# Integral $\int_0^{\infty}\frac{\operatorname{arctg}^2x}{x^2}dx$

What's the way to find the following integral $$\int_0^{\infty}\frac{\operatorname{arctg}^2x}{x^2}\,dx$$? I know the answer is $$\pi \ln2$$. I tried to integrate by parts and got $$2\int_0^{\infty}\frac{\operatorname{arctg}x}{x(1+x^2)}\,dx.$$ I don't know what to do next, maybe another integration by parts could help or this was a wrong path at all?

• You can use differentiating by the parameter if you consider $\text{arctg} ax$ under the integral sign. Apr 6, 2020 at 12:37

Since $$\displaystyle \frac{\arctan{x}}{x} = \int_0^1 \frac{1}{1+x^2y^2}\, \mathrm{dy}$$, we have:

\begin{aligned} I & = \int_0^{\infty} \frac{\arctan{x}}{x(1+x^2)}\,\mathrm{dx} \\& = \int_0^{\infty} \int_0^1 \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dy}\,\mathrm{dx} \\& = \int_0^{1} \int_0^\infty \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dx}\,\mathrm{dy} \\& =\frac{\pi}{2} \int_0^{1} \frac{1}{1+y}\,\mathrm{dy} \\& =\pi \log \sqrt{2}. \end{aligned}

Let $$I(a,b)=\int_0^{\infty}\frac{\arctan(ax)\arctan(bx)}{x^2}\,dx.$$ Clearly $$I(a,0)=I(0,b)=I(0,0)=0$$ and $$I(1,1)=0$$. Note $$\begin{eqnarray} \frac{\partial^2I(a,b)}{\partial a\partial b}&=&\int_0^{\infty}\frac1{(1+a^2x^2)(1+b^2x^2)}\,dx\\ &=&\int_0^{\infty}\bigg(\frac{a^2}{a^2-b^2}\frac{1}{1+a^2x^2}-\frac{b^2}{a^2-b^2}\frac{1}{1+b^2x^2}\bigg)\,dx\\ &=&\frac{a^2}{a^2-b^2}\frac{\pi}{2a}-\frac{b^2}{a^2-b^2}\frac{\pi}{2b}\\ &=&\frac{\pi}{2(a+b)}. \end{eqnarray}$$ So $$I=\int_0^1\int_0^1\frac{\pi}{2(a+b)}dadb=\pi\ln2.$$

• thank you for solution. I don't understand one moment. In the case a=b we get division by zero. Could you please explain why we don't examine the integral in the case a=b seperately from the case $a\neq b$. Apr 6, 2020 at 15:01
• @baranka, Work on this region $R=\{(a,b): 0\le a, b\le 1, 0<|a-b|<\epsilon\}$ and let $\epsilon\to0$. Apr 6, 2020 at 15:19
• $\int_{0}^{+\infty}\frac{\arctan\left(at\right)\arctan\left(bt\right)}{t^{2}}dt=\frac{\pi}{2}\left(a\ln\left(1+\frac{b}{a}\right)+b\ln\left(1+\frac{a}{b}\right)\right)$ Jun 28, 2021 at 14:37

As you said :

\begin{aligned}I=\int_{0}^{+\infty}{\frac{\arctan^{2}{x}}{x^{2}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{\arctan{x}}{x\left(1+x^{2}\right)}\,\mathrm{d}x}\end{aligned}

Using the substitution \small \left\lbrace\begin{aligned}y&=\frac{1}{x}\\ \mathrm{d}x&=-\frac{\mathrm{d}x}{x^{2}}\end{aligned}\right., we get : \begin{aligned} I=\int_{0}^{+\infty}{\frac{2y}{1+y^{2}}\arctan{\left(\frac{1}{y}\right)}\,\mathrm{d}y}&=\left[\ln{\left(1+y^{2}\right)}\arctan{\left(\frac{1}{y}\right)}\right]_{0}^{+\infty}+\int_{0}^{+\infty}{\frac{\ln{\left(1+y^{2}\right)}}{1+y^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{+\infty}{\frac{\ln{\left(1+y^{2}\right)}}{1+y^{2}}\,\mathrm{d}y} \end{aligned}

Using another substitution \small \left\lbrace\begin{aligned}y&=\tan{x}\\ \mathrm{d}x&=\frac{\mathrm{d}y}{1+y^{2}}\end{aligned}\right., we get : \begin{aligned}I=-2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\cos{x}\right)}\,\mathrm{d}x}=-2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}\end{aligned}

And since :

\begin{aligned} \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{x}\right)}\,\mathrm{d}x}=2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{\left(2t\right)}\right)}\,\mathrm{d}t}&=2\int_{0}^{\frac{\pi}{4}}{\ln{\left(2\sin{t}\cos{t}\right)}\,\mathrm{d}t}\\&=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\cos{t}\right)}\,\mathrm{d}t}\\ &=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{4}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{\left(\sin{u}\right)}\,\mathrm{d}u} \\ \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}&=\frac{\pi}{2}\ln{2}+2\int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}\\ \iff \int_{0}^{\frac{\pi}{2}}{\ln{\left(\sin{t}\right)}\,\mathrm{d}t}&=-\frac{\pi}{2}\ln{2}\end{aligned}

We get that : $$I=\pi\ln{2}$$