Evaluate $\int\limits_0^{\sqrt 2 } {\frac{1}{{3{a^2} + 2}}\frac{{\arctan \left( {\sqrt {{a^2} + 1} } \right)}}{{\sqrt {{a^2} + 1} }}da}$

Question

I am trying to evaluate this: $$I=\int\limits_0^{\sqrt 2 } {\frac{1}{{3{a^2} + 2}}\frac{{\arctan \left( {\sqrt {{a^2} + 1} } \right)}}{{\sqrt {{a^2} + 1} }}da} $$ It looks like Ahmed's integral, but I don't know how to solve this. Here is what I tried: $$\eqalign{ & I\left( b \right) = \int\limits_0^{\sqrt 2 } {\frac{1}{{3{a^2} + 2}}\frac{{\arctan \left( {b\sqrt {{a^2} + 1} } \right)}}{{\sqrt {{a^2} + 1} }}da} ,I\left( 0 \right) = 0 \cr & I'\left( b \right) = \int\limits_0^{\sqrt 2 } {\frac{1}{{\left( {3{a^2} + 2} \right)\left( {1 + {b^2}\left( {{a^2} + 1} \right)} \right)}}da} = \left. {\left( {\frac{{\sqrt 6 }}{{2{b^2} + 6}}{{\tan }^{ - 1}}\left( {\sqrt {\frac{3}{2}} a} \right) - \frac{b}{{\sqrt {{b^2} + 1} \left( {{b^2} + 3} \right)}}{{\tan }^{ - 1}}\left( {\frac{{ab}}{{\sqrt {{b^2} + 1} }}} \right)} \right)} \right|_0^{\sqrt 2 } \cr & = \frac{{\pi \sqrt {\frac{2}{3}} }}{{2{b^2} + 6}} - \frac{b}{{\left( {{b^2} + 3} \right)\sqrt {{b^2} + 1} }}{\tan ^{ - 1}}\left( {\frac{{\sqrt 2 b}}{{\sqrt {{b^2} + 1} }}} \right),{\text{ integrate both sides from 0 to }}1 \cr & I\left( 1 \right){\text{ = }}\int\limits_0^1 {\left( {\frac{{\pi \sqrt {\frac{2}{3}} }}{{2{b^2} + 6}} - \frac{b}{{\left( {{b^2} + 3} \right)\sqrt {{b^2} + 1} }}{{\tan }^{ - 1}}\left( {\frac{{\sqrt 2 b}}{{\sqrt {{b^2} + 1} }}} \right)} \right)db = \frac{{{\pi ^2}}}{{18\sqrt 2 }} - } {\text{ }}\int\limits_0^1 {\frac{b}{{\left( {{b^2} + 3} \right)\sqrt {{b^2} + 1} }}{{\tan }^{ - 1}}\left( {\frac{{\sqrt 2 b}}{{\sqrt {{b^2} + 1} }}} \right)db} \cr} $$

At this point, I don't know how to process further. May I ask for help? Thank you very much.

