A calculus problem by Tunk-Fey Ariawan

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { \left( { x }^{ 4 }-2{ x }^{ 2 }+5 \right) } -\ln { \left( 5{ x }^{ 4 }-2{ x }^{ 2 }+1 \right) } }{ 1-{ x }^{ 2 } } dx } = \pi \arctan { \sqrt { \frac { \sqrt { 5 } -1 }{ 2 } } }$

The method I am going to use is not the best method to do it hence any help in simplification is appreciated.

Proof :

$\displaystyle I = \int _{ 0 }^{ 1 }{ \frac { \log { \left( { x }^{ 4 }-2{ x }^{ 2 }+5 \right) } -\log { \left( 5{ x }^{ 4 }-2{ x }^{ 2 }+1 \right) } }{ 1-{ x }^{ 2 } } dx }$

Start with the substitution $y = \dfrac{1+{x}^{2}}{1-{x}^{2}}$ , to get :

$\displaystyle I = \frac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \frac { \ln { \left( \frac { { y }^{ 2 }+2y+2 }{ { y }^{ 2 }-2y+2 } \right) } }{ \sqrt { { y }^{ 2 }-1 } } dx }$

Integrating by parts : $\displaystyle \int { uv' } =uv-\int { u'v } , u = \ln { \left( \frac { { y }^{ 2 }+2y+2 }{ { y }^{ 2 }-2y+2 } \right) } , v = \frac { 1 }{ \sqrt { { y }^{ 2 }-1 } }$ we get :

$\displaystyle I=\ln { \left( \frac { { y }^{ 2 }+2y+2 }{ { y }^{ 2 }-2y+2 } \right) } \ln { \left( y+\sqrt { { y }^{ 2 }-1 } \right) } { | }_{ 1 }^{ \infty }+2\int _{ 1 }^{ \infty }{ \frac { { y }^{ 2 }-2 }{ { y }^{ 4 }+4 } \ln { \left( y+\sqrt { { y }^{ 2 }-1 } \right) } dx }$

$\displaystyle \Rightarrow I=2\int _{ 1 }^{ \infty }{ \frac { { y }^{ 2 }-2 }{ { y }^{ 4 }+4 } \ln { \left( y+\sqrt { { y }^{ 2 }-1 } \right) } dx }$

Now we introduce a parameter here :

$\displaystyle f(a) = a\int _{ 1 }^{ \infty }{ \frac { { y }^{ 2 }-{ a }^{ 2 } }{ { y }^{ 4 }+{ a }^{ 4 } } \ln { \left( y+\sqrt { { y }^{ 2 }-1 } \right) } dx }$

Put $y=ax$ , to get :

$\displaystyle f(a) = \int _{ 1/a }^{ \infty }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 4 }+1 } \ln { \left( ax+\sqrt { { (ax) }^{ 2 }-1 } \right) } dx }$

Differentiate with respect to $a$ we have and using newton leibnitz rule :

$\displaystyle f'(a)={\left(\dfrac{1}{a}\right)}^{2}\frac { { \left(1/a \right) }^{ 2 }-1 }{ { (1/a) }^{ 4 }+1 } \ln { \left( a(\frac { 1 }{ a } )+\sqrt { { \left(a\left(\frac { 1 }{ a } \right)\right) }^{ 2 }-1 } \right) } +\int _{ 1/a }^{ \infty }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 4 }+1 } \frac { x }{ \sqrt { { (ax) }^{ 2 }-1 } } dx }$

$\Rightarrow \displaystyle f'(a) = \int _{ 1/a }^{ \infty }{ \frac { { x }^{ 2 }-1 }{ { x }^{ 4 }+1 } \frac { x }{ \sqrt { { (ax) }^{ 2 }-1 } } dx }$

Put ${(ax)}^{2}-1={y}^{2}$ to get :

$\displaystyle f'(a) = \int _{ 0 }^{ \infty }{ \frac { ({ y }^{ 2 }+1-{ a }^{ 2 }) }{ { { (y }^{ 2 }+1) }^{ 2 }+{ a }^{ 4 } } dy }$

With the substitution $y \mapsto\dfrac{\sqrt{{a}^4+1}}{y}$ we get :

$\displaystyle f'(a) = \frac { 1 }{ \sqrt { { a }^{ 4 }+1 } } \int _{ 0 }^{ \infty }{ \frac { (1-{ a }^{ 2 }){ y }^{ 2 }+({ a }^{ 4 }+1) }{ { y }^{ 4 }+2{ y }^{ 2 }+({ a }^{ 2 }+1) } dy }$

