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I want to calculate $\displaystyle \int \limits_0^1\dfrac{x^3}{\sqrt{x^2-1}}\dfrac{1}{1-a^2x^2}\dfrac{1}{1-b^2x^2}\dfrac{1}{c-x}\mathrm dx$

$a$ and $b$ are real parameters, c could be complex and is the solution of a cubic equation.

I tried to find an appropriate contour in the complex plane but failed because it seems impossible to go around the cut (integration from $-1$ to $1$ would be easier). Have anyone an idea? Or is any other integration technique better suited?

tired
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    Take a partial fraction expansion of $${\frac {1}{ \left( 1-{a}^{2}{x}^{2} \right) \left( 1-{b}^{2}{x}^{2} \right) \left( c-x \right) }}$$ then integrals of the form $$\int _{0}^{1}!{\frac {{x}^{3}}{\sqrt {{x}^{2}-1} \left( bx-1 \right) }}{dx}$$ are returned in Maple in terms of logs and square roots with assumptions on $b$. – Graham Hesketh Oct 16 '13 at 16:39
  • There is a partial fraction command in Maple.... – Graham Hesketh Oct 16 '13 at 16:40
  • The Maple code $$int(x^3/(sqrt(x^2-1)(-a^2x+1)(-b^2x+1)(c-x)), x = 0 .. 1) assuming c<0 or c>1 $$ produces the answer which is too long to be cited here. Its fragment is $$-2a^6cln(-Isqrt(-(b^4-1)/b^4)b^2+b^2)sqrt(c^2-1)sqrt(-(a^4-1)/a^4)

    .$$

     – user64494 Oct 16 '13 at 18:33

1 Answers1

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Let us denote$A:=\arccos(a)$, $B:=\arccos(b)$ and $C:=\arccos(1/c)$. By firstly substituting $x=\sin(y)$ and then $u=\tan(y/2)$ we get: \begin{eqnarray} &&\int\limits_0^1 \frac{x^3}{\sqrt{1-x^2}} \frac{1}{1-a^2 x^2} \frac{1}{1-b^2 x^2} \frac{1}{c-x} dx= \\ &&\frac{16}{c} \int\limits_0^{\tan(\pi/4)} \frac{u^3+u^5}{\prod\limits_{\xi_2=\pm}(u-\exp(\xi_2 \imath C))\prod\limits_{\xi_1=\pm}(u+\xi_1 \exp(\xi_2 \imath A))(u+\xi_1 \exp(\xi_2 \imath B))} du \end{eqnarray}

Now the only thing we need to do is to use the following partial fraction decomposition: \begin{eqnarray} \frac{u^p}{\prod\limits_{j=1}^d (u-u_j)} = \sum\limits_{i=1}^d \frac{u_i^p}{(u-u_i)} \cdot \prod\limits_{j\ne i} \frac{1}{u_i-u_j} \end{eqnarray} which is valid for $d\ge 1$ and $p=0,1,2,\cdots,d-1$ and the computation is completed.

Przemo
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