Answer 1

Given the questioner's knowledge regarding techniques of integration, I shall omit the calculation of the following integrals. They should be easy for OP. \begin{align*} &I_1=\int_0^{\sqrt{2}}{\frac{1}{\left( 3x^2+2 \right) \sqrt{x^2+1}}}\mathrm{d}x=\frac{\pi}{6\sqrt{2}} \\ &I_2=\int_0^{\frac{\pi}{2}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta}=\frac{\pi ^2}{12\sqrt{2}} \end{align*} So HERE WE GO. \begin{align*} I=&\int_0^{\sqrt{2}}{\frac{\mathrm{arctan} \left( \sqrt{x^2+1} \right)}{\left( 3x^2+2 \right) \sqrt{x^2+1}}}\mathrm{d}x \\ =&\frac{\pi}{2}\int_0^{\sqrt{2}}{\frac{1}{\left( 3x^2+2 \right) \sqrt{x^2+1}}}\mathrm{d}x-\int_0^{\sqrt{2}}{\frac{1}{\left( 3x^2+2 \right) \sqrt{x^2+1}}}\mathrm{arctan} \left( \frac{1}{\sqrt{x^2+1}} \right) \mathrm{d}x \\ =&\frac{\pi}{2}I_1-\int_0^{\sqrt{2}}{\int_0^1{\frac{1}{\left( t^2+x^2+1 \right) \left( 3x^2+2 \right)}}\mathrm{d}t\mathrm{d}x} \\ \stackrel{x\rightsquigarrow \sqrt{2}x}{=}&\frac{\pi ^2}{12\sqrt{2}}-\frac{1}{\sqrt{2}}\int_0^1{\int_0^1{\frac{1}{\left( t^2+2x^2+1 \right) \left( 3x^2+1 \right)}}\mathrm{d}t\mathrm{d}x} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{1}{\sqrt{2}}\int_0^1{\int_0^1{\frac{1}{\left( t^2+2x^2+1 \right) \left( 3x^2+1 \right)}\mathrm{d}x}\mathrm{d}t} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{1}{\sqrt{2}}\int_0^1{\frac{1}{3t^2+1}\int_0^1{\left( \frac{3}{3x^2+1}-\frac{2}{2x^2+t^2+1} \right) \mathrm{d}x}\mathrm{d}t} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{1}{\sqrt{2}}\int_0^1{\frac{1}{3t^2+1}\int_0^1{\left( \frac{1}{x^2+\frac{1}{3}}-\frac{1}{x^2+\frac{t^2+1}{2}} \right) \mathrm{d}x}\mathrm{d}t} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{1}{\sqrt{2}}\int_0^1{\frac{1}{3t^2+1}\left( \sqrt{3}\mathrm{arctan} \left( \sqrt{3} \right) -\frac{\sqrt{2}}{\sqrt{t^2+1}}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sqrt{t^2+1}} \right) \right) \mathrm{d}t} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{\pi}{3\sqrt{6}}\int_0^1{\frac{1}{t^2+\frac{1}{3}}\mathrm{d}t}+\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sqrt{t^2+1}} \right) \mathrm{d}t} \\ =&\frac{\pi ^2}{12\sqrt{2}}-\frac{\pi ^2}{9\sqrt{2}}+\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sqrt{t^2+1}} \right) \mathrm{d}t} \\ =&\frac{-\pi ^2}{36\sqrt{2}}+\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sqrt{t^2+1}} \right) \mathrm{d}t} \end{align*} Now, write \begin{align*} J=&\int_0^1{\frac{1}{\left( 3t^2+1 \right) \sqrt{t^2+1}}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sqrt{t^2+1}} \right) \mathrm{d}t} \\ \stackrel{t=\tan \left( \theta \right)}{=}&\int_0^{\frac{\pi}{4}}{\frac{\sec \left( \theta \right)}{3\tan ^2\left( \theta \right) +1}\mathrm{arctan} \left( \frac{\sqrt{2}}{\sec \left( \theta \right)} \right) \mathrm{d}\theta} \\ =&\int_0^{\frac{\pi}{4}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta} \\ =&\int_0^{\frac{\pi}{4}}{\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}}\left( \frac{1}{\sqrt{2}}\mathrm{arctan} \left( \sqrt{2}\sin \left( \theta \right) \right) \right) \\ =&\left. \frac{\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{arctan} \left( \sqrt{2}\sin \left( \theta \right) \right)}{\sqrt{2}} \right|_{0}^{\frac{\pi}{4}}+\int_0^{\frac{\pi}{4}}{\frac{\sin \left( \theta \right)}{1+2\cos ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\sin \left( \theta \right) \right) \mathrm{d}\theta} \\ =&\frac{\pi ^2}{16\sqrt{2}}+\int_0^{\frac{\pi}{4}}{\frac{\sin \left( \theta \right)}{1+2\cos ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\sin \left( \theta \right) \right) \mathrm{d}\theta} \\ \stackrel{\theta \rightsquigarrow \frac{\pi}{2}-\theta}{=}&\frac{\pi ^2}{16\sqrt{2}}+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta} \end{align*} This means that \begin{align*} J=&\frac{1}{2}\left( J+J \right) \\ =&\frac{1}{2}\int_0^{\frac{\pi}{4}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta}+\frac{\pi ^2}{32\sqrt{2}}+\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta} \\ =&\frac{\pi ^2}{32\sqrt{2}}+\frac{1}{2}\int_0^{\frac{\pi}{2}}{\frac{\cos \left( \theta \right)}{1+2\sin ^2\left( \theta \right)}\mathrm{arctan} \left( \sqrt{2}\cos \left( \theta \right) \right) \mathrm{d}\theta} \\ =&\frac{\pi ^2}{32\sqrt{2}}+\frac{\pi ^2}{24\sqrt{2}} \end{align*} At the end, we have $$ I=\frac{-\pi ^2}{36\sqrt{2}}+\frac{\pi ^2}{32\sqrt{2}}+\frac{\pi ^2}{24\sqrt{2}}=\frac{13\pi ^2}{288\sqrt{2}} $$

