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The following integral is simple to compute

$$I=\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx \,\,\tag{1}$$

letting $$u=\arctan x \Rightarrow du=\frac{dx}{1+x^2}$$

$$\int\frac{\arctan(x)}{1+x^2}dx=\int u \, du = \frac{u^2}{2}$$

Therefore

$$\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx=\frac{\arctan^2 (x)}{2}\Big|_{0}^{\infty}=\frac{\pi^2}{8}$$

But if we introduce a parameter $a$ in (1) such that the integral becomes

$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx \,\,\tag{2}$$

the same technique above does not apply anymore.

My attempt for this integral is differentiating (2) w.r. to a

$$I^{\prime}(a)=\int_{0}^{\infty}\frac{x}{\left(1+x^2\right)\left(1+(ax)^2 \right)}dx \,\,$$

From this point I got stuck, altough I tried one more step using the fact that

$$\int_{0}^{\infty}e^{-xt}\cos(t)dt=\frac{x}{1+x^2}$$

$$I^{\prime}(a)=\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-xt}\cos(t)}{\left(1+(ax)^2 \right)}\,dt\,dx $$

$$I^{\prime}(a)=\int_{0}^{\infty}\cos(t)\int_{0}^{\infty}\frac{e^{-xt}}{\left(1+(ax)^2 \right)}\,dx\,dt $$

But it seems too hard.

Alternatively, integrating by parts we obtain

$$I(a)=\int_{0}^{\infty}\frac{\arctan(ax)}{1+x^2}dx=-a\int_{0}^{\infty}\frac{\arctan(x)}{1+(ax)^2}dx$$

now using the following integral representation

$$\arctan (x)=\int_{0}^{1}\frac{x}{1+x^2y^2}dy$$

$$I(a)=-a\int_{0}^{\infty}\int_{0}^{1}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dy \,dx$$

$$I(a)=-a\int_{0}^{1}\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx \, dy$$

I dont know the solution of this integral, however I saw these results below and I wanted to proof them

$$\int_{0}^{\infty} \frac{\arctan \frac{x}{\phi}}{1+x^{2}} d x=\frac{\pi^{2}}{12}+\frac{3 \ln ^{2} \phi}{4} $$

$$\int_{0}^{\infty} \frac{\arctan \phi x}{1+x^{2}} d x =\frac{\pi^{2}}{6}-\frac{3 \ln ^{2} \phi}{4}$$

Ricardo770
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    about the integral in the title, the substitution $z=x^2$ should work and give the answer $\frac{\log \left(a/y\right)}{a^2-y^2}$ – Giulio R Jun 18 '21 at 15:19

2 Answers2

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Note that

$$I’(a)=\int_{0}^{\infty}\frac{xdx }{\left(1+x^2\right)\left(1+a^2x^2 \right)} \overset{t=x^2}= \frac12\int_{0}^{\infty}\frac{dt}{\left(1+t\right)\left(1+a^2 t\right)} = \frac{\ln a}{a^2-1} $$ Then

\begin{align} I (a)=&\int_0^a I’(s) ds= \int_0^a \frac{\ln s}{s^2-1} ds \overset{s=ay}=\int_0^1 \frac{a\ln a}{a^2y^2-1}dy +\int_0^1 \frac{a\ln y}{a^2y^2-1}dy\\ =& \frac12 \ln a\ln\frac{1-a}{1+a} +\frac12 (\text{Li}_2(a)- \text{Li}_2(-a)) \end{align}

