Find this integral $$\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$$

My try: let $$f(x)=x^4-2x^3+2x^2-x+1$$ I found $$f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)$$ so $$I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ so $$2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ then I can't,Thank you very much

  • The denominator has four complex roots. May be, performing a decomposition of the denominator in partial fractions would lead to four much simpler integrals. This is just an idea I submit to your sagacity. – Claude Leibovici Feb 20 '14 at 12:23
  • f(1)=1 but your decomposition implies that 1-x divides f(x), so it should be wrong. – blues66 Feb 20 '14 at 12:28

This is a very challenging integral which can be solved with brute force. The final answer is

$$\pi\sqrt{\frac{1+\sqrt{13}}{78}} \log \left(\frac{1+\sqrt{13}+\sqrt{2 \sqrt{13}-2}}{4} \right)+\pi\sqrt{\frac{\sqrt{13}-1}{78}}\tan ^{-1}\left(\frac{\sqrt{5+2 \sqrt{13}}}{3} \right)$$


Proof

Let $I$ denote the integral.

$$I=\int_0^1 \frac{\sin^{-1}\sqrt{x}}{x^4-2x^3+2x^2-x+1}dx$$

Substituting $x\mapsto 1-x$ and after averaging both integrals we arrive at

$$I=\frac{\pi}{4} \int_0^1 \frac{1}{x^4-2x^3+2x^2-x+1}dx$$

Finally substitute $x\mapsto x+\frac{1}{2}$ to get

$$I = \frac{\pi}{4}\int_{-1/2}^{1/2}\frac{1}{\left( x^2+\frac{1}{4}\right)^2+\frac{3}{4}}dx=\frac{\pi}{2}\int_{0}^{1/2}\frac{1}{\left( x^2+\frac{1}{4}\right)^2+\frac{3}{4}}dx$$

Then we may write $$ \begin{align*} I &= \frac{\pi}{2}\cdot \frac{2}{\sqrt{3}}\text{Im}\int_0^{1/2}\frac{1}{x^2+\frac{1}{4}-i\frac{\sqrt{3}}{2}}dx \\ &= \frac{\pi}{\sqrt{3}}\text{Im}\left\{ \frac{1}{\sqrt{\frac{1}{4}-i\frac{\sqrt{3}}{2}}}\tan^{-1}\frac{x}{\sqrt{\frac{1}{4}-i\frac{\sqrt{3}}{2}}}\right\}_{x=0}^{x=1/2} \\ &= \frac{2\pi}{\sqrt{3}}\text{Im} \left\{ \frac{1}{\sqrt{1-i2\sqrt{3}}}\tan^{-1}\frac{1}{\sqrt{1-i2\sqrt{3}}}\right\} \end{align*} $$

Note that $$\frac{1}{\sqrt{1-i2\sqrt{3}}} = \sqrt{\frac{1+\sqrt{13}}{26}}+i\sqrt{\frac{-1+\sqrt{13}}{26}}$$ So, we can further simplify our previous expression using the following identities

$$ \begin{align*} \log z &= \log|z|+i\text{arg}z \\ \tan^{-1}z &= \frac{i}{2}\log\left( \frac{1-iz}{1+iz}\right) \end{align*} $$ Don't forget to take care over the multivaluedness of logarithms. $$ \begin{align*} I &= \frac{2\pi}{\sqrt{3}}\text{Im}\left\{\left( \sqrt{\frac{1+\sqrt{13}}{26}}+i\sqrt{\frac{-1+\sqrt{13}}{26}}\right)\tan^{-1}\left( \sqrt{\frac{1+\sqrt{13}}{26}}+i\sqrt{\frac{-1+\sqrt{13}}{26}}\right)\right\} \\ &= \frac{\pi}{\sqrt{3}}\text{Im}\Bigg\{ \Bigg( i\sqrt{\frac{1+\sqrt{13}}{26}}-\sqrt{\frac{-1+\sqrt{13}}{26}}\Bigg) \\ &\quad \log\left(\frac{1}{4} \left(3+\sqrt{-5+2 \sqrt{13}}\right)-\frac{i}{2} \sqrt{\frac{1}{2} \left(4+\sqrt{13}+\sqrt{15+6 \sqrt{13}}\right)} \right)\Bigg\} \\ &= \frac{\pi}{\sqrt{3}}\text{Im}\Bigg\{ \Bigg( i\sqrt{\frac{1+\sqrt{13}}{26}}-\sqrt{\frac{-1+\sqrt{13}}{26}}\Bigg) \\ &\quad \Bigg( \log \left(\frac{1+\sqrt{13}+\sqrt{2 \sqrt{13}-2}}{4} \right)-i \tan ^{-1}\left(\frac{1}{3} \sqrt{5+2 \sqrt{13}}\right)\Bigg)\Bigg\} \end{align*} $$ Finally separating the imaginary part one gets:

$$\begin{align*} I &=\pi\sqrt{\frac{1+\sqrt{13}}{78}} \log \left(\frac{1+\sqrt{13}+\sqrt{2 \sqrt{13}-2}}{4} \right)+\pi\sqrt{\frac{\sqrt{13}-1}{78}}\tan ^{-1}\left(\frac{\sqrt{5+2 \sqrt{13}}}{3} \right) \\ &= 0.909520809086566\cdots \end{align*}$$