Answer 2

$$I=\frac{\pi ^2}{18 \sqrt{2}}-\int_0^1 \frac{b \tan ^{-1}\left(\frac{\sqrt{2} b}{\sqrt{b^2+1}}\right)}{\sqrt{b^2+1} \left(b^2+3\right)}\,db$$

Let $\frac{t}{\sqrt{2-t^2}}$ and the integral becomes $$J=\int_0^1 \frac{t \tan ^{-1}(t)}{\sqrt{4-2 t^2} \left(3-t^2\right)}$$ The series expansion of the arctangent does not lead to anything since, if $$K_n=\int_0^1 \frac{t^{2 n+2}}{(2 n+1) \sqrt{4-2 t^2} \left(3-t^2\right)}\,dt$$ $$(2 n+1)^2\,K_n=\, _2F_1\left(\frac{1}{2},\frac{2n+1}{2};\frac{2n+3}{2};\frac{1}{2}\right) -$$ $$F_1\left(\frac{2n+1}{2};-\frac{1}{2},1;\frac{2n+3}{2};\frac{1}{2 },\frac{1}{3}\right)$$ where appear the Gaussian hypergeometric function and the Appell hypergeometric function of two variables.

A full series expansion converges relatively fast to a value close to the value of $$\sum_{n=1}^\infty \frac{2^{-n}}{9 n^2-3 n+2}$$ which is $$\frac{1}{112} \left(\left(7+5 i \sqrt{7}\right) \, _2F_1\left(1,1;\frac{11+i \sqrt 7}{6};-1\right)+\left(7-5 i \sqrt{7}\right) \, _2F_1\left(1,1;\frac{11-i \sqrt{7}}{6} ;-1\right)\right)$$ The absolute error is $3.405\times 10^{-10}$.

Many thanks to the $\text{ISC}$

Edit

Starting from @Black Emperor's answer

$$J_n=\int_{0}^{\sqrt{2}}\frac{du}{\left(1+u^{2}\right)^{n}\left(3u^{2}+2\right)}=\frac 1 {\sqrt 2}F_1\left(\frac{1}{2};n,1;\frac{3}{2};-2,-3\right)$$ We also have that $$S_n=\sum_{l=1}^{n}\left(\frac{\left(2l-2\right)!}{12^{l}\left(l-1\right)!\left(l-1\right)!}\right)$$

$$S_n=\frac{1}{4 \sqrt{6}}-\frac{\Gamma \left(n+\frac{1}{2}\right)}{4 \sqrt{\pi } \,\, 3^{n+1} \,\,\Gamma (n+1)} \, \, \,_2F_1\left(1,n+\frac{1}{2};n+1;\frac{1}{3}\right)$$

Answer 3

For the fun of approximation

Write $${\frac{{\tan^{-1} \left( {\sqrt {{a^2} + 1} } \right)}}{(3a^2+2){\sqrt {{a^2} + 1} }}}=\frac \pi 8+a^2 P_n(a)$$ where $P_n(a)$ is the $[2n,2n]$ Padé approximant, the simplest being $$P_1=\frac{ -24 \left(20-67 \pi +54 \pi ^2\right)+\left(304-416 \pi +123 \pi ^2\right) a^2}{64 (81 \pi -60)+64 (129 \pi -112) a^2 }$$ whose error is $\sim\frac{a^6}{110}$.