Quanto
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  • thank you for your answer. Can you please expand how did you get $$\frac{1}{2}\int_{0}^{\infty}\frac{dt}{\left(1+t\right)\left(1+a^2 t\right)}=\frac{\ln a}{a^2-1}$$? I don´t see it. – Ricardo770 Jun 18 '21 at 16:09
  • @Ricardo770 - Decompose $$\frac{1}{\left(1+t\right)\left(1+a^2 t\right)}=\frac1{a^2-1}\left( \frac{a^2}{1+a^2t}- \frac1{1+t}\right)$$ and integrate the two terms – Quanto Jun 18 '21 at 16:19
  • Thank you I got it $$\frac1{a^2-1} \cdot \frac{1}{2} \lim_{t \rightarrow \infty} \ln\left(\frac{1+a^2t}{1+t}\right)=\frac{\ln a }{a^2-1}+\frac1{a^2-1} \cdot \frac{1}{2}\lim_{t \rightarrow \infty} \ln\left(\frac{a^2+t}{a^2(1+t)}\right)$$ $$\frac1{a^2-1} \cdot \frac{1}{2} \ln\left(\lim_{t \rightarrow \infty}\frac{a^2+t}{a^2(1+t)}\right)$$

    by L´Hopital goes to zero

     – Ricardo770 Jun 18 '21 at 17:13
  • thank you very much! $+1$ – Ricardo770 Jun 18 '21 at 17:53
  • sorry for being annoying but for the Dilog I get. I don´t get the $1/2$ term in front$$Li_2(a)=-\int_0^1\frac{a \ln u}{1-au}du, \tag{1}$$

    $$Li_2(-a)=\int_0^1\frac{a \ln u}{1+au}du , \tag{2}$$

    (1)-(2)

    $$Li_2(a)-Li_2(-a)=-\int_0^1 a \ln u \left( \frac{1}{1-au}+\frac{1}{1+au}\right) du$$

    $$Li_2(a)-Li_2(-a)=-\int_0^1 \frac{a \ln u}{1-a^2u^2} du$$

    $$Li_2(a)-Li_2(-a)=\int_0^1 \frac{a \ln u}{a^2u^2-1} du$$

     – Ricardo770 Jun 18 '21 at 18:53
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    @Ricardo770 - Note $\text{Li}_2(a)- \text{Li}_2(-a)=2 \int_0^1 \frac{a\ln y}{a^2y^2-1}dy$, or $\frac1{1-ay}+\frac1{1+ay}= \frac2{1-a^2y^2}$ – Quanto Jun 18 '21 at 19:04
  • Yeah, my mistake. Thank you. – Ricardo770 Jun 19 '21 at 21:38
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I've left some ideas below. However I'm having difficulty confirming them numerically.

I'll consider the integral in your title, namely $$F(x,y)=\int_0^\infty \frac{t}{(1+x^2t^2)(1+y^2t^2)}\mathrm{d}t$$ What we do first is use a substitution $s=t^2\implies \mathrm ds/2=t~\mathrm dt$. So, $$F(x,y)=\frac{1}{2}\int_0^\infty\frac{\mathrm ds}{(1+x^2s)(1+y^2s)}$$ Though it's tempting to use partial fractions to split the integral up here, it fails, since the individual pieces will fail to converge. So we need to get a little more creative. What we do is expand the denominator, then complete the square. $$(1+x^2s)(1+y^2s)=x^2y^2\left(\frac{1}{x^2y^2}+\frac{x^2+y^2}{x^2y^2}s+s^2\right)=x^2y^2\left[\left(s+\frac{x^2+y^2}{2x^2y^2}\right)^2+\left(\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}\right)\right]$$ Which leads us to make the substitution $$z=s+\frac{x^2+y^2}{2x^2y^2}$$ Hence our integral is now $$F(x,y)=\frac{1}{2x^2y^2}\int\limits_{\frac{x^2+y^2}{2x^2y^2}}^\infty \frac{\mathrm dz}{z^2+\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}$$ Finally now this is a well known integral: $$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\arctan\left(\frac{z}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)~\Bigg|^{z=\infty}_{z=\frac{x^2+y^2}{2x^2y^2}}$$ So $$F(x,y)=\frac{1}{2x^2y^2\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\left[\frac{\pi}{2}-\arctan\left(\frac{(x^2+y^2)/2x^2y^2}{\sqrt{\frac{1}{x^2y^2}-\frac{(x^2+y^2)^2}{4x^4y^4}}}\right)\right]$$

K.defaoite
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