How to evaluate $\int_0^{\infty}\frac{\arctan{x}}{x^{1/4}(1+x)}\mathrm{d}x$?

Question

I want to evaluate

$ \displaystyle \int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } \mathrm{d}x \tag*{} $

I know that it equals

$\displaystyle \frac{\pi^2}{2 \sqrt{2} } + \frac{\pi}{\sqrt{2}} \ln \left( \frac{2+\sqrt{2-\sqrt{2}} }{2-\sqrt{2-\sqrt{2}}}\right) \tag*{} $

But I can't seem to find a way to prove it.

After substituting $x=t^4$ , the integral is transformed to

$ \displaystyle I= 4\int_{0}^{\infty} \frac{t^2 \arctan(t^4) }{1+t^4} \mathrm{d}t \tag*{} $

After this I tried differentiating under the integral, but the resulting integrand is very complicated, and I couldn't make much progress futher on.

I'm not too familiar with contour integration, maybe that can help here?

Answer 1

More generally, let $R(x)$ be a rational function without out poles on $\mathbb{R}$, $R(x) = O(1/x^2),x\to\infty$. Let us do $$I= \int_{-\infty}^{\infty} R(x)\arctan(x^4) dx$$ OP's question corresponds to $R(x) = x^2/(1+x^4)$. Write $\zeta = e^{\pi i /8}$, then $$\begin{aligned} I &= \frac{i}{2}\int_{-\infty}^\infty R(x)[\log(1-ix^4)-\log(1+ix^4)] dx \\ &= \frac{i}{2}\sum_{r=0}^3 \int_{-\infty}^\infty R(x)[\log(1-\zeta^{4r+1} x)-\log(1-\zeta^{4r+3} x)] \end{aligned}$$ So we are left with integral of form $J(k) = \int_{-\infty}^\infty R(x) \log(1-\zeta^k x)dx$, with $k\nmid 8$. Using the conventional branch cut of $\log$ at negative axis, integrand of $J(k)$ has a branch cut at $\{u \zeta^{-k} | u>1\} \subset \mathbb{C}$, which is disjoint to $\mathbb{R}$ if $k\nmid 8$. Therefore when $k=1,..,7$, there is no branch cut of integrand on upper half-plane, and for $k=9,...,15$, there are no branch cut of integrand on lower half-plane. Hence $$\tag{*}J(k) = \begin{cases} 2\pi i \sum_{\substack{\Im(z_j)>0 \\ z_j \text{ poles of }R}} \text{Res}\left[R(z)\log(1-\zeta^k z),z=z_j\right] & k =1,\cdots,7\\ \\ -2\pi i \sum_{\substack{\Im(z_j)<0 \\ z_j \text{ poles of }R}} \text{Res}\left[R(z)\log(1-\zeta^k z),z=z_j\right] & k =9,\cdots,15\end{cases}$$ since $$I = \frac{i}{2}(J(1)+J(5)+J(9)+J(13)-J(3)-J(7)-J(11)-J(15))$$ each term can be found by taking residues on upper or lower half plane according to $(*)$, this is the formula on calculating $R$.

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For OP's example, $R(x) = x^2/(1+x^4)$, poles of $R$ are $\zeta^2,\zeta^6,\zeta^{10},\zeta^{14}$. Since $R$ has real coefficient, we have $\Im J(k) = -\Im J(16-k)$, thus $$I = -\Im(J(1)+J(5)+J(9)+J(13))$$ so we only need to calculate $4$ terms instead of $8$ terms.

I am too lazy to do an explicit calculation by hand, here is some Mathematica code, input the following into Mathematica,

\[Zeta]=Exp[Pi*I/8];j1=2Pi*I*Residue[z^2*Log[1-\[Zeta]*z]/(1+z^4),{z,\[Zeta]^2}]+2Pi*I*Residue[z^2*Log[1-\[Zeta]*z]/(1+z^4),{z,\[Zeta]^6}];j5=2Pi*I*Residue[z^2*Log[1-\[Zeta]^5*z]/(1+z^4),{z,\[Zeta]^2}]+2Pi*I*Residue[z^2*Log[1-\[Zeta]^5*z]/(1+z^4),{z,\[Zeta]^6}];j9=-2Pi*I*Residue[z^2*Log[1-\[Zeta]^9*z]/(1+z^4),{z,\[Zeta]^10}]+-2Pi*I*Residue[z^2*Log[1-\[Zeta]^9*z]/(1+z^4),{z,\[Zeta]^14}];j13=-2Pi*I*Residue[z^2*Log[1-\[Zeta]^13*z]/(1+z^4),{z,\[Zeta]^10}]+-2Pi*I*Residue[z^2*Log[1-\[Zeta]^13*z]/(1+z^4),{z,\[Zeta]^14}];

then type TrigToExp@ToRadicals@FullSimplify@Im@ComplexExpand[-j1-j5-j9-j13], it then outputs $$\frac{\pi ^2}{4 \sqrt{2}}+\frac{\pi \log \left(\frac{\sqrt{2-\sqrt{2}}}{2}+1\right)}{2 \sqrt{2}}-\frac{\pi \log \left(1-\frac{\sqrt{2-\sqrt{2}}}{2}\right)}{2 \sqrt{2}}$$ which is equivalent to given result.

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This approach can be generalized to any $\int_{-\infty}^\infty R(x) \arctan(x^p) dx$.

Another easy corollary: $$\int_0^\infty \frac{\arctan(x^4)}{1-x^2+x^4} dx = \frac{\pi ^2}{8}+\frac{\pi \log \left(\sin \left(\frac{7 \pi }{48}\right) \cos \left(\frac{5 \pi }{48}\right) \sec \left(\frac{\pi }{48}\right) \sec \left(\frac{11 \pi }{48}\right)\right)}{2\sqrt{3}}$$

Answer 2

Let $I(a)=\int_{0}^{\infty} \frac{\arctan{(ax)}}{x^{1/4}(1+x) } dx $ and evaluate \begin{align} I’(a)=& \int_{0}^{\infty} \frac{x^{3/4}}{(1+x)(1+a^2 x^2) } dx \\ =& \>\frac1{1+a^2}\int_0^\infty \left(\frac{a^2 x^{3/4}}{1+a^2 x^2} + \frac{x^{-1/4}}{1+a^2 x^2} - \frac{x^{-1/4}}{1+x}\right)dx\\ =& \>\frac1{1+a^2}\left(\frac\pi2 \csc\frac\pi8\>a^{1/4}+ \frac\pi2 \sec\frac\pi8\>a^{-3/4} - \sqrt2\pi \right)dx\\ \end{align} Then \begin{align} \int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } dx =& \>I(1) =\int_0^1\overset{a=y^4}{I’(a)\>da}\\ = &\>2\pi\int_0^1 \frac{\csc\frac\pi8 \>y^4 +\sec\frac\pi8 }{1+y^8}\>dy- \frac{\pi^2}{2\sqrt2}\\ = &\>2\pi\cdot \left(\frac\pi{2\sqrt2}-\frac1{\sqrt2}\ln \tan\frac{3\pi}{16}\right)- \frac{\pi^2}{2\sqrt2}\\ = &\> \frac{\pi^2}{2\sqrt2} -\sqrt2\pi\ln \tan\frac{3\pi}{16} \ \end{align}