Adding these two we have :

$\displaystyle f'(a) = \left( \frac { (1-{ a }^{ 2 })+\sqrt { { a }^{ 4 }+1 } }{ 2\sqrt { { a }^{ 4 }+1 } } \right) \int _{ 0 }^{ \infty }{ \frac { y^{ 2 }+\sqrt { { a }^{ 4 }+1 } }{ y^{ 4 }+2{ y }^{ 2 }+({ a }^{ 2 }+1) } dy }$

$\displaystyle f'(a) = \left( \frac { (1-{ a }^{ 2 })+\sqrt { { a }^{ 4 }+1 } }{ 2\sqrt { { a }^{ 4 }+1 } } \right) \int _{ 0 }^{ \infty }{ \frac { 1+\frac { \sqrt { { a }^{ 4 }+1 } }{ { y }^{ 2 } } }{ { \left( y-\frac { \sqrt { { a }^{ 4 }+1 } }{ y } \right) }^{ 2 }+2(1+\sqrt { { a }^{ 4 }+1 } ) } dy }$

Use the substitution $\displaystyle \left( y-\frac { \sqrt { { a }^{ 4 }+1 } }{ y } =x \right)$ to get :

$\displaystyle f'(a) = \left( \frac { (1-{ a }^{ 2 })+\sqrt { { a }^{ 4 }+1 } }{ 2\sqrt { { a }^{ 4 }+1 } } \right) \int _{ -\infty }^{ \infty }{ \frac { dx }{ { x }^{ 2 }+2(1+\sqrt { { a }^{ 4 }+1 } ) } }$

$\displaystyle \Rightarrow f'(a) = \left( \frac { (1-{ a }^{ 2 })+\sqrt { { a }^{ 4 }+1 } }{ 2\sqrt { { a }^{ 4 }+1 } } \right) \frac { \pi }{ \sqrt { 2(1+\sqrt { { a }^{ 4 }+1 } ) } }$

Simplifying it we have :

$\displaystyle f'(a) = \frac { \pi }{ 2 } \left( \frac { 1 }{ \sqrt { { a }^{ 4 }+1 } } \right) \frac { \sqrt { 2({ a }^{ 4 }+1)-2{ a }^{ 2 }+2(1-{ a }^{ 2 })\sqrt { { a }^{ 4 }+1 } } }{ \sqrt { 2(1+\sqrt { { a }^{ 4 }+1 } ) } }$

$\Rightarrow \displaystyle f'(a) = \frac { \pi }{ 2 } \sqrt { \frac { \sqrt { { a }^{ 4 }+1 } -{ a }^{ 2 } }{ { a }^{ 4 }+1 } }$

Now integrating both sides we have :

$\displaystyle \int _{ 0 }^{ a }{ f'(x)dx } =\frac { \pi }{ 2 } \int _{ 0 }^{ a }{ \sqrt { \frac { \sqrt { { x }^{ 4 }+1 } -{ x }^{ 2 } }{ { x }^{ 4 }+1 } } dx }$

We will evaluate $\displaystyle G = \int _{ 0 }^{ a }{ \sqrt { \frac { \sqrt { { x }^{ 4 }+1 } -{ x }^{ 2 } }{ { x }^{ 4 }+1 } } dx }$

Put $\sqrt{{x}^{4}+1} = y+x^{2}$ , to get :

$\displaystyle G = \frac { 1 }{ \sqrt { 2 } } \int _{ \sqrt { { a }^{ 4 }+1 } -{ a }^{ 2 } }^{ 1 }{ \frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } }$

$\displaystyle \Rightarrow G = \sqrt { 2 } \frac { \arccos { \left( \sqrt { { a }^{ 4 }+1 } -{ a }^{ 2 } \right) } }{ 2 }$

Hence $\displaystyle G = \sqrt { 2 } \arctan { \left( \sqrt { \frac { \sqrt { { a }^{ 4 }+1 } -1 }{ { a }^{ 2 } } } \right) }$

Finally we have :

$f(a) = f(0)+\frac { \pi }{ \sqrt { 2 } } \arctan { \left( \sqrt { \frac { \sqrt { { a }^{ 4 }+1 } -1 }{ { a }^{ 2 } } } \right) }$

It is obvoius that $f(0)=0$

Finally we have our result :

$f(a) = \frac { \pi }{ \sqrt { 2 } } \arctan { \left( \sqrt { \frac { \sqrt { { a }^{ 4 }+1 } -1 }{ { a }^{ 2 } } } \right) }$

