# Using complex analysis, show that $\int_0^\infty\frac{\arctan(x)}{1+x^2}\,dx=\frac{\pi^2}8$

The result is of course confirmed by substituting $$x\mapsto\tan(x)$$:

$$I = \int_0^\infty \frac{\arctan(x)}{1+x^2} \, dx = \int_0^{\frac\pi2} x \, dx = \frac{\pi^2}8$$

Or integrating by parts:

$$I = \lim_{x\to\infty} \arctan^2(x) - I \implies 2I = \left(\frac\pi2\right)^2 \implies I = \frac{\pi^2}8$$

Or splitting the integral at $$x=1$$ and substituting $$x\mapsto\frac1x$$ on the integral over $$[1,\infty)$$:

$$I= \int_0^1 \frac{\arctan(x) + \arctan\left(\frac1x\right)}{1 + x^2} \, dx = \frac\pi2 \int_0^1 \frac{dx}{1+x^2} = \frac{\pi^2}8$$

Or differentiating under the integral sign:

$$I(a) = \int_0^\infty \frac{\arctan(ax)}{1+x^2} \, dx \implies I'(a) = \int_0^\infty \frac x{(1+x^2)(1+a^2x^2)} \, dx = \frac{\ln(a)}{a^2-1} \\ I(0) = 0 \implies I(1) = \int_0^1 \frac{\ln(x)}{x^2-1} \, dx = \frac{\pi^2}8$$

Or getting the same integral of $$\frac{\ln(x)}{x^2-1}$$ by converting $$\arctan(x)$$ to an integral representation and computing the resulting double integral (per @Dr.WolfgangHintze's suggestion) using the same substitution as in the third method above:

\begin{align*} I &= \int_0^\infty \int_0^x \frac x{(1+x^2)(1+x^2y^2)} \, dy \, dx \\[1ex] &= \int_0^\infty \int_y^\infty \frac x{(1+x^2)(1+x^2y^2)} \, dx \, dy \\[1ex] &= \frac12 \int_0^\infty \frac{\ln\left(\frac{y^2+y^4}{1+y^4}\right)}{y^2-1} \, dy \\[1ex] &= \int_0^\infty \frac{\ln(y)}{y^2-1} \, dy + \frac12 \int_0^\infty \frac{\ln(1 + y^2)}{y^2-1} \, dy - \frac12 \int_0^\infty \frac{\ln(1+y^4)}{y^2-1} \, dy \\[1ex] &= (1 + 2 - 2) \int_0^1 \frac{\ln(y)}{y^2-1} \, dy = \frac{\pi^2}8 \end{align*}

I was wondering how, if at all possible, one might approach it with the residue theorem? I see that $$z=\pm i$$ are simple poles of $$\frac1{1+z^2}$$, but they're also the branch points of $$\arctan(z)$$, since

$$\arctan(z) = -\frac i2 \log\left(\frac{i-z}{i+z}\right)$$

so I don't believe the theorem can be readily applied here. The integrand is odd so I don't think there's much to infer from symmetry. Maybe there's a way to massage the integrand to get closer to something with which we can use a contour integral.

• Similar question here using complex methods, but one answer exploits symmetry that I don't think we have here, and the other I have yet to fully understand Commented Oct 25, 2022 at 21:47
• We can "remove" one of the branch points by writing as $$I=\Im\int_0^1\frac{\log(1+(i-1)u)}{2u^2-2u+1}\,du$$ but we still have $u=(1+i)/2$ occurring on both numerator and denominator. Commented Oct 25, 2022 at 22:37
• Just added the closed form for the case $a=1; b\in[0;1)$ Commented Oct 27, 2022 at 9:17
• @Dr.WolfgangHintze Impractical or not, my curiosity remains satisfied :) Commented Oct 27, 2022 at 16:40
• Yet another method: the integrand's antiderivatives are $\frac12\arctan^2x+C$.
– J.G.
Commented Oct 29, 2022 at 15:34

Complex integration can also be used to evaluate the integral; often, this is a very power tool for finding general solutions - if the system has an appropriate symmetry.

