An incorrect answer for an integral

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 $$(*)$$.

Do you have any idea on how to evaluate the above integral?

• What you have tried? – Mohammad W. Alomari Jul 27 '14 at 22:23
• @mwomath I have transformed the initial integral to a trigonometric one which is still difficult to untangle. Thanks. – Olivier Oloa Jul 27 '14 at 22:30
• 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. – Mohammad W. Alomari Jul 27 '14 at 22:50
• I did a few trig identities to get it a bit nicer. Not convinced it really helps, though... – Semiclassical Jul 27 '14 at 23:35
• 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... – user170231 Mar 23 '16 at 21:17

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}

• Congratulations! This is a tour de force! – Olivier Oloa Jun 5 '16 at 6:47
• @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. – David H Jun 5 '16 at 7:17
• I will check it later :) – Olivier Oloa Jun 5 '16 at 7:49