For solving our problem we know that $I = \sqrt{2}f(\sqrt{2})$

$\large \Rightarrow \displaystyle \boxed{ I = \pi \arctan { \left( \sqrt { \dfrac { \sqrt { 5 } -1 }{ 2 } } \right) } }$

Picking up quite early on in Ronak's proof, we have $I \; = \; 2\int_1^\infty \frac{y^2-1}{y^4 + 4}\ln\big(y + \sqrt{y^2-1}\big),dy \;.$ Substituting $y = \cosh u$ , and later $v = e^u$ gives the integral $\begin{array}{rcl} I & = & \displaystyle2\int_0^\infty \frac{\cosh^2u - 2}{\cosh^4u + 4} u \sinh u \,du \; = \; \int_{\mathbb{R}}\frac{\cosh^2u - 2}{\cosh^4u + 4} u \sinh u \,du \\ & = & \displaystyle 2\int_{\mathbb{R}} \frac{\big((e^u + e^{-u})^2 - 8\big)\big(e^u - e^{-u}\big) u}{\big(e^u + e^{-u}\big)^4 + 64}\,du \\ & = & \displaystyle 2\int_{\mathbb{R}} \frac{\big((e^{2u} + 1)^2 - 8e^{2u}\big)\big(e^{2u} - 1\big)u}{\big(e^{2u} + 1\big)^4 + 64e^{4u}}\,e^u\,du \\ & = & \displaystyle 2\int_0^\infty \frac{((v^2+1)^2 -8v^2)(v^2-1)\ln v}{(v^2+1)^4 + 64v^4}\,dv \; = \; 2\int_0^\infty F(v)\ln v\,dv \;. \end{array}$ Now $F$ is a rational function of $v$ whose denominator has degree $2$ more than its numerator. If we take the principal branch of the logarithm, cutting the plane along the negative real axis, and consider the contour which runs

• from $\epsilon$ to $R$ along the positive real axis,
• along the semicircular arc $z \,=\, Re^{i\theta}$ for $0 \le \theta \le \pi$ ,
• along the negative real axis (just above the cut) from $-R$ to $-\epsilon$ ,
• along the semicircular arc $z \,=\, \epsilon e^{i\theta}$ for $\tfrac12\pi \ge \theta \ge 0$

then standard contour integration (letting $R \to \infty$ and $\epsilon \to 0$ ) tells us that $I + \pi i \int_0^\infty F(v)\,dv \; = \; 2\pi i \sum_{q \in R(+)} \mathrm{Res}_{z=q} F(z) \log z \;,$ where $R(+)$ is the set of poles of $F(z)$ with positive imaginary part. The order $8$ polynomial which is the denominator of $F$ can be factorized quite easily, and we obtain the fact that $R(+) \,=\, \{a,b,c,d\}$ , where $\begin{array}{rclcrcl} a & = & -1 + i - \sqrt{-1-2i} & \qquad & b & = & 1 - i - \sqrt{-1 - 2i} \\ c & = & -1 - i + \sqrt{-1+2i} & \qquad & d & = & 1 + i + \sqrt{-1 + 2i} \end{array}$ (the square roots are chosen with positive real part), and we note that $a = -d^\star$ and $c = -b^\star$ . From here on in it is just number-crunching. The sum of the residues is $\begin{array}{rcl} \displaystyle \sum_{q \in R(+)} \mathrm{Res}_{z=q} F(z) \log z & = & \tfrac18\big[(\log d-\log a) + (\log b - \log c)\big] \\ & = & \tfrac18i\big[\delta - (\pi - \delta) + \beta - (\pi - \beta)\big] \\ & = & \tfrac14i(\beta + \delta - \pi) \end{array}$ (since $|a| = |d|$ and $|b| = |c|$ ), from which we deduce that $I \; = \; \tfrac12\pi(\pi - \beta - \delta)$ where $0 \le \beta, \delta \le \tfrac12\pi$ are the arguments of $b,d$ respectively. After a fair amount of trigonometric calculation, we can show that $\tan\tfrac12(\beta+\delta) \; = \; \sqrt{\frac{\sqrt{5}+1}{2}} \;,$ and hence $I \; =\; \pi\big(\tfrac12\pi - \tfrac12(\beta+\delta)\big) \; = \; \pi \tan^{-1}\sqrt{\frac{\sqrt{5}-1}{2}} \;,$ as required.

This is still not a perfect solution, just an alternative one. I feel pretty certain that a more subtle piece of contour integration will get the answer more efficiently...