Frullani's Integral