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The question is how to show the identity

$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$

I checked this numerically for several cases, but even Mathematica 11 could not manage this symbolically for general $a$, except for some special cases like $a = 0, 1, 2$.

Addendum. Here are some backgrounds and my ideas:

  • This integral came from my personal attempt to find the pattern for the integral

    $$ J(a, b) := \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{1 + ax^2 + bx^4}}. $$

    This drew my attention as we have the following identity

    $$ \int_{0}^{\infty} \frac{x}{x+1} \cdot \frac{dx}{\sqrt{4x^4 + 8x^3 + 12x^2 + 8x + 1}} = J(6,-3), $$

    where the LHS is the integral from this question. So establishing the claim in this question amounts to showing that $J(6,-3) = \frac{1}{2}\log 3 - \frac{1}{3}\log 2$, though I am skeptical that $J(a, b)$ has a nice closed form for every pair of parameters $(a, b)$.

  • A possible idea is to write

    \begin{align*} &\int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} \\ &\hspace{5em}= \int_{0}^{1} \frac{(x^{-2} + 1) - 2x^{-1}}{x^{-1} - x} \cdot \frac{dx}{\sqrt{(x^{-1} - x)^2 + a + 2}} \end{align*}

    This follows from a simple algebraic manipulation. This suggests that we might be able to apply Glasser's master theorem, though in a less trivial way.

I do not believe that this is particularly hard, but I literally have not enough time to think about this now. So I guess it is a good time to seek help.

Community
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Sangchul Lee
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  • The generating function for Legendre polynomials give $$\frac{1}{\sqrt{1-2\kappa t+t^2}}=\sum_{n\geq 0}P_n(\kappa) t^n\tag{1}$$ hence by enforcing the substitutions $x=\sqrt{t}$, $\kappa=-\frac{a}{2}$ we get $$\begin{eqnarray} \int_{0}^{1}\frac{1-x}{1+x}\cdot\frac{1}{\sqrt{x^4+ax^2+1}},dx&=&\frac{1}{2}\sum_{n\geq 0}P_n\left(-\frac{a}{2}\right)\int_{0}^{1}\frac{1-\sqrt{t}}{1+\sqrt{t}},t^{n-1/2},dt\end{eqnarray} $$ – Jack D'Aurizio Feb 05 '17 at 00:42
  • where the involved integrals depend on harmonic numbers and the resulting series should be simple to compute for special values of $a$ related with the zeroes of Legendre polynomials. This identity easily solves the cases $a=0$ and $a=2$, for instance. – Jack D'Aurizio Feb 05 '17 at 00:42
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    @JackD'Aurizio, Thank you for the comment. Considering that $$P_n\left(-\frac{a}{2}\right) = \sum_{k=0}^{n} \binom{n}{k}\binom{n+k}{k} \left( - \frac{a+2}{4} \right)^k $$ exhibits the quantity $\sqrt{a+2}$ in the proposed closed form, it is quite tempting to use your representation. On the other hand, the harmonic number term discourages me... – Sangchul Lee Feb 05 '17 at 01:10
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    Don't get discouraged so easily: if we are able to sum $P_n(-a/2)$ over $n=0,1,\ldots,N$, we can turn the harmonic numbers into something way more manageable through summation by parts. – Jack D'Aurizio Feb 05 '17 at 01:12
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    @JackD'Aurizio, Oh, I see. And that may explain why we expect to see logarithm in the end. Thank you, I will give a shot when I have time. – Sangchul Lee Feb 05 '17 at 01:17
  • We may also recall that harmonic numbers arise from the binomial transform of a simple function, hence we may attempt an inverse binomial transform, hoping to get something less distasteful. – Jack D'Aurizio Feb 05 '17 at 01:18

2 Answers2

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Enforcing the substitution $x^{-1}-x=u$ in your last integral we get:

$$ \int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+u^2}}\right)\frac{du}{u\sqrt{u^2+a+2}} $$ and by setting $u=\sqrt{a+2}\sinh\theta$ we get: $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+(a+2)\sinh^2\theta}}\right)\frac{d\theta}{\sinh\theta}$$ We may get rid of the last term through the "hyperbolic Weierstrass substitution" $$ \theta = 2\,\text{arctanh}(e^{-v}) = \log\left(\frac{e^v+1}{e^v-1}\right)$$ that wizardly gives $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+\frac{a+2}{\sinh^2 v}}}\right)\,dv$$ i.e., finally, a manageable integral through differentiation under the integral sign.
This proves OP's initial identity. Beers on me.

Jack D'Aurizio
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    It is a nice solution! I am happy that I was not far from a shortcut and learned a new substitution $$ \sinh\theta = \frac{1}{\sinh v}, \qquad \frac{d\theta}{\sinh \theta} = dv. $$ Also, the last integral can be solved directly from $$ \int_{0}^{R} \left( 1 - \frac{\sinh v}{\sqrt{\sinh^2 v + \alpha^2}} \right) , dv = \log(1+\alpha) + R - \log\left(\cosh R + \sqrt{\sinh^2 R + \alpha^2}\right) $$ where $\alpha^2 = \frac{a+2}{4}$. Finally, beers on you! – Sangchul Lee Feb 05 '17 at 02:07
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    If you set $\displaystyle u\ \mapsto\ {1 \over u}$ in your first integral you get a nice simplification. – Felix Marin Feb 05 '17 at 18:36
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$$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}\overset{\large \frac{1-x}{1+x}=t}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$ $$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right)$$

Zacky
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