Evaluate the integral $\int_{0}^{+\infty}\frac{\arctan \pi x-\arctan x}{x}dx$

Compute improper integral : $\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx$.

Solution 1:

$$I(a)=\int\limits_{0}^{+\infty}\dfrac{\arctan (a x)-\arctan x}{x}dx$$

$$I'(a) = \int^{\infty}_0 \frac{1}{1+a^2x^2}\, dx$$

$$I'(a) =\frac{\pi}{2 a}\, $$

Hence by integrating

$$I(a) =\frac{\pi}{2} \log(a) +C $$

by $a=1$ we have $C=0$

$$I(a) = \frac{\pi}{2} \log(a) $$

$$I(\pi)=\int\limits_{0}^{+\infty}\dfrac{\arctan (\pi x)-\arctan x}{x}dx = \frac{\pi}{2} \log(\pi) $$

Solution 2:

You can write $$I=\int_0^\infty dx\int_1^\pi dy\ \frac{1}{1+x^2y^2}.$$ Change the order of integration: $$I=\int_1^\pi dy\int_0^\infty dx\ \frac{1}{1+x^2y^2}=\int_1^\pi dy\ \frac{\pi}{2y}=\frac{\pi}{2}\ln\pi.$$ If memory serves, this is an old Putnam problem.

Addendum:

Eric Auld correctly points out that the change in the order of integration should be justified. By thm. 6.3 of these notes, we can change the order of integration if $$\int_0^\infty dx\ \frac{1}{1+x^2y^2}$$ converges uniformly for $y\in[1,\pi]$. That is, we want to show that for any $\epsilon>0$ there exists $x_0\in[0,\infty)$ with $$\epsilon>\left|\int_0^\infty dx\ \frac{1}{1+x^2y^2}-\int_0^\xi dx\ \frac{1}{1+x^2y^2}\right|=\left|\int_\xi^\infty dx\ \frac{1}{1+x^2y^2}\right|$$ ("convergence") for all $\xi\ge x_0$ and $y\in[1,\pi]$ ("uniform"). In other words, the unbounded region of the integral can be made arbitrarily small for all values of $y$ simultaneously.

In this case, $$\int_\xi^\infty dx\ \frac{1}{1+x^2y^2}\le\int_\xi^\infty dx\ \frac{1}{1+x^2}=\frac{\pi}{2}-\tan^{-1}\xi$$ so if we choose $$x_0>\tan\left(\frac{\pi}{2}-\epsilon\right)$$ then we are done.

Solution 3:

We have $$\int_a^b \dfrac{\arctan \pi x-\arctan x}{x}dx=\int_{\pi a}^{\pi b}\dfrac{\arctan x}{x}dx-\int_{ a}^{ b}\dfrac{\arctan x}{x}dx\\=\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx-\int_{ a}^{ \pi a}\dfrac{\arctan x}{x}dx$$ and since the function $\arctan$ is increasing so $$\arctan( b)\log\pi=\arctan( b)\int_b^{\pi b}\frac{dx}{x}\leq\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx\\ \leq\arctan(\pi b)\int_b^{\pi b}\frac{dx}{x}=\arctan(\pi b)\log\pi$$ so if $b\to\infty$ we have $$\int_{ b}^{\pi b}\dfrac{\arctan x}{x}dx\to\frac{1}{2}\pi\log\pi$$ and by a similar method we prove that $$\int_{ a}^{ \pi a}\dfrac{\arctan x}{x}dx\to 0,\quad a\to0$$ hence we conclude $$\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx=\frac{1}{2}\pi\log\pi$$

Solution 4:

With \begin{eqnarray*} \lim_{x \to \infty}\left\lbrack\arctan\left(\pi x\right) - \arctan\left(x\right)\right\rbrack \ln\left(x\right) & = & \lim_{x \to \infty} \left\lbrack\arctan\left(1 \over x\right) - \arctan\left(1 \over \pi x\right)\right\rbrack \ln\left(x\right) \\ & = & \left(1 - {1 \over \pi}\right)\lim_{x \to \infty}{\ln\left(x\right) \over x} \\ & = & \left(1 - {1 \over \pi}\right)\lim_{x \to \infty}{1/x \over 1} = 0 \end{eqnarray*}

and \begin{eqnarray*} \lim_{x \to 0}\left\lbrack\arctan\left(\pi x\right) - \arctan\left(x\right)\right\rbrack \ln\left(x\right) & = & \left(\pi - 1\right)\lim_{x \to 0} \left\lbrack x\ln\left(x\right)\right\rbrack \\ & = & \left(\pi - 1\right)\lim_{x \to 0}{\ln\left(x\right) \over 1/x} \\ & = & \left(\pi - 1\right)\lim_{x \to 0}{1/x \over -1/x^{2}} = 0 \end{eqnarray*}

we have

\begin{eqnarray*}\color{#66f}{\large\int_{0}^{\infty}% {\arctan\left(\pi x\right) - \arctan\left(x\right) \over x}\,{\rm d}x} & = & -\int_{0}^{\infty}\ln\left(x\right)\left\lbrack {\pi \over \left(\pi x\right)^{2} + 1} - {1 \over x^{2} + 1} \right\rbrack\,{\rm d}x \\ & = & -\int_{0}^{\infty}\left\lbrack\ln\left(x \over \pi\right) - \ln\left(x\right)\right\rbrack {1 \over x^{2} + 1}\,{\rm d}x \\ & = & \ln\left(\pi\right) \underbrace{\quad\int_{0}^{\infty} {1 \over x^{2} + 1}\,{\rm d}x\quad}_{=\ \pi/2} = \color{#66f}{\large{\pi \over 2}\,\ln\left(\pi\right)} \approx {\tt 1.7981} \end{eqnarray*}