This leads to an explicit expression which, converted to decimals, is $0.31632$ (relative error of $0.4$%).

The process can continue using (tedious) fraction decompositions but still giving (nasty) explicit formulae.

$$\left( \begin{array}{cc} n & \text{result using } P_n\\ 1 & 0.3163196512 \\ 2 & 0.3150283511 \\ 3 & 0.3150183239 \\ 4 & 0.3150181786 \\ 5 & 0.3150181759 \\ 6 & 0.3150181758 \\ \end{array} \right)$$

Edit

One thing you could do is to write $${\frac{{\tan^{-1} \left( {\sqrt {{a^2} + 1} } \right)}}{(3a^2+2){\sqrt {{a^2} + 1} }}}=\sum_{n=0}^\infty \frac{\tan ^{-1}\left(\sqrt{a^2+1}\right)}{3^{n+1} \left(a^2+1\right)^{n+\frac{3}{2}} }$$ I did not find the formula for general $n$ (they are quite simple) but, if $$J_n=\int_0^{\sqrt 2} \frac{\tan ^{-1}\left(\sqrt{a^2+1}\right)}{3^{n+1} \left(a^2+1\right)^{n+\frac{3}{2}} }\,da$$ the summation converges quite fast since $$\frac 3{11}<\frac{J_{n+1}}{J_n}<\frac 13$$

Summing the first ten terms leads to $0.31502$

Answer 4

\begin{align} &\int_{0}^{\sqrt2}\frac{\tan^{-1}\sqrt{x^2+1}}{(3x^2+2)\sqrt{x^2+1}}\,dx\\ =& \ \int_0^{\sqrt2} \tan^{-1}\left(\sqrt{x^2+1}\right)\ d\bigg(-\frac1{\sqrt2}\cot^{-1}\frac x{\sqrt{2x^2+2}} \bigg)\\ \overset{ibp}=& \ \frac{\pi^2}{72\sqrt2}+\frac1{\sqrt2}\int_0^{\sqrt2} \frac{x\cot^{-1}\frac x{\sqrt{2x^2+2}} }{(x^2+2)\sqrt{x^2+1}} \ dx\\ =& \ \frac{\pi^2}{72\sqrt2}+\frac1{\sqrt2}\int_0^{\sqrt2} \int_0^{\sqrt2} \frac{x^2 }{(x^2+2)(x^2+y^2 +x^2y^2 )} \ dy \ dx\\ =& \ \frac{\pi^2}{72\sqrt2}+\frac1{2\sqrt2}\int_0^\sqrt2 \int_0^\sqrt2 \bigg( \frac{x^2 }{x^2+2}+\frac{y^2}{y^2+2} \bigg)\frac1{x^2+y^2 +x^2y^2 }\ dy \ dx\\ =& \ \frac{\pi^2}{72\sqrt2}+\frac1{\sqrt2}\int_0^\sqrt2 \int_0^\sqrt2 \frac{1}{(x^2+2)(y^2+2)}\ dy \ dx\\ =& \ \frac{\pi^2}{72\sqrt2}+\frac1{\sqrt2}\left( \frac\pi{4\sqrt2}\right)^2 =\frac{13\pi^2}{288\sqrt2} \end{align}

Answer 5

Not a full answer but hopefully someone takes it from here.

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Using, $$\arctan \left(x\right)=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$$

$$I=\frac{\pi}{2}\underbrace{\int_{0}^{\sqrt{2}}\frac{1}{\left(3t^{2}+2\right)\sqrt{t^{2}+1}}dt}_{\displaystyle \frac{\pi}{6\sqrt{2}}}-\int_{0}^{\sqrt{2}}\frac{1}{3t^{2}+2}\cdot\frac{\arctan\left(\frac{1}{\sqrt{t^{2}+1}}\right)}{\sqrt{t^{2}+1}}dt$$