We will consider a general case: $$I(a,b)=\int_0^\infty\frac{x^b\arctan x}{a^2+x^2}dx;\,b\in[0;1);\,a>0\tag{0}$$ We will also choose $$a<1$$ - it allows us to separate poles from branch points; as the integrand is a continuous function of $$a$$, our solution will be valid for all $$a>0$$ (after an appropriate rearrangement if $$a>1$$).

We define $$\arctan z$$ in a standart way: $$\displaystyle\arctan z=-\frac{i}{2}\ln\frac{i-z}{i+z}$$, and the regular cuts are $$[-i;-i\infty);\,[i;i\infty)$$. The modified keyhole contour looks

To define the integrand on the banks of the cut (upper half-plane), we choose $$z=r<1$$ on the axis $$X$$, then make a turn counter-clockwise around $$\displaystyle z=0$$: $$\displaystyle\,r\to re ^\frac{\pi i}{2}; \ln z\to-\frac{i}{2}\ln\frac{i-ir}{i+ir}=-\frac{i}{2}\ln\frac{1-r}{1+r}$$. To move on the bank of the cut (area A), we make another turn (angle $$\pi$$ around $$z=i$$; counter-clockwise): $$\displaystyle 1-r\to (1-r)e ^{\pi i};\,-\frac{i}{2}\ln\frac{1-r}{1+r}\to -\frac{i}{2}\ln\Big(\frac{1-r}{1+r}e ^{\pi i}\Big)=-\frac{i}{2}\ln\frac{1-r}{1+r}+\frac{\pi}{2}$$

If we move to the opposite bank (region C), we will get $$-\frac{i}{2}\ln\frac{1-r}{1+r}-\frac{\pi}{2}$$. On the both banks of the cut $$z= re ^\frac{\pi i}{2}$$ and $$z^b= re^\frac{\pi ib}{2}$$.

Integral along our close contour $$\oint=I(a,b)(1-e^{2\pi ib})+I_A+I_C+I_D+I_E+I_R+I_{r_k}$$ It is not difficult to show that the integrals along a big circle and small circles (around $$z=0,\pm i$$) $$I_R$$ and $$I_{r_k}$$ tend to zero at $$R\to\infty$$ and $$r_k\to0$$. Making the similar analysis for the bottom part of the contour (as was done above for the regions A, C) and bearing in mind that in this case $$z= re ^\frac{3\pi i}{2}$$ (our contour goes counter-clockwise) $$I(a,b)(1-e^{2\pi ib})+I_A+I_C+I_D+I_E=2\pi i\underset {z=ae ^\frac{\pi i}{2}, \,ae ^\frac{3\pi i}{2}}{\operatorname{Res}}-\frac{i}{2}\frac{z^b}{z^2+a^2}\ln\frac{i-z}{i+z}\tag{1}$$ To evaluate $$\displaystyle I_{A, C, D, T}$$ we notice that the terms with $$\ln\frac{i-z}{i+z}$$ cancel (we integrate in the opposite directions), and we are left, for example $$I_A+I_C=-\frac{\pi}{2}e ^\frac{\pi i}{2}e ^\frac{\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx+\frac{\pi}{2}e ^\frac{\pi i}{2}e ^\frac{\pi ib}{2}\int_\infty^1\frac{x^b}{a^2-x^2}dx$$ $$=-\pi ie ^\frac{\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx$$ In the similar way, $$I_D+I_E=-\pi ie ^\frac{3\pi ib}{2}\int_1^\infty\frac{x^b}{a^2-x^2}dx$$ The residues evaluation gives $$2\pi i\sum\operatorname {Res}=-\frac{\pi i}{2}a^{b-1}\ln\frac{1-a}{1+a}\Big(e ^\frac{\pi ib}{2}+e ^\frac{3\pi ib}{2}\Big)$$ Putting all in (1) $$I(a,b)=\frac{\pi}{2\sin\frac{\pi b}{2}}\int_1^\infty\frac{x^b}{x^2-a^2}dx-\frac{\pi}{4\sin\frac{\pi b}{2}}a^{b-1}\ln\frac{1+a}{1-a}$$ Integrating the first term by part $$\boxed{\,\,I(a,b)=\frac{\pi}{4a\sin\frac{\pi b}{2}}\bigg((1-a^b)\ln\frac{1+a}{1-a}+b\int_1^\infty x^{b-1}\ln\frac{x+a}{x-a}dx\bigg);\, a\in(0;1]\,\,}\tag{2}$$ Now we can consider different cases.

