Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem

Question

I have to solve the following double integral

$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$

with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.

$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} = \int_0^\infty\frac{1}{1+x^2}\int_0^1\frac{dy}{1+x^2 y^2} \, dx. $$

I've solved the internal integral by substitution, remembering that $\int\frac{du}{1+u^2}=\arctan u$ Substitution:

$$x^2 y^2= u^2 \to y=\frac{1}{x}u \to dy=\frac{1}{x}du.$$ $$y=0 \to u=0, \qquad y=1 →u=x$$

So:

\begin{align} & \int_0^∞\frac{1}{1+x^2} \left( \int_0^x\frac{1}{x}\frac{1}{1+u^2}\,du \right) \, dx \\[8pt] = {} & \int_0^\infty\frac{1}{x}\frac{1}{1+x^2}[\arctan u] \, dx \\[8pt] = {} & \int_0^\infty\frac{\arctan x}{x(1+x^2)} \, dx. \end{align}

Now I have to solve this last integral with the Fubini's theorem but I don't know how to do it.

Answer 1

Solution 1. Let $I$ denote the double integral. Then

\begin{align*} I &= \int_0^1 \int_0^\infty \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}x\,\mathrm{d}y \\ &= \int_0^1 \int_0^\infty \frac{1}{1-y^2} \left( \frac{1}{1+x^2} - \frac{y^2}{1+x^2y^2} \right) \, \mathrm{d}x\,\mathrm{d}y \\ &= \int_{0}^{1} \frac{1}{1-y^2} \left( \frac{\pi}{2} - \frac{\pi y}{2} \right) \, \mathrm{d}y \\ &= \frac{\pi}{2} \int_0^1 \frac{1}{1+y} \, \mathrm{d}y \\ &= \frac{\pi}{2} \log 2. \end{align*}

Solution 2.

\begin{align*} I = \int_0^\infty \int_0^1 \frac{1}{(1+x^2)(1+x^2 y^2)} \, \mathrm{d}y \, \mathrm{d}x = \int_0^\infty \frac{\arctan x}{(1+x^2)x} \, \mathrm{d}x. \end{align*}

Substituting $x = \tan\theta$, then

\begin{align*} \require{cancel} I &= \int_0^{\frac{\pi}{2}} \frac{\theta}{\tan\theta} \, \mathrm{d}\theta = \cancel{\left[ \theta \log \sin\theta \right]_0^{\frac{\pi}{2}}} - \int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta. \end{align*}

Regarding the last integral, we observe that the following holds:

$$ I = -\int_0^{\frac{\pi}{2}} \log \sin\theta \, \mathrm{d}\theta = -\int_0^{\frac{\pi}{2}} \log \cos\theta \, \mathrm{d}\theta = -\int_0^{\frac{\pi}{2}} \log \sin(2\theta) \, \mathrm{d}\theta. $$

Using this, we get

$$ 2I = -\int_0^{\frac{\pi}{2}} \log (\sin\theta\cos\theta) \, \mathrm{d}\theta = -\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin2\theta}{2}\right) \, \mathrm{d}\theta = I + \frac{\pi}{2}\log 2. $$

Therefore

$$ I = \frac{\pi}{2}\log 2. $$

Answer 2

Following Michael Hardy's suggestion, I will use partial fractions. We will have

$$\frac{1}{(1+x^2)(1+x^2y^2)}=\frac{1}{1-y^2}\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)$$

So

$$ \begin{eqnarray} \iint_{A=[0,\infty[\times[0,1]}\frac{dx\,dy}{(1+x^2)(1+x^2y^2)}&=&\int_0^1\frac{1}{1-y^2}\left[\int_0^\infty\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)dx\right]dy \\ &=&\int_0^1\frac{1}{1-y^2}\left[\arctan x|_0^\infty-y\arctan (xy)|_0^\infty\right]dy \\ &=&\int_0^1\frac{1}{1-y^2}\left[\frac{\pi}{2}-y\frac{\pi}{2}\right]dy \\ &=&\frac{\pi}{2}\int_0^1\frac{dy}{1+y}=\left.\frac{\pi}{2}\ln|1+y|\right|_0^1 \\ &=&\frac{\pi}{2}\ln 2 \end{eqnarray} $$

Answer 3

\begin{align} & \iint\limits_{[0,+\infty) \times [0,1]} \frac{d(x,y)}{(1+x^2)(1+x^2 y^2)} \\[8pt] = {} & \int_0^1 \left( \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \right) \, dy \end{align}

So we need partial fractions: \begin{align} & \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {Ax+B} {1+x^2} + \frac{Cx+D}{1+x^2y^2} \\[10pt] 1 & = (Ax+B)(1+x^2y^2) + (Cx+D)(1+x^2) \\[8pt] & = (Ay^2+C)x^3 + (By^2+D)x^2 + (A+C)x + (B+D) \\[8pt] \text{So } & Ay^2 + C=0 \\ & By^2+D=0 \\ & A+C=0 \\ & B+D=1 \\[8pt] \text{and so } & A=0, \quad C=0, \quad B=1/(1-y^2), \quad D= y^2/(y^2-1). \\[8pt] & \frac 1 {(1+x^2)(1+x^2 y^2)} = \frac {1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2} \end{align} And so the first integral: \begin{align} & \int_0^\infty \frac{dx}{(1+x^2)(1+x^2y^2)} \\[8pt] = {} & \int_0^\infty \left( \frac{1/(1-y^2)} {1+x^2} + \frac{y^2/(y^2-1)}{1+x^2y^2} \right) \, dx \\[8pt] = {} & \frac \pi {2(1+y)} \end{align} and so on.

Corollary: In view of one of my comments under the question, we conclude that $$ \int_0^{\pi/2} \frac{w}{\tan w} \, dw = \frac \pi 2 \log 2. $$