This problem's taking me a lot longer than I like (probably doing things the hard way...). I have 9 different terms to integrate, some of which are messier than others. This one, though, is messier than most. $$-{\Large\int}_{\!0}^{a} \!\raise 0.8ex {at^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,} $$

I tried a whole series of substitutions on this. First $t=a\sin\left(u\right)$, then $v=\cos\left(u\right)$ and split the resulting integral into two, only to find (after substitution number three) that the first half diverged (haven't tried the second half). In any case, I'm assuming there has to be a better way to attempt this. Any suggestions appreciated.

  • 2
    $\begingroup$ This seems rather similar to an elliptic integral. $\endgroup$
    – egreg
    Dec 20, 2013 at 23:10
  • $\begingroup$ No, it can be computed (I think it's what is called an abelian integral?). At least, Maxima can do it, using $\log$ or $\arcsin$ according to the sign of $b-a$. $\endgroup$ Dec 20, 2013 at 23:12
  • 2
    $\begingroup$ A possibly useful start is $a^2-t^2=u$, though I mildly prefer $a^2-t^2=u^2$. $\endgroup$ Dec 20, 2013 at 23:16
  • $\begingroup$ Any range on $a$? (e.g. $a \in [-\pi/2, \pi/2]$?) $\endgroup$
    – apnorton
    Dec 21, 2013 at 0:20
  • $\begingroup$ Andre Nicolas' comment works very well with u = $a^2-t^2$. $\endgroup$
    – Betty Mock
    Dec 21, 2013 at 0:31

2 Answers 2


Using the substitution $u=a^2-t^2$ as suggested by André Nicolas, we get:

$\dfrac{\mathrm du}{\mathrm dt}=-2t$. So,

$\begin{align}&\int-\frac{at^3}{\sqrt{(a^2-t^2)(b^2t^2+a^4-a^2t^2)}}\mathrm dt\\ &=\frac{a}2\int\frac{t^2(-2t)}{\sqrt{(a^2-t^2)((b^2-a^2)t^2+a^4)}}\mathrm dt\\ &=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u((b^2-a^2)(a^2-u)+a^4)}}\mathrm du\\ &=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u(m-nu)}}\mathrm du \end{align}$

where $m=a^2b^2$ and $n=(b^2-a^2)$

$\begin{align} &=\frac{a^3}2\int\frac{1}{\sqrt{u(m-nu)}}\mathrm du - \frac{a}2\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du\\ &=\frac{a^3}2I_1 - \frac{a}2I_2. \end{align}$

where I have substituted $I_1$ and $I_2$ for the first and second integral respectively.

For $I_1$, substitute $v=\sqrt{u}$, so that $\dfrac{\mathrm dv}{\mathrm du}=\dfrac1{2\sqrt{u}}$. Then,

$\displaystyle I_1=2\int \frac1{\sqrt{m-nv^2}}\mathrm dv =\frac2{\sqrt{m}}\int \frac1{\sqrt{1-{\left(\sqrt{\frac{n}{m}}v\right)}^2}}\mathrm dv =\dfrac{2}{\sqrt{m}}\sqrt{\dfrac{m}{n}}\arcsin \sqrt{\dfrac{n}{m}}v+C$.

Now for $I_2$,

$\begin{align} I_2=&\int\frac{\sqrt{u}}{\sqrt{m-nu}}\mathrm du =\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\int\frac{\sqrt{\frac{n}{m}u}}{\sqrt{1-\frac{n}{m}u}}\mathrm du\\ &=\frac1{\sqrt{m}}\sqrt{\frac{m}{n}}\frac{m}{n}\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx \qquad\qquad{\left(x=\dfrac{n}{m}u\right)} \end{align}$

To solve $\displaystyle\int\frac{\sqrt{x}}{\sqrt{1-x}}\mathrm dx$, let $\theta=\arcsin \sqrt{x} (\implies x=\sin^2 \theta)$. We get, $\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}$.


$\begin{align}&\int \sqrt{\frac{x}{1-x}} \mathrm dx =\int 2x \dfrac1{\sqrt{1-x}}\cdot\dfrac1{2\sqrt{x}}\mathrm dx\\ &=\int 2\sin^2\theta \;\mathrm d\theta =\theta-\dfrac{\sin2\theta}{2}+C\\ &=\arcsin \sqrt{x}-\dfrac12\sin(2\arcsin \sqrt{x})+C=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\cos (\arcsin \sqrt{x})+C\\ &=\arcsin \sqrt{x}-\sin(\arcsin \sqrt{x})\sqrt{1-\sin^2 (\arcsin \sqrt{x})}+C\\ &=\arcsin \sqrt{x}-\sqrt{x}\sqrt{1-x}+C. \end{align}$

