Interesting integral 5.1

Question

For $a,b>0$ find an integral: $$ \int\limits_0^{\infty}\frac{\arctan ax \cdot \arctan bx}{x^2} d x $$ Looks like we may use Frullani's Integral here. But how?

Answer 1

Take note: The function is completely symmetric in $a,b$

Differentiate the function twice, once w.r.t. $a$, once to w.r.t. $b$.

Then you can easily integrate w.r.t. $x$ using partial fractions.

Observe $f(0,b) = f(a,0) = 0$

Thus, you integrate both sides from $0$ to $a$ and $0$ to $b$ to obtain the original function.

No Frullani.

I leave the details as an exercise to the reader.

Answer 2

On Jack Lam's path:

Define on $[0;+\infty[\times [0;+\infty[$,

$\displaystyle F(a,b)=\int\limits_0^{\infty}\frac{\arctan(ax)\arctan (bx)}{x^2} d x$

Observe that for $(a,b)\in [0;+\infty[\times [0;+\infty[$,

$F(0,b)=F(a,0)=0$.

If $a>b>0$,

$\displaystyle \left(\dfrac{\partial}{\partial u}F(u,v)\right)(a,b)=\int_0^{\infty}\dfrac{\arctan(bx)}{x(1+a^2x^2)}dx$

Observe that for $a\in [0;+\infty[$,

$\displaystyle \left(\dfrac{\partial}{\partial u}F(u,v)\right)(a,0)=0$

$\begin{align} \left(\dfrac{\partial^2}{\partial u\partial v}F(u,v)\right)(a,b)&=\int_0^{\infty}\dfrac{1}{(1+a^2x^2)(1+b^2x^2)}dx\\& =\dfrac{1}{b^2-a^2}\Big[b\arctan(bx)-a\arctan(ax)\Big]_0^{\infty}\\ &=\dfrac{\pi}{2(a+b)} \end{align}$

Therefore,

$\begin{align} \left(\dfrac{\partial}{\partial u}F(u,v)\right)(a,b)-\left(\dfrac{\partial}{\partial u}F(u,v)\right)(a,0)&=\dfrac{\pi}{2}\int_0^b\dfrac{1}{a+u}du\\ &=\dfrac{\pi}{2}\ln\left(1+\dfrac{b}{a}\right) \end{align}$

Therefore,

$\displaystyle \left(\dfrac{\partial}{\partial u}F(u,v)\right)(a,b)=\dfrac{\pi}{2}\ln\left(1+\dfrac{b}{a}\right)$

Therefore,

$\begin{align} F(a,b)-F(0,b)&=\dfrac{\pi}{2}\int_0^a \ln\left(1+\dfrac{b}{u}\right)du\\ &=\dfrac{\pi}{2}\Big[u\ln(u+b)+b\ln(u+b)-u\ln u\Big]_0^a\\ &=\dfrac{\pi}{2}\Big((a+b)\ln(a+b)-a\ln a-b\ln b\Big) \end{align}$

Therefore,

For $a,b>0$ and $a\neq b$,

$\boxed{\displaystyle F(a,b)=\dfrac{\pi}{2}\Big((a+b)\ln(a+b)-a\ln a-b\ln b\Big)}$

Case $a=b>0$.

$\displaystyle F(a,a)=\int\limits_0^{\infty}\frac{(\arctan(ax))^2}{x^2} dx$

Perform the change of variable $y=ax$,

$\begin{align}F(a,a)&=a\int_0^{\infty} \dfrac{(\arctan x)^2}{x^2}dx\\ &=-a\Big[\dfrac{(\arctan x)^2}{x}\Big]_0^{\infty}+2a\int_0^{\infty} \dfrac{\arctan x}{x(1+x^2)}dx\\ &=2a\int_0^{\infty} \dfrac{\arctan x}{x(1+x^2)}dx\\ &=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+2a\int_1^{\infty} \dfrac{\arctan x}{x(1+x^2)}dx \end{align}$