Problem reduces to, $$H=\int_{0}^{\sqrt{2}}\frac{1}{3t^{2}+2}\cdot\frac{\arctan\left(\frac{1}{\sqrt{t^{2}+1}}\right)}{\sqrt{t^{2}+1}}dt=\frac{11\pi^{2}}{288\sqrt{2}}$$

$t\to \tan t$ $$H=\int_{0}^{\arctan\left(\sqrt{2}\right)}\frac{\cos\left(t\right)}{\sin^{2}t+2}\cdot\arctan\left(\cos t\right)dt$$ Using Series Expansion for $\arctan \left(x\right)$, $$H=\sum_{r=0}^{\infty}\frac{\left(-1\right)^{r}}{2r+1}\int_{0}^{\arctan\left(\sqrt{2}\right)}\frac{\left(\cos t\right)^{2r+2}}{\sin^{2}t+2}dt$$

Let, $$J_n=\int_{0}^{\arctan\left(\sqrt{2}\right)}\frac{\left(\cos t\right)^{2n}}{\sin^{2}t+2}dt$$ Which is equal to, $$J_n=\int_{0}^{\sqrt{2}}\frac{1}{\left(1+u^{2}\right)^{n}\left(3u^{2}+2\right)}du$$ Also, $$\frac{1}{\left(1+x^{2}\right)^{n}\left(3x^{2}+2\right)}=\frac{3^{n}}{3x^{2}+2}-\sum_{l=1}^{n}\frac{3^{n-l}}{\left(x^{2}+1\right)^{l}}$$ $$J_n=\int_{0}^{\sqrt{2}}\left(\frac{3^{n}}{3u^{2}+2}-\sum_{l=1}^{n}\frac{3^{n-l}}{\left(u^{2}+1\right)^{l}}\right)du$$ $$=\frac{3^{n-1}}{\sqrt{6}}\pi-\int_{0}^{\sqrt{2}}\left(\sum_{l=1}^{n}\frac{3^{n-l}}{\left(u^{2}+1\right)^{l}}\right)du$$ $$=\frac{3^{n-1}}{\sqrt{6}}\pi-\sum_{l=1}^{n}3^{n-l}\int_{0}^{\sqrt{2}}\frac{1}{\left(u^{2}+1\right)^{l}}du$$ Also, $$\int_{0}^{\sqrt{2}}\frac{1}{\left(u^{2}+1\right)^{l}}du=\int_{0}^{\arctan\left(\sqrt{2}\right)}\left(\cos^{2}s\right)^{l-1}ds$$ Using, $$\left(\cos x\right)^{m}=\frac{1}{2^{m-1}}\sum_{k=0}^{\frac{m}{2}-1}\operatorname{nCr}\left(m,k\right)\cos\left(\left(m-2k\right)x\right)+\frac{1}{2^{m}}\operatorname{nCr}\left(m,\frac{m}{2}\right)$$ For even $m$. Therefore, $$\int_{0}^{\arctan\left(\sqrt{2}\right)}\left(\cos x\right)^{2a-2}dx=\frac{1}{2^{2a-3}}\sum_{k=0}^{a-2}\operatorname{nCr}\left(2a-2,k\right)\frac{\sin\left(\left(2a-2-2k\right)\arctan\left(\sqrt{2}\right)\right)}{\left(2a-2-2k\right)}+\frac{1}{2^{2a-2}}\operatorname{nCr}\left(2a-2,a-1\right)\arctan\left(\sqrt{2}\right)$$

Hence, $$J_n=\frac{3^{n-1}}{\sqrt{6}}\pi-\sum_{l=1}^{n}\frac{3^{n-l}}{2^{2l-3}}\sum_{k=0}^{l-2}\operatorname{nCr}\left(2l-2,k\right)\frac{\sin\left(\left(2l-2-2k\right)\arctan\left(\sqrt{2}\right)\right)}{\left(2l-2-2k\right)}-4\cdot3^{n}\arctan\left(\sqrt{2}\right)\sum_{l=1}^{n}\left(\frac{\left(2l-2\right)!}{12^{l}\left(l-1\right)!\left(l-1\right)!}\right)$$

Cannot take it from here.