  • $\begingroup$ I'll take a closer look at this later. It looks like you dropped a radical in $I_1$ though. $\endgroup$
    – Mike
    Dec 21, 2013 at 2:45
  • $\begingroup$ @Mike I've fixed it. Thank you. $\endgroup$
    – Alraxite
    Dec 21, 2013 at 3:03
  • $\begingroup$ Well, this looks like the cleanest way of doing this. I messed up and did the substitution $u=\sqrt{a^2-t^2}$, but apparently, this might have just helped me skip a step. :) One of these times I'll have to sign up for WolframAlpha. Not being able to view the step-by-step solution, I assume $x=\sin^2\theta$ causes $I_2$ to unravel? I just hope this definite integral converges or I'll have a bigger mess later... $\endgroup$
    – Mike
    Dec 21, 2013 at 18:27
  • $\begingroup$ @Mike The substitution, $x=\sin^2 \theta$ works great (I've edited my answer). I got a bit tired while typing when I reached the second integral, so instead of trying to solve it, I checked for its solution on WolframAlpha and on seeing a complicated one, I thought that this integral wouldn't be easily solvable. But it would seem that I'm wrong! $\endgroup$
    – Alraxite
    Dec 21, 2013 at 22:05

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\fermi\pars{a,b} \equiv -\int_{0}^{a}{at^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,} \,,\quad \fermi\pars{0,b} = 0}$

When $a \not= 0$: \begin{align} \\[3mm] \fermi\pars{a,b} &= -\sgn\pars{a}\int_{0}^{\verts{a}}{\verts{a}t^{3}\,{\rm d}t \over \,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,} \\[3mm]&= -a\verts{a}{\rm F}\pars{\mu}\quad\mbox{where}\quad{\rm F}\pars{\mu} \equiv\int_{0}^{1}{t^{3}\,{\rm d}t \over \,\sqrt{\,\left(1 - t^{2}\right)\left(\mu t^{2} + 1\right)\,}\,}\,,\quad \mu \equiv \pars{b \over a}^{2} - 1 \end{align}

\begin{align} {\rm F}\pars{\mu} &= \half\int_{0}^{1}{t\,{\rm d}t \over \root{\pars{1 - t}\pars{\mu t + 1}}} \end{align}

It's clear that $\color{#00f}{{\rm F}\pars{0} = 2/3}$. When $\color{#00f}{\mu \not= 0}$ we perform the change of variables $t = 1 - x^{2}$ $\quad\iff\quad$ $x = \pars{1 - t}^{1/2}$: \begin{align} {\rm F}\pars{\mu} &= \half\int_{1}^{0}{1 - x^{2} \over x\bracks{\mu\pars{1 - x^{2}} + 1}}\, \pars{-2x\,\dd x} = \int^{1}_{0}{1 - x^{2} \over \mu + 1 - \mu x^{2}}\,\dd x \\[3mm]&= {1 \over \mu}\int^{1}_{0}{\pars{\mu + 1- \mu x^{2}} - 1 \over \mu + 1 - \mu x^{2}}\,\dd x ={1 \over \mu} - {1 \over \mu^{2}}\int_{0}^{1}{\dd x \over \pars{1 + 1/\mu} - x^{2}} \end{align}

  1. $\color{#00f}{\large \mu < 0}$: \begin{align} {\cal F}\pars{\mu} &=\color{#00f}{\large{1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1} {\dd x \over x^{2} + \pars{\root{-1 - 1/\mu}}^{2}}} \\[3mm]&={1 \over \mu} + {1 \over \mu^{2}}\,\root{-\mu \over \mu + 1} \arctan\pars{\root{-\mu \over \mu + 1}} \end{align}
  2. $\color{#00f}{\large \mu > 0}$: \begin{align} {\cal F}\pars{\mu} &={1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1} {\dd x \over x^{2} - \pars{\root{1 + 1/\mu}}^{2}} \\[3mm]&={1 \over \mu} + {1 \over \mu^{2}} \int_{0}^{1}\pars{{1 \over x - \root{1 + 1/\mu}} - {1 \over x + \root{1 + 1/\mu}}} {1 \over 2\root{1 + 1/\mu}} \\[3mm]&=\color{#00f}{\large{1 \over \mu} + {1 \over 2\mu^{2}}\, \root{\mu \over 1 + \mu} \ln\pars{\verts{\root{\mu} - \root{1 + \mu} \over \root{\mu} + \root{1 + \mu}}}} \end{align}
  3. $\color{#00f}{\large \mu = 0}$: $${\cal F}\pars{\mu} = \color{#00f}{\large{\cal F}\pars{0} = {2 \over 3}} $$
  • $\begingroup$ I'm getting the distinct feeling there's no clean way of doing this. Looks like it would be a nightmare spotting any potential mistakes. $\endgroup$
    – Mike
    Dec 21, 2013 at 18:22
  • $\begingroup$ @Mike It is a common place to make some typo in long calculations. Sometimes I return to one of my answers and check it again. I checked this one twice but I guess sometimes I will check it again. Thanks. $\endgroup$ Dec 21, 2013 at 19:12

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