Integral $\int_0^\infty \frac{\phi^2 x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}\mathrm dx$

Question

I am trying to evaluate the definite integral $$\int_0^\infty \frac{\phi^2 x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}\mathrm dx$$

with $\phi$ being the golden ratio.

I tried completing the square $$2\left(x+\frac12\right)^2+ \frac12=2y^2+\frac12$$ and then applying the substitution $y= \frac12 \tan t$. The resulting trigonometric integral may require additional half-angle substitution. It would be quite messy and I do not know how to proceed. I know that I could also apply one of the Euler substitutions. But, it feels brute-force and may not be necessary.

Answer 1

Integrals of the form

$$I=\int_0^\infty \frac{\phi^2 x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}dx$$ result in $\tan^{-1}(\cdot)$ or $ \tanh^{-1}(\cdot)$ in their antiderivatives after substituting $$t=\frac{x +p}{\sqrt{2x^2+2x+1}}$$ Thus, setting $$\frac{A}{t^2+B}dt = \frac{\phi^2 x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}dx$$ to match the coefficients in the given integrand, and obtaining $p=\phi^2$, $A= -\phi^{2}$ and $B= \phi$. As a result $$I=\int_{\frac1{\sqrt2}}^{\phi^2} \frac{\phi^2}{t^2+\phi}dt=\phi^{3/2}\tan^{-1}\left[\phi^{1/2}-(\sqrt2-1)^2\phi^{5/2} \right] $$

Answer 2

The Euler substitution path isn't so bad. Let's break down the steps as follows, making frequent use of the identity $\phi^2=\phi+1$ along the way. (NB: many of these coefficients could be simplified more optimally.)

Substitute $x=\dfrac{\phi^{-2}-\phi^2t}{t-1}$ to eliminate the linear terms from the quadratics; both end up with a coefficient of $2\left(\phi^2+\phi^{-2}\right)-6\equiv0$.

$$\begin{align*} I &= \int_0^\infty \frac{1 + \phi^2x}{\left(2+2x+x^2\right) \sqrt{1+2x+2x^2}} \, dx \\ &= \int_\tfrac1{\phi^4}^1 \frac{\sqrt5\left(\phi^4-1\right)t}{\left(2-\frac2{\phi^2}+\frac1{\phi^4} + \left(2-2\phi^2+\phi^4\right)t^2\right) \sqrt{1-\frac2{\phi^2}+\frac2{\phi^4}+\left(1-2\phi^2+2\phi^4\right)t^2}} \, dt \\ &= \int_\tfrac1{\phi^4}^1 \frac{\sqrt{3-\phi}\,t}{\left(2-\phi+t^2\right) \sqrt{5-\frac8\phi+t^2}} \, dt \end{align*}$$

Next substitute $u=\sqrt{7\phi-11}\,\sqrt{5-\dfrac8\phi+t^2}$ to significantly tidy up the limits:

$$\begin{align*} I &= \int_{\sqrt{5-\tfrac8\phi+\tfrac1{\phi^8}}}^{\sqrt{6-\tfrac8\phi}} \frac{\sqrt{3-\phi}}{7\phi-11+u^2} \, du & u=\sqrt{5-\dfrac8\phi+t^2} \\ &= \int_{\sqrt{2\phi-3}}^{\sqrt{2\phi}} \frac{\sqrt{1+2\phi}}{1+u^2} \, du & u\to\sqrt{7\phi-11}\,u \\ &= \phi^{3/2} \left(\arctan\sqrt{2\phi} - \arctan\left(\sqrt\phi-\frac1{\sqrt\phi}\right)\right) \\ &= \boxed{\phi^{3/2} \arctan\left(\sqrt\phi \frac{1+\left(\sqrt2-1\right)\phi}{\sqrt2+\phi}\right)} \end{align*}$$

Answer 3

Using the technique mentioned in this post as the given integral is of the form $\displaystyle\int\frac{L}{Q_1\sqrt{Q_2}}\mathrm dx$ $($where $L$ is linear and $Q_1$ and $Q_2$ are quadratic polynomials in $x)$,

$$\begin{align}\int_0^\infty\frac{\phi^2x+1}{(x^2+2x+2)\sqrt{2x^2+2x+1}}\mathrm dx&\overset{x\to x-\frac32}{\underset{(1)}{=}}\frac{\phi^2}{\sqrt2}\int_\frac32^\infty\frac{\frac1{\sqrt x}-\frac{\sqrt5}{2x\sqrt x}}{\left(x+\frac5{4x}-1\right)\sqrt{x+\frac5{4x}-2}}\mathrm dx\\&\overset{\sqrt x+\frac{\sqrt5}{2x}\to x}{=}\sqrt2\phi^2\int_{\sqrt{\frac23}\phi^2}^\infty\frac{\mathrm dx}{\left(x^2-1-\sqrt5\right)\sqrt{x^2-2-\sqrt5}}\\&\overset{x\to\frac1x}{\underset{(2)}{=}}\frac1{\sqrt{2\phi}}\int_0^{\sqrt{\frac32}\phi^{-2}}\frac{x\,\mathrm dx}{\left(\frac1{\sqrt5+1}-x^2\right)\sqrt{\sqrt5-2-x^2}}\\&\overset{\sqrt{\sqrt5-2-x^2}\to x}{\underset{(3)}{=}}\frac1{\sqrt{2\phi}}\int_{\sqrt{\frac{\phi^{-7}}2}}^{\phi^{-\frac32}}\frac{\mathrm dx}{x^2+\frac{\phi^{-4}}2}\\&=\phi^\frac32\left(\tan^{-1}\sqrt{2\phi}-\tan^{-1}\phi^{-\frac32}\right) \\&=\boxed{\phi^\frac32\tan^{-1}\left(\frac{\sqrt2\phi^2-1}{\phi^\frac32+\sqrt{2\phi}}\right)}\end{align}$$

Identities related to $\phi$ that were used:

$(1):1-\dfrac32\phi^2=\dfrac{-\sqrt5}2\phi^2$ since $\phi^2=\dfrac{3+\sqrt5}2$

$(2):\dfrac{\phi}{\sqrt{4+2\sqrt5}}=\dfrac1{\sqrt{2\phi}}$ since $\phi^{-3}=\sqrt5-2$

$(3):\dfrac{13\sqrt5-29}4=\dfrac{\phi^{-7}}2$ and $\dfrac{\sqrt5-2}{\sqrt5+1}=\dfrac{\phi^{-4}}2$

Answer 4

Consider the indefinite integral form: $$I=\int\frac{\phi^2 x+1}{\left(x^2+2x+2\right)\sqrt{2x^2+2x+1}}dx$$ $$=\sqrt{2}\int\frac{\phi^2 x+1}{\left(x^2+2x+2\right)\sqrt{(2x+1)^2+1}}dx$$ Let $u=2x+1$ $$\implies I/\sqrt{2}=\int\frac{\phi^2 u-\phi^2+2}{\left(u^2+2u+5\right)\sqrt{u^2+1}}du$$ Let $u=\tan v\implies du=\sec^2 v\,dv$ $$\implies I/\sqrt{2}=\int\frac{\sec v\left(\phi^2\tan v-\phi^2+2\right)}{\tan^2 v+2\tan v+5}dv$$ Let $t=\tan\frac{v}{2}\implies\sec v=\frac{1+t^2}{1-t^2},v=2\arctan t\implies dv=\frac{2}{1+t^2}\,dt$ $$\implies I/\sqrt{2}=2\int\frac{\left(\phi^2-2\right)t^2+2\phi^2t-\phi^2+2}{5t^4-4t^3-6t^2+4t+5}\,dt$$ $$\implies I/\sqrt{8}=\int\frac{\left(\phi^2-2\right)t^2+2\phi^2t-\phi^2+2}{5t^4-4t^3-6t^2+4t+5}\,dt$$ Using this formula, the integral can be splitted into: $$I/\sqrt{8}=\int\sum_{w:5w^4-4w^3-6w^2+4w+5=0}\frac{\left(\phi^2-2\right)w^2+2\phi^2w-\phi^2+2}{\left(20w^3-12w^2-12w+4\right)(t-w)}\,dt$$ $$=\sum_{w:5w^4-4w^3-6w^2+4w+5=0}\frac{\left(\phi^2-2\right)w^2+2\phi^2w-\phi^2+2}{20w^3-12w^2-12w+4}\ln(t-w)+C$$ $$\implies I=\sqrt{2}\sum_{w:5w^4-4w^3-6w^2+4w+5=0}\frac{\left(\phi^2-2\right)w^2+2\phi^2w-\phi^2+2}{10w^3-6w^2-6w+2}\ln(t-w)+C$$ If instead of reverting the transformation, we transform the limits of integration, then the original problem reduces to $$\sqrt{2}\sum_{w:5w^4-4w^3-6w^2+4w+5=0}\frac{\left(\phi^2-2\right)w^2+2\phi^2w-\phi^2+2}{10w^3-6w^2-6w+2}\ln(t-w)|_{t=\sqrt{2}-1}^{t=1}$$ This would took quite some time to simplify, but it should be equivalent to the value provided by KStar.