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As the authors pointed out in this paper (p. 2), the following evaluation which was in Gradshteyn and Ryzhik (sixth edition) is incorrect (and has been removed). $$ ''\int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)^{3/2}} \frac{1}{\sqrt{1+ \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}+\sqrt{1+ \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}}}} \mathrm{d}x = \frac{\pi}{2\sqrt{6}}'' \qquad (*) $$ A numerical evaluation gives $0.6663771 \cdots$ on the left hand side and $0.64127491 \cdots $ on the right hand side.

I have not succeeded to correct $(*)$.

Would you have any idea how evaluating the above integral?

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  • $\begingroup$ What you have tried? $\endgroup$ – mwomath Jul 27 '14 at 22:23
  • $\begingroup$ @mwomath I have transformed the initial integral to a trigonometric one which is still difficult to untangle. Thanks. $\endgroup$ – Olivier Oloa Jul 27 '14 at 22:30
  • $\begingroup$ Let us try some ideas: try to entre $\frac{1}{(1+x^2)^{3/2}} $ to both of $\sqrt{ . . . }$. I like to post however i use from smart phone so it is not easy to print. $\endgroup$ – mwomath Jul 27 '14 at 22:50
  • $\begingroup$ I did a few trig identities to get it a bit nicer. Not convinced it really helps, though... $\endgroup$ – Semiclassical Jul 27 '14 at 23:35
  • $\begingroup$ I have no way of knowing what @Semiclassical tried back when s/he did, but I managed to rewrite the integral via $u=\sin(\arctan x)$ as $$\int_0^1\frac{\mathrm{d}u}{\sqrt{f(u)+\sqrt{f(u)}}}$$ where $f(u)=\frac{4}{3}-\frac{4}{3}\left(u^2-\frac{1}{2}\right)^2$. Whether this only makes the integral appear more tractable, I'm not sure... $\endgroup$ – user170231 Mar 23 '16 at 21:17
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Let $\mathcal{I}$ denote the value of the algebraic integral,

$$\mathcal{I}:=\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)^{3/2}\sqrt{1+\frac{4x^{2}}{3\left(1+x^{2}\right)^{2}}+\sqrt{1+\frac{4x^{2}}{3\left(1+x^{2}\right)^{2}}}}}\approx0.666377.$$

We show below that

$$\mathcal{I}=\frac{1+\sqrt{3}}{2\sqrt{2}}K{\left(\frac{1}{\sqrt{3}}\right)}-\frac{\sqrt{3}}{4\sqrt{2}}\Pi{\left(\frac14,\frac{1}{\sqrt{3}}\right)}-\frac{1}{2\sqrt{2}}F{\left(\frac{\pi}{3},\frac{1}{\sqrt{3}}\right)}.$$


A good first step to clarifying $\mathcal{I}$ would be to rewrite it as a hyper-elliptic integral.

