Evaluating $\int_{0}^{1}\frac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$

Question

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

Answer 1

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

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

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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:

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

Answer 2

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*}

Answer 3

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.

Answer 4

Let $$I=\int_0^1 \frac{\arcsin \sqrt{x}}{x^4-2 x^3+2 x^2-x+1} d x$$ Then $x\mapsto 1-x$ transforms the integral into $$ I=\int_0^1 \frac{\arcsin \sqrt{1-x}}{x^4-2 x^3+2 x^2-x+1} d x $$ Averaging these two versions of $I$ yields

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

Noting that $$\arcsin \sqrt{x}+\arcsin \sqrt{1-x} =\frac{\pi}{2}, $$ we have

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

Letting $u=x-\frac{1}{2} $ simplifies the denominator of the integrand as

\begin{aligned} I & =4\pi \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{16 u^4+8 u^2+13} d u \\ & = \frac{\pi}{2} \int_0^{\frac{1}{2}} \frac{1}{u^4+\frac{1}{2} u^2+\frac{13}{16}} d u \\ & =\frac{\pi}{2} \int_0^{\frac{1}{2}} \frac{\frac{1}{u^2}}{u^2+\frac{13}{16 u^2}+\frac{1}{2}} d u \\ & =\frac{2 \pi}{\sqrt{13}} \int_0^{\frac{1}{2}} \frac{\left(1+\frac{\sqrt{13}}{4 u^2}\right)-\left(1-\frac{\sqrt{13}}{4 u^2}\right)}{u^2+\frac{13}{16 u^2}+\frac{1}{2}} d u \\&= \frac{\pi}{\sqrt{13}}\left[\int_0^{\frac{1}{2}} \frac{d\left(u-\frac{\sqrt{13}}{4}\right)}{\left(u-\frac{\sqrt{13}}{4}\right)^2+\frac{\sqrt{13}+1}{2}}-\int_0^{\frac{1}{2}} \frac{d\left(u+\frac{\sqrt{13}}{4}\right)}{\left(u+\frac{\sqrt{13}}{4}\right)^2-\frac{\sqrt{13}-1}{2}}\right]\\&= \frac{\pi}{\sqrt{13}}\left[\sqrt{\frac{2}{\sqrt{13}+1}}\left(\tan ^{-1}\left(\frac{u-\frac{\sqrt{13}}{4}}{\sqrt{\frac{\sqrt{13}+1}{2}}}\right)-\frac{1}{2} \sqrt{\frac{2}{\sqrt{13}-1}} \ln \left|\frac{u+\frac{\sqrt{13}}{4}-\sqrt{\frac{\sqrt{13}-1}{2}}}{u+\frac{\sqrt{13}}{4}+\sqrt{\frac{\sqrt{3}-1}{2}}}\right|\right)\right]_0^{\frac 12} \\ & =\pi \sqrt{\frac{2}{13+\sqrt{13}}}\left[\tan ^{-1}\left(\frac{2 \sqrt{2}-\sqrt{26}}{\sqrt{\sqrt{13}+1}}\right)-\tan ^{-1}\left(\frac{\sqrt{26}}{\sqrt{\sqrt{13}+1}}\right)\right] +\frac{\pi}{\sqrt{13-\sqrt{13}}} \ln \left| \frac{2 \sqrt{2}+\sqrt{26}+4 \sqrt{\sqrt{13}-1}}{2 \sqrt{2}+\sqrt{26}-4 \sqrt{\sqrt{13}-1}} \cdot \frac{\sqrt{26}-4 \sqrt{\sqrt{13}-1}}{\sqrt{26}+4 \sqrt{\sqrt{13}-1}}\right| \end{aligned}

Answer 5

Here are some residue-theorem-friendly forms, though how friendly is debatable.

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Rewrite $\arcsin x\to\arctan\dfrac{\sqrt{1-x^2}}x$, then substitute $y=\sqrt{\dfrac{1-x}x}$:

$$\begin{align*} I &= \int_0^1 \frac{\arctan \sqrt{\frac{1-x}x}}{x^4-2x^3+2x^2-x+1} \, dx \\ &= 2 \int_0^\infty \frac{\arctan y}{\frac1{(1+y^2)^4}-\frac2{(1+y^2)^3}+\frac2{(1+y^2)^2}-\frac1{1+y^2}+1} \, \frac{y}{(1+y^2)^2} \, dy \\ &= 2 \int_0^\infty \frac{y (1+y^2)^2 \arctan y}{1+3y^2+5y^4+3y^6+y^8} \, dy \tag{$\star$} \end{align*}$$

Now we can theoretically approach $(\star)$ as a more intricate flavor of integral $J(a,b,c)$ and integrate a similar $f(z)$ along the same contour.

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On the other hand, substituting $y\to\dfrac1y$ to exploit $\arctan y+\arctan\dfrac1y=\dfrac\pi2$, then $y\to\sqrt y$, reduces $(\star)$ to

$$\begin{align*} I &= \frac\pi2 \int_0^\infty \frac{y (y^2+1)^2}{y^8+3y^6+5y^4+3y^2+1} \, dy \\ &= \frac\pi4 \int_0^\infty \frac{(y+1)^2}{y^4+3y^3+5y^2+3y+1} \, dy \end{align*}$$

which can be computed with the typical keyhole contour, considering the integral of

$$f(z) = \frac{(z+1)^2}{z^4+3z^3+5z^2+3z+1} \log z$$