1. Leading $$\displaystyle b\to 0$$ $$I(a,0)=\int_0^\infty\frac{\arctan x}{a^2+x^2}dx=\int_1^\infty\frac{\ln x}{x^2-a^2}dx-\frac{\ln a}{2a}\ln\frac{1+a}{1-a}\tag{a}$$
2. Putting in turn $$\displaystyle a=1$$ $$I(1,0)=\int_0^\infty\frac{\arctan x}{1+x^2}dx=\int_1^\infty\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}\tag{b}$$
3. Taking $$a\to1$$ (the first term in (2) is zero), integrating the second term by part, using expressions for gamma (duplication formula) and digamma functions, we can get an interesting result $$I(1,b)=\int_0^\infty\frac{x^b\arctan x}{1+x^2}dx=\frac{\pi}{4\sin\frac{\pi b}{2}}\bigg(\psi\Big(\frac{1}{2}\Big)-\psi\Big(\frac{1-b}{2}\Big)\bigg);\,b\in[0;1)\tag{c}$$ When $$b\to 0$$, we get again $$I(1;0)=\int_0^\infty\frac{\arctan x}{1+x^2}=\frac{1}{4}\psi'\Big(\frac{1}{2}\Big)=\sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}\tag{d}$$
4. The integral has also a closed form for $$\displaystyle b=\frac{1}{2}$$ (and arbitrary $$a>0$$): $$I\Big(a, \frac{1}{2}\Big)=\frac{\pi}{2\sqrt2 a}\bigg((1-\sqrt a)\ln\frac{1+a}{1-a}+\frac{1}{2}\int_1^\infty\ln\frac{x+a}{x-a}\frac{dx}{\sqrt x}\bigg)$$ Making in the second term the substitution $$t=\sqrt x$$, integrating by part and using $$\arctan\frac{1}{s}=\frac{\pi}{2}-\arctan s$$ $$= \frac{\pi}{2\sqrt2 a}\bigg((1-\sqrt a)\ln\frac{1+a}{1-a}+\ln\frac{1-a}{1+a}+\sqrt a\ln\frac{1+\sqrt a}{1-\sqrt a}+2\sqrt a\arctan\sqrt a\bigg)$$ and we get the known result: $$I\Big(a, \frac{1}{2}\Big)=\int_0^\infty\frac{\sqrt x\arctan x}{a^2+x^2}dx=\frac{\pi}{\sqrt {2a}}\bigg(\arctan\sqrt a+\ln\frac{1+\sqrt a}{\sqrt{1+a}}\,\bigg),\,a>0\tag{e}$$
• Thanks so much for taking the time to write this out! I had a similar contour in mind, with keyhole paths around the branch cuts, but I had not considered artificially moving the poles then letting them approach the branch points - neat technique! I was also having some trouble figuring out the arguments along the banks of the cuts but your solution helped clear up that confusion. Commented Oct 27, 2022 at 16:38
• doesn't this get much easier if you use $\arctan(x)=\Im \log(1+i x)$ so you can avoid integrating over nasty branch cuts altogether? Commented Oct 28, 2022 at 18:08
• @asgeige I'm afraid we cannot avoid them: if we use some kind of keyhole contour, we have to deal with branch points. And we use this contour, because the integrand is odd (we cannot extend integration to the whole axis X). Commented Oct 28, 2022 at 18:50
• @Svyatoslav sorry i misread the integration interval :( Commented Nov 3, 2022 at 14:21