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.

  • $\begingroup$ The [generating function for Legendre polynomials][1] 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 $$\begin{eqnarray*} \int_{0}^{1}\frac{1-x}{1+x}\cdot\frac{1}{\sqrt{x^4+ax^2+1}}\,dx&=&\frac{1}{2}\sum_{n\geq 0}P_n\left(-\frac{a}{2}\right)\int_{0}^{1}\frac{1-\sqrt{t}}{1+\sqrt{t}}\,t^{n-1/2}\,dt\end{eqnarray*} $$ [1]: en.wikipedia.org/wiki/Legendre_polynomials $\endgroup$ – Jack D'Aurizio Feb 5 '17 at 0:42
  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Feb 5 '17 at 0:42
  • 1
    $\begingroup$ @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... $\endgroup$ – Sangchul Lee Feb 5 '17 at 1:10
  • 1
    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Feb 5 '17 at 1:12
  • 1
    $\begingroup$ @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. $\endgroup$ – Sangchul Lee Feb 5 '17 at 1:17

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.

  • 1
    $\begingroup$ 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! $\endgroup$ – Sangchul Lee Feb 5 '17 at 2:07
  • 1
    $\begingroup$ If you set $\displaystyle u\ \mapsto\ {1 \over u}$ in your first integral you get a nice simplification. $\endgroup$ – Felix Marin Feb 5 '17 at 18:36

$$\text{let } \ \frac{1-x}{1+x}=t \Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt\ \text{ then:}$$ $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$ $$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\boxed{\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right),\quad a>-2}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.