $$\Pi$$ in the usual way without worrying about Cauchy principal values since $$0<1-\eta\sin^{2}{\left(\varphi\right)}$$.

Substitute $$x=\sqrt t,dx=\frac1{2\sqrt t}\,dt$$ to get $$I=\frac12\int_1^{y^2}\sqrt{\frac t{(t+\beta^2)(t-\gamma^2)(t-1)}}\,dt$$ This is an elliptic integral. Byrd & Friedman 258.13 gives $$I=\frac1{\sqrt{\gamma^2+\beta^2}}\int_0^{F(\varphi,m)}\frac{\operatorname{dn}^2u}{1-n\operatorname{sn}^2u}\,du$$ where $$m=\frac{\gamma^2(1+\beta^2)}{\gamma^2+\beta^2}$$, $$\sin^2\varphi=\frac{(\gamma^2+\beta^2)(y^2-1)}{(1+\beta^2)(y^2-\gamma^2)}$$ and $$n=\frac{1+\beta^2}{\gamma^2+\beta^2}$$. In turn, B&F 339.01 solves the second integral, yielding $$I=\frac1{\sqrt{\gamma^2+\beta^2}}\cdot\frac1n(mF(\varphi,m)+(n-m)\Pi(n,\varphi,m))$$ $$=\frac1{\sqrt{\gamma^2+\beta^2}}(\gamma^2F(\varphi,m)+(1-\gamma^2)\Pi(n,\varphi,m))$$