# Integrals involving reciprocal square root of a quartic

For two integration below, what is the ratio of them? $$I_1 = \int_0^1\frac{dt}{\sqrt{1-t^4}}$$ $$I_2 = \int_0^1 \frac{dt}{\sqrt{1+t^4}}$$ What is the ration $\frac{I_1}{I_2}$?
I do not have any thoughts for solving this question so could anyone give me some hints?
Thank you!!

• These are both elliptic integrals. $I_{1}$ is a pretty standard integral with value $\dfrac{\pi}{2M(1, \sqrt{2})}$ where $M(a, b)$ denotes arithmetic-geometric mean of two numbers $a, b$. – Paramanand Singh Feb 12 '14 at 5:45
• The first one is the beta function in disguise, just let $x=t^4$. – Lucian Feb 12 '14 at 6:18
• @Paramanand What do you mean by that? Sorry..I am a college student. Maybe some theorems are not known to me... – Utaha Tahimi Feb 12 '14 at 16:51
• @BoanBowenTAN: Well the integrals involving square roots of third and fourth degree polynomials are called elliptic integrals. They arise while calculating the perimeter of an ellipse (hence such name). You can read more about them in en.wikipedia.org/wiki/Elliptic_integral About Arithmetic Geometric Mean you can read my blog post paramanands.blogspot.com/2009/08/… – Paramanand Singh Feb 13 '14 at 3:50
• @Lucian: Almost. We need to show that the integral on $[0,1]$ is equal to the integral on $[1,\infty)$. See this answer. – robjohn Mar 3 '16 at 22:50

The first one:let $t^2=\sin{x}$,then $$I_{1}=\dfrac{1}{2}\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{\sqrt{\sin{x}}}$$

The second one: let $t^2=\tan{x}$

then $$I_{2}=\dfrac{1}{2}\int_{0}^{\frac{\pi}{4}}\dfrac{dx}{\sqrt{\sin{x}\cos{x}}}=\dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\dfrac{dx}{\sqrt{\sin{x}}}$$ so $$\dfrac{I_{1}}{I_{2}}=\sqrt{2}$$

• To me this solution seems to be the simplest because it involves very simple substitutions. +1 – Paramanand Singh Feb 13 '14 at 4:03
• This is cool. Normally, one would avoid $\sqrt{\sin(x)}$, but here, it is the thing to do! (+1) – robjohn Feb 13 '14 at 8:42

Hint:

$$I_1(x)=\int_x^1\frac{dt}{\sqrt{1-t^4}}=\frac{1}{\sqrt{2}}\int_{0}^{\arccos x}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$

$$I_2(x)=\int_0^x \frac{dt}{\sqrt{1+t^4}}=\frac{1}{2}\int_{0}^{\arccos\frac{1-x^2}{1+x^2}}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$

and since $$\arccos 0 = \arccos \frac{1-1^2}{1+1^2}$$

$$\frac{I_1(0)}{I_2(1)}=\sqrt{2}$$

From the hint the substitutions should be pretty straightforward.

Added due to the comment below:

The first one: let $t=\cos u$, $u=\arccos t$.

$$\frac{dt}{\sqrt{1-t^4}}=\frac{-\sin udu}{\sqrt{1-\cos^4 u}}=\frac{-\sin udu}{\sqrt{\sin^2 u(1+\cos^2 u)}}$$

and you get the RHS almost immediately.

The second one: let $t=\tan\frac{u}{2}$, $t^2=\frac{1}{\cos^2 \frac{u}{2}}-1=\frac{1-\cos u}{1+\cos u}$, $u=\arccos \frac{1-t^2}{1+t^2}$. Plug it in, and get the RHS also (almost) immediately.

• Here what substitutions did you do?? – Utaha Tahimi Feb 12 '14 at 16:45
• Well, I thought you should be able to figure it out yourself. I will extend the answer showing the first one soon. – Vadim Feb 13 '14 at 0:01

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\ol}{\overline{#1}}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#00f}{\large I_{1}}&\equiv\int_{0}^{1}{\dd t \over \root{1 - t^{4}}} =\int_{0}^{1}\pars{1 - t}^{-1/2}\,{1 \over 4}\,t^{-3/4}\,\dd t ={1 \over 4}\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t ={1 \over 4}\,{\rm B}\pars{{1 \over 4},\half} \\[3mm]&={1 \over 4}\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}} ={1 \over 4}\,{\Gamma\pars{1/4}\root{\pi} \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}}= \color{#00f}{\large{1 \over 4\root{2\pi}}\,\Gamma^{\,2}\pars{1 \over 4}} \approx 1.3110 \end{align}

${\rm B}\pars{x,y}$ and $\Gamma\pars{z}$ are the Beta and Gamma functions, respectively. We used well known properties of them.

\begin{align} \color{#00f}{\large I_{2}}&\equiv\int_{0}^{1}{\dd t \over \root{1 + t^{4}}} =\half\int_{0}^{\infty}{\dd t \over \root{1 + t^{4}}} \end{align} Lets $\ds{x \equiv {1 \over 1 + t^{4}}\quad\iff\quad t = \pars{{1 \over x} - 1}^{1/4}}$ \begin{align} \color{#00f}{\large I_{2}}&=\half\int_{0}^{\infty}{\dd t \over \root{1 + t^{4}}} =\half\int_{1}^{0}x^{1/2}\,{1 \over 4}\,\pars{1 - x \over x}^{-3/4} \pars{-\,{\dd x \over x^{2}}} \\[3mm]&={1 \over 8}\int_{0}^{1}x^{-3/4}\pars{1 - x}^{-3/4}\,\dd x ={1 \over 8}\,{\rm B}\pars{{1 \over 4},{1 \over 4}} ={1 \over 8}\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/2}} \\[3mm]&=\color{#00f}{\large{1 \over 8\root{\pi}}\,\Gamma^{\,2}\pars{1 \over 4}} \approx 0.9270 \end{align}

$$\color{#00f}{\large{I_{1} \over I_{2}}} = {1/\pars{4\root{2}} \over 1/8} =\color{#00f}{\large\root{2}}$$

As shown in this answer, $$\int_0^1t^{\alpha-1}\,(1-t)^{\beta-1}\,\mathrm{d}t =\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$ and $$\int_0^\infty t^{\alpha-1}\,(1+t)^{-\beta}\,\mathrm{d}t =\frac{\Gamma(\alpha)\Gamma(\beta-\alpha)}{\Gamma(\beta)}$$ Substituting $t\mapsto1/t$ and then $t\mapsto t^{1/4}$, we get \begin{align} \int_0^1\frac{\mathrm{d}t}{\sqrt{1+t^4}} &=\int_1^\infty\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\ &=\frac12\int_0^\infty\frac{\mathrm{d}t}{\sqrt{1+t^4}}\\ &=\frac18\int_0^\infty t^{-3/4}(1+t)^{-1/2}\,\mathrm{d}t\\ &=\frac18\frac{\Gamma(1/4)^2}{\Gamma(1/2)} \end{align} Furthermore, \begin{align} \int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}} &=\frac14\int_0^1t^{-3/4}(1-t)^{-1/2}\,\mathrm{d}t\\ &=\frac14\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)} \end{align} As shown in this answer, $\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)$. Therefore, \begin{align} \frac{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}}}{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1+t^4}}} &=2\frac{\Gamma(1/2)^2}{\Gamma(1/4)\Gamma(3/4)}\\ &=2\frac{\pi\csc(\pi/2)}{\pi\csc(\pi/4)}\\[12pt] &=\sqrt2 \end{align}