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    Indeed, $$int(arcsin(sqrt(x))/(x^4-2*x^3+2*x^2-x+1), x = 0 .. 1, numeric) $$ produces $0.9095208091 $. – user64494 Feb 21 '14 at 9:36
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    @Nasser: think about what the answer could be. Do any of the values of the integrand lend themselves to being complex? Then why would the integral be? And if Maple is outputting garbage, why attempt to inject doubt in this answer (which is clearly correct and skillfully demonstrated) with such garbage? – Ron Gordon Feb 21 '14 at 10:15
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    @RonGordon I am simply posting the result from Maple. Do not shoot the messenger if Maple has bugs in it. This is rude of you to do. – Nasser Feb 21 '14 at 10:30
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    @Nasser: Your original language seemed designed to inject doubt in this answer; that deserves a harsh rebuttal. If not and you were just pointing out a bug in Maple (which apparently user64494 does not seem to suffer), then there are other forums for that. – Ron Gordon Feb 21 '14 at 10:45
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    @RonGordon I do not play these games if you do. I was simply saying FYI this is what Maple gave. simply an informational message. I did not even follow the solutions shown. I have no idea why you think I want to inject doubt into any one's answer. Why would I do that for? What will I gain? I am simply pointing what another program was finding for this computation. Pure and simple. – Nasser Feb 21 '14 at 10:54

Firstly, as shown in @Ron Gordon's answer, we should eliminate the $\arcsin(\sqrt{x})$ in the numerator. Take the substitution that $\arcsin(\sqrt{x})=t$, then \begin{equation} I=\int_0^{\pi/2}\frac{t\sin(2t)}{\sin(t)^8-2\sin(t)^6+2\sin(t)^4-\sin(t)^2+1}dt\\ =\int_0^{\pi/2}\frac{t\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\ \end{equation} By taking $t'=\pi/2-t$, we can check \begin{equation} I=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin(2t)}{115/128+1/128\cos(8t)+3/32\cos(4t)}dt\\ =\frac{\pi}{4}\int_0^1\frac{1}{x^4-2x^3+2x^2-x+1}dx\\ =\frac{\pi}{4}\int_0^1\frac{1}{(x^2-x+\frac{1}{2})^2+\frac{3}{4}}dx\\ =\frac{\pi}{4}\int_0^1\frac{1}{((x-\frac{1}{2})^2+\frac{1}{4})^2+\frac{3}{4}}dx\\ \stackrel{y=x-1/2}{=}\frac{\pi}{4}\int_{-1/2}^{1/2}\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}}dy \end{equation} Also, I do not have some good manner to solve it but to factorize the denominator in complex. We can easy to get the 4 roots of the denominator that \begin{equation} x_{1,2}=\pm\frac{1}{2}\sqrt{-1-2i\sqrt{3}}\\ x_{3,4}=\pm\frac{1}{2}\sqrt{-1+2i\sqrt{3}} \end{equation} Then, you can factorize the last fraction by solve the following problem: \begin{equation} \frac{C_1x+C_2}{x-x_1}+\frac{C_3x+C_4}{x-x_2}+\frac{C_5x+C_6}{x-x_3}+\frac{C_7x+C_8}{x-x_4}=\frac{1}{(y^2+\frac{1}{4})^2+\frac{3}{4}} \end{equation} By doing this, you can get the solution of $I$. Since the result is too long I ignore them. The approximation of the result is $0.9095208091$. I think there must be some easier method to deal with the integral of rational fraction.

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    You beat me to the punch. The Weierstrass stuff I posted is ill-advised. – Ron Gordon Feb 20 '14 at 16:04
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    @RonGordon You are modesty. I think my method to deal with the integral of rational fraction is but a general way. There seem like be a more easier method to calculate that such as using residue theorem. But I have no idea how to achieve that now. – Lion Feb 20 '14 at 16:27
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    Why down my answer? – Lion Feb 21 '14 at 10:16

The way you have transformed the integral $I$ is an important step to finding the solution. Using @shobhit.iands's substitution I managed to expand the integrand into two partial fractions whose denominator is a quadratic polynomial. Let $J$ denote

\begin{equation*} J=2I=\int_{0}^{1}\frac{\arcsin \sqrt{x}+\arcsin \sqrt{1-x}}{x^{4}-2x^{3}+2x^{2}-x+1}dx.\tag{1} \end{equation*}

For $0\leq x\leq 1$ we have

\begin{equation*} \arcsin \sqrt{x}+\arcsin \sqrt{1-x}=\frac{\pi }{2}.\tag{2} \end{equation*}

