Integrals involving reciprocal square root of a quartic

Question

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!!

Answer 1

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 $$\dfrac{I_{1}}{I_{2}}=\sqrt{2}$$

Answer 2

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.

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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.

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

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

Answer 4

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, $$ \begin{align} \frac{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^4}}}{\displaystyle\int_0^1\frac{\mathrm{d}t}{\sqrt{1+t^4}}} &=2\frac{\Gamma(1/2)^2}{\Gamma(1/4)\Gamma(3/4)}\\ &=2\frac{\pi\csc(\pi/2)}{\pi\csc(\pi/4)}\\[12pt] &=\sqrt2 \end{align} $$