Perform the change of variable $y=\dfrac{1}{x}$ in the latter integral,

$\begin{align}F(a,a)&=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+2a\int_0^{1}\dfrac{x\arctan\left(\tfrac{1}{x}\right)}{1+x^2}dx\\ &=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+2a\int_0^{1}\dfrac{x\left(\tfrac{\pi}{2}-\arctan x\right)}{1+x^2}dx\\ &=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+\dfrac{\pi}{2}a\int_0^1 \dfrac{2x}{1+x^2}dx-2a\int_0^{1}\dfrac{x\arctan x}{1+x^2}dx\\ &=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+\dfrac{\pi}{2}a\Big[\ln(1+x^2)\Big]_0^1-a\Big[\ln(1+x^2)\arctan x\Big]_0^1+\\ &a\int_0^1\dfrac{\ln(1+x^2)}{1+x^2}dx\\ &=2a\int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx+\dfrac{\pi}{4}a\ln 2+a\int_0^1\dfrac{\ln(1+x^2)}{1+x^2}dx\\ \end{align}$

Since,

It's well-known that,

$\displaystyle \int_0^1\dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{1}{2}\pi\ln 2-G$

$G$ being the Catalan constant.

$\begin{align} \int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx&=\int_0^1\int_0^1 \dfrac{1}{(1+x^2)(1+x^2y^2)}dxdy\\ &=\int_0^1 \Big[\dfrac{x\arctan(xy)}{y^2-1}-\dfrac{\arctan(x)}{y^2-1}\Big]_{x=0}^{x=1}dy\\ &=\int_0^1 \dfrac{4y\arctan y-\pi}{4(y^2-1)}dy\\ &=\dfrac{\pi}{8}\int_0^1 \dfrac{1}{(1+y)}dy+\dfrac{1}{2}\int_0^1 \dfrac{\arctan y}{1+y}dy+\int_0^1 \dfrac{4\arctan y-\pi}{8(y-1)}dy\\ &=\dfrac{\pi}{8}\ln 2+\dfrac{1}{2}\Big[\ln(1+x)\arctan x\Big]_0^1-\dfrac{1}{2}\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx+\\ &\dfrac{1}{2}\int_0^1\dfrac{\arctan\left(\tfrac{1-x}{x+1}\right)}{1-x}dx\\ &=\dfrac{\pi}{4}\ln 2-\dfrac{1}{2}\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx+\dfrac{1}{2}\int_0^1\dfrac{\arctan\left(\tfrac{1-x}{x+1}\right)}{1-x}dx \end{align}$

Perform the change of variable $y=\dfrac{1-x}{1+x}$ in the latter integral,

$\begin{align} \int_0^{1} \dfrac{\arctan x}{x(1+x^2)}dx&=\dfrac{\pi}{4}\ln 2-\dfrac{1}{2}\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx+\dfrac{1}{2}\int_0^{1} \dfrac{\arctan x}{x(1+x)}dx\\ &=\dfrac{\pi}{4}\ln 2-\dfrac{1}{2}\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx+\dfrac{1}{2}\int_0^1 \dfrac{\arctan x}{x}dx-\dfrac{1}{2}\int_0^1\dfrac{\arctan x}{1+x}dx\\ &=\dfrac{\pi}{4}\ln 2-\dfrac{1}{2}\int_0^1 \dfrac{\ln(1+x)}{1+x^2}dx+\dfrac{1}{2}\int_0^1 \dfrac{\arctan x}{x}dx-\\ &\dfrac{1}{2}\Big[\ln(1+x)\arctan x\Big]_0^1+\dfrac{1}{2}\int_0^1\dfrac{\ln(1+x)}{1+x^2}dx\\ &=\dfrac{\pi}{8}\ln 2+\dfrac{1}{2}G \end{align}$

Therefore,

$a>0$,

$\boxed{\displaystyle F(a,a)=a\pi\ln 2}$

Therefore,

For $a,b>0$,

$\boxed{\displaystyle F(a,b)=\dfrac{\pi}{2}\Big((a+b)\ln(a+b)-a\ln a-b\ln b\Big)}$