So

\begin{eqnarray*} J &=&\frac{\pi }{2}\int_{0}^{1}\frac{1}{x^{4}-2x^{3}+2x^{2}-x+1}dx, \\ &=&8\pi \int_{-1/2}^{1/2}\frac{1}{16u^{4}+8u^{2}+13}\,du,\qquad u=x-\frac{1}{ 2} \\ &=&16\pi \int_{0}^{1/2}\frac{1}{16u^{4}+8u^{2}+13}\,du.\tag{3} \end{eqnarray*}

The polynomial in the denominator can be factored as follows

\begin{eqnarray*} 16u^{4}+8u^{2}+13 &=&16\left[ \left( u-u_{1}\right) (u-u_{2})\right] \left[ \left( u-u_{3}\right) (u-u_{4})\right] \\ &=&\left[ 4\left( u-u_{1}\right) (u-u_{2})\right] \left[ 4\left( u-u_{3}\right) (u-u_{4})\right] , \end{eqnarray*}

where $u_{k},k=1,2,3,4$, are its roots

\begin{eqnarray*} u_{1} &=&\frac{1}{2}\sqrt{-1+i\sqrt{12}},u_{3}=-\frac{1}{2}\sqrt{-1+i\sqrt{12}} \\ u_{2} &=&\frac{1}{2}\sqrt{-1-i\sqrt{12}},u_{4}=-\frac{1}{2}\sqrt{-1-i\sqrt{12}}. \end{eqnarray*}

Since

\begin{eqnarray*} 4\left( u-u_{1}\right) (u-u_{2}) &=&4u^{2}-2\sqrt{-2+2\sqrt{13}}u+\sqrt{13} \\ 4\left( u-u_{3}\right) (u-u_{4}) &=&4u^{2}+2\sqrt{-2+2\sqrt{13}}u+\sqrt{13}, \end{eqnarray*}

we obtain

\begin{equation*} 16u^{4}+8u^{2}+13=(4u^{2}-2\sqrt{-2+2\sqrt{13}}u+\sqrt{13})(4u^{2}+2\sqrt{ -2+2\sqrt{13}}u+\sqrt{13}) \end{equation*}

and are thus able to expand the integrand into partial fractions

\begin{eqnarray*} \frac{1}{16u^{4}+8u^{2}+13} &=&\frac{1}{(4u^{2}-2\sqrt{-2+2\sqrt{13}}u+\sqrt{ 13})(4u^{2}+2\sqrt{-2+2\sqrt{13}}u+\sqrt{13})} \\ &=&-\frac{1}{312}\frac{\sqrt{-2+2\sqrt{13}}(\sqrt{13}+13)u-12\sqrt{13}}{ 4u^{2}-2\sqrt{-2+2\sqrt{13}}u+\sqrt{13}} \\ &&+\frac{1}{312}\frac{\sqrt{-2+2\sqrt{13}}(\sqrt{13}+13)u+12\sqrt{13}}{ 4u^{2}+2\sqrt{-2+2\sqrt{13}}u+\sqrt{13}}. \end{eqnarray*}

Hence

\begin{eqnarray*} J &=&-\frac{2\pi }{39}\int_{0}^{1/2}\frac{\sqrt{-2+2\sqrt{13}}(\sqrt{13} +13)u-12\sqrt{13}}{4u^{2}-2\sqrt{-2+2\sqrt{13}}u+\sqrt{13}}\,du \\ &&+\frac{2\pi }{39}\int_{0}^{1/2}\frac{\sqrt{-2+2\sqrt{13}}(\sqrt{13}+13)u+12 \sqrt{13}}{4u^{2}+2\sqrt{-2+2\sqrt{13}}u+\sqrt{13}}\,du.\tag{4} \end{eqnarray*}

In this way we have reduced the evaluation of $J$ to the evaluation (see this entry of Wikipedia ) of two integrals of rational functions of the form

\begin{equation*} \int_{0}^{1/2}\frac{mu+n}{au^{2}+bu+c}du=\left. \frac{m}{2a}\ln \left\vert au^{2}+bu+c\right\vert +\frac{2an-bm}{a\sqrt{4ac-b^{2}}}\arctan \frac{2au+b}{ \sqrt{4ac-b^{2}}}\right\vert _{0}^{1/2}, \end{equation*}

where

\begin{equation*} 4ac-b^{2}=4\times 4\times \sqrt{13}-(2\sqrt{-2+2\sqrt{13}})^{2}=8\sqrt{13} +8>0. \end{equation*}

ADDED. I've got

\begin{eqnarray*} I =\frac{J}{2}&=&\frac{\sqrt{-2+2\sqrt{13}}\left( \sqrt{13}+13\right) \pi }{ 312}\ln \frac{1+\sqrt{-2+2\sqrt{13}}+\sqrt{13}}{1-\sqrt{-2+2\sqrt{13}}+\sqrt{ 13}} \\ &&-\frac{2\left( -2+2\sqrt{13}\right) \left( \sqrt{13}+13\right) \pi }{312 \sqrt{2+2\sqrt{13}}}\arctan \frac{12}{\sqrt{2+2\sqrt{13}}\left( -7+\sqrt{13} \right) }\tag{5} \\ &\approx &0.90952. \end{eqnarray*}

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