$$\begin{align} \mathcal{I} &=\int_{0}^{\infty}\frac{\mathrm{d}x}{\left(1+x^{2}\right)^{3/2}\sqrt{1+\frac{4x^{2}}{3\left(1+x^{2}\right)^{2}}+\sqrt{1+\frac{4x^{2}}{3\left(1+x^{2}\right)^{2}}}}}\\ &=\int_{0}^{\infty}\frac{y}{\left(1+y^{2}\right)^{3/2}\sqrt{1+\frac{4y^{2}}{3\left(1+y^{2}\right)^{2}}+\sqrt{1+\frac{4y^{2}}{3\left(1+y^{2}\right)^{2}}}}}\,\mathrm{d}y;~~~\small{\left[x=y^{-1}\right]}\\ &=\frac12\int_{0}^{\infty}\frac{\mathrm{d}t}{\left(1+t\right)^{3/2}\sqrt{1+\frac{4t}{3\left(1+t\right)^{2}}+\sqrt{1+\frac{4t}{3\left(1+t\right)^{2}}}}};~~~\small{\left[y^{2}=t\right]}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}u}{\sqrt{1-u}\sqrt{1+\frac43u\left(1-u\right)+\sqrt{1+\frac43u\left(1-u\right)}}};~~~\small{\left[\frac{t}{1+t}=u\right]}\\ &=\frac12\int_{2}^{\frac23}\frac{1}{\sqrt{\frac{9v^{2}-4}{2\left(4+3v^{2}\right)}}\sqrt{\frac{64v^{2}}{\left(4+3v^{2}\right)^{2}}+\frac{8v}{\left(4+3v^{2}\right)}}}\cdot\frac{\left(-1\right)48v}{\left(4+3v^{2}\right)^{2}}\,\mathrm{d}v;~~~\small{\left[\sqrt{\frac{4\left(3-2u\right)}{3\left(2u+1\right)}}=v\right]}\\ &=12\int_{\frac23}^{2}\frac{v}{\left(4+3v^{2}\right)^{2}\sqrt{\frac{9v^{2}-4}{2\left(4+3v^{2}\right)}}\sqrt{\frac{2v\left(4+8v+3v^{2}\right)}{\left(4+3v^{2}\right)^{2}}}}\,\mathrm{d}v\\ &=12\int_{\frac23}^{2}\frac{\sqrt{v}}{\sqrt{4+3v^{2}}\sqrt{9v^{2}-4}\sqrt{4+8v+3v^{2}}}\,\mathrm{d}v\\ &=12\int_{\frac23}^{2}\frac{\sqrt{v}}{\sqrt{3v^{2}+4}\sqrt{\left(3v+2\right)\left(3v-2\right)}\sqrt{\left(v+2\right)\left(3v+2\right)}}\,\mathrm{d}v\\ &=12\int_{\frac23}^{2}\frac{\sqrt{v}}{\left(3v+2\right)\sqrt{\left(3v^{2}+4\right)\left(v+2\right)\left(3v-2\right)}}\,\mathrm{d}v\\ &=\frac{\sqrt{2}}{\sqrt[4]{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\sqrt{w}}{\left(w+\frac{1}{\sqrt{3}}\right)\sqrt{\left(w^{2}+1\right)\left(w+\sqrt{3}\right)\left(w-\frac{1}{\sqrt{3}}\right)}}\,\mathrm{d}w;~~~\small{\left[v=\frac{2w}{\sqrt{3}}\right]}\\ &=\frac{\sqrt{2}}{\sqrt[4]{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{w}{\left(w+\frac{1}{\sqrt{3}}\right)\sqrt{w\left(w^{2}+1\right)\left(w+\sqrt{3}\right)\left(w-\frac{1}{\sqrt{3}}\right)}}\,\mathrm{d}w.\\ \end{align}$$


Consider a linear fractional transformation of the form

$$w=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}x}{1+x};~~~\small{w>-\frac{1}{\sqrt{3}}},$$

$$\implies\frac{\sqrt{3}-w}{\frac{1}{\sqrt{3}}+w}=x;~~~\small{x>-1}.$$

Under this substitution our integral transforms as,

$$\begin{align} \mathcal{I} &=\frac{\sqrt{2}}{\sqrt[4]{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{w}{\left(w+\frac{1}{\sqrt{3}}\right)\sqrt{w\left(w^{2}+1\right)\left(w+\sqrt{3}\right)\left(w-\frac{1}{\sqrt{3}}\right)}}\,\mathrm{d}w\\ &=\frac{\sqrt{2}}{\sqrt[4]{3}}\int_{1}^{0}\frac{3-x}{4\sqrt{\frac{2\left(1-x\right)}{\sqrt{3}\left(1+x\right)}\cdot\frac{2\left(9-x^{2}\right)}{3\left(1+x\right)^{2}}\cdot\frac{4\left(3+x^{2}\right)}{3\left(1+x\right)^{2}}}}\cdot\frac{\left(-1\right)4\,\mathrm{d}x}{\sqrt{3}\left(1+x\right)^{2}};~~~\small{\left[w=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}x}{1+x}\right]}\\ &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{3-x}{\sqrt{\left(\frac{1-x}{1+x}\right)\left(9-x^{2}\right)\left(3+x^{2}\right)}}\,\mathrm{d}x\\ &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{\left(3-x\right)\left(1+x\right)}{\sqrt{\left(1-x^{2}\right)\left(9-x^{2}\right)\left(3+x^{2}\right)}}\,\mathrm{d}x\\ &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{3+2x-x^{2}}{\sqrt{\left(1-x^{2}\right)\left(9-x^{2}\right)\left(3+x^{2}\right)}}\,\mathrm{d}x\\ &=\frac{\sqrt{3}}{4\sqrt{2}}\int_{0}^{1}\frac{3-y}{\sqrt{y\left(1-y\right)\left(9-y\right)\left(3+y\right)}}\,\mathrm{d}y\\ &~~~~~+\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{\left(1-y\right)\left(9-y\right)\left(3+y\right)}};~~~\small{\left[x^{2}=y\right]}\\ &=:\mathcal{E}_{1}+\mathcal{E}_{2}.\\ \end{align}$$

Thus, the hyper-elliptic integral $\mathcal{I}$ can apparently be reduced to elliptic integrals.


$$\begin{align} \mathcal{E}_{1} &=\frac{\sqrt{3}}{4\sqrt{2}}\int_{0}^{1}\frac{3-y}{\sqrt{\left(y+3\right)y\left(1-y\right)\left(9-y\right)}}\,\mathrm{d}y\\ &=\small{\frac{\sqrt{3}}{4\sqrt{2}}\int_{0}^{1}\frac{6\left(2-t^{2}\right)}{\left(4-t^{2}\right)}\cdot\frac{1}{\sqrt{\frac{12}{\left(4-t^{2}\right)}\cdot\frac{3t^{2}}{\left(4-t^{2}\right)}\cdot\frac{4\left(1-t^{2}\right)}{\left(4-t^{2}\right)}\cdot\frac{12\left(3-t^{2}\right)}{\left(4-t^{2}\right)}}}\cdot\frac{24t}{\left(4-t^{2}\right)^{2}}\,\mathrm{d}t};~~~\small{\left[\frac{2\sqrt{y}}{\sqrt{3+y}}=t\right]}\\ &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{\left(2-t^{2}\right)}{\left(4-t^{2}\right)}\cdot\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}\\ &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}-\frac{\sqrt{3}}{4\sqrt{2}}\int_{0}^{1}\frac{\mathrm{d}t}{\left(1-\frac14t^{2}\right)\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}\\ &=\frac{\sqrt{3}}{2\sqrt{2}}K{\left(\frac{1}{\sqrt{3}}\right)}-\frac{\sqrt{3}}{4\sqrt{2}}\Pi{\left(\frac14,\frac{1}{\sqrt{3}}\right)}.\\ \end{align}$$


$$\begin{align} \mathcal{E}_{2} &=\frac{\sqrt{3}}{2\sqrt{2}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{\left(1-y\right)\left(9-y\right)\left(3+y\right)}}\\ &=\frac{2\sqrt{3}}{\sqrt{2}}\int_{\frac{\sqrt{3}}{2}}^{1}\frac{\mathrm{d}t}{\sqrt{16\left(1-t^{2}\right)\left(3-t^{2}\right)}};~~~\small{\left[\sqrt{\frac{y+3}{4}}=t\right]}\\ &=\frac{1}{2\sqrt{2}}\int_{\sin{\left(\frac{\pi}{3}\right)}}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}\\ &=\frac{1}{2\sqrt{2}}\left[\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}-\int_{0}^{\sin{\left(\frac{\pi}{3}\right)}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(1-\frac13t^{2}\right)}}\right]\\ &=\frac{1}{2\sqrt{2}}\left[K{\left(\frac{1}{\sqrt{3}}\right)}-F{\left(\frac{\pi}{3},\frac{1}{\sqrt{3}}\right)}\right].\\ \end{align}$$


Hence,

$$\begin{align} \mathcal{I} &=\mathcal{E}_{1}+\mathcal{E}_{2}\\ &=\frac{\sqrt{3}}{2\sqrt{2}}K{\left(\frac{1}{\sqrt{3}}\right)}-\frac{\sqrt{3}}{4\sqrt{2}}\Pi{\left(\frac14,\frac{1}{\sqrt{3}}\right)}+\frac{1}{2\sqrt{2}}\left[K{\left(\frac{1}{\sqrt{3}}\right)}-F{\left(\frac{\pi}{3},\frac{1}{\sqrt{3}}\right)}\right]\\ &=\frac{1+\sqrt{3}}{2\sqrt{2}}K{\left(\frac{1}{\sqrt{3}}\right)}-\frac{\sqrt{3}}{4\sqrt{2}}\Pi{\left(\frac14,\frac{1}{\sqrt{3}}\right)}-\frac{1}{2\sqrt{2}}F{\left(\frac{\pi}{3},\frac{1}{\sqrt{3}}\right)}.\\ \end{align}$$

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  • $\begingroup$ Congratulations! This is a tour de force! $\endgroup$ – Olivier Oloa Jun 5 '16 at 6:47
  • $\begingroup$ @OlivierOloa Thank you! Regarding defs: I use the same definitions as DLMF, dlmf.nist.gov/19.2#ii . These definitions happen to be consistent with Gradshteyn 's, but not with wolphramalpha. As far as I can tell, my value checks out numerically. $\endgroup$ – David H Jun 5 '16 at 7:17
  • $\begingroup$ I will check it later :) $\endgroup$ – Olivier Oloa Jun 5 '16 at 7:49
  • $\begingroup$ Hi David, do you have a personal adress? $\endgroup$ – Olivier Oloa Nov 21 '16 at 22:07

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