# A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$

Is it possible to evaluate the following integral in a closed form? $$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx,$$ where $\phi$ is the golden ratio: $$\phi=\frac{1+\sqrt{5}}{2}.$$

• Numerical integration and isc.carma.newcastle.edu.au suggest $\frac\pi{4\,\phi}$. Commented Oct 9, 2013 at 18:41

Since $\frac1\phi=\phi-1$, we get \begin{align} \int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x &=\int_0^\infty\frac{x^\phi\arctan(x)}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{1}\\ &=\int_0^\infty\frac{x^\phi(\frac\pi2-\arctan(x))}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{2} \end{align} Average $(1)$ and $(2)$ to get \begin{align} \int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x &=\frac\pi4\int_0^\infty\frac{x^\phi}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{3}\\ &=\frac\pi{4\phi}\int_0^\infty\frac{x}{\left(x+1\right)^2}\frac{\mathrm{d}x}{x}\tag{4}\\ &=\frac\pi{4\phi}\tag{5} \end{align} Explanation:

$(1)$: $\frac1\phi=\phi-1$
$(2)$: Substitute $x\mapsto\frac1x$
$(3)$: Average $(1)$ and $(2)$
$(4)$: Substitute $x\mapsto x^{1/\phi}$
$(5)$: $\int_0^\infty\frac{\mathrm{d}x}{(x+1)^2}=\left[-\frac1{x+1}\right]_0^\infty=1$

• wow you're really fast . (+1) Commented Oct 9, 2013 at 19:38
• Won't the limits be reversed? Or if remaining the same, there should be a minus sign. Did the inside simplification of the integral require a minus sign? (I'm not able to reach step 2 from step 1 after suggested substitution; could you please elaborate?) Commented Dec 28, 2016 at 9:02
• \begin{align} \int_0^\infty f(x)\,\frac{\mathrm{d}x}{x} &= \int_\infty^0 f\left(\frac1x\right)\left(-\frac{\mathrm{d}x}{x}\right)\\ &=\int_0^\infty f\left(\frac1x\right)\frac{\mathrm{d}x}{x} \end{align}
– robjohn
Commented Dec 28, 2016 at 10:48
• Sigh.. Can't believe I forgot that.. thank you :) Commented Dec 31, 2016 at 8:56

robjohn's result can be generalized to all real $a\ne0$: $$\int_0^\infty\frac{x^{a-1}\arctan x}{(x^a+1)^2}dx=\frac\pi{4\,|a|}.$$

• The original question would translate as $$\int_0^\infty\frac{x^{1/a}\arctan(x)}{(x^a+1)^2}\mathrm{d}x$$ It is $a=\phi$ that gives $1/a=a-1$. However, if you start from my second step, then your statement is true.
– robjohn
Commented Oct 10, 2013 at 13:01

Integration by Parts

Using $$\phi^2=\phi+1 \Rightarrow \frac{1}{\phi}=\phi-1$$, we have \begin{aligned} \int_0^{\infty} \frac{x^{\frac{1}{\phi}} \arctan x}{\left(x^\phi+1\right)^2} d x &=\int_0^{\infty} \frac{x^{\phi-1} \arctan x}{\left(x^\phi+1\right)^2} d x \\ &=-\frac{1}{\phi} \int_0^{\infty} \arctan x d\left(\frac{1}{x^\phi+1}\right) \\ &=- \underbrace{ \left[\frac{\arctan x}{\phi\left(x^\phi+1\right)}\right]_0^{\infty}}_{=0} +\frac{1}{\phi} \int_0^{\infty} \frac{1}{\left(1+x^2\right)\left(1+x^\phi\right)} d x \quad \textrm{ via IBP} \\ \end{aligned}

\begin{aligned}\int_0^{\infty} \frac{1}{\left(x^2+1\right)\left(x^\phi+1\right)} d x &\stackrel{x\mapsto\frac{1}{x}}{=} \int_0^{\infty} \frac{x^\phi}{\left(1+x^2\right)\left(1+x^\phi\right)} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{1+x^\phi}{\left(1+x^2\right)\left(1+x^\phi\right)} d x =\frac{\pi}{4}\\ \end{aligned} Now we can conclude that $$\boxed{\int_0^{\infty} \frac{x^{\frac{1}{\phi}} \arctan x}{\left(x^\phi+1\right)^2} d x =\frac{\pi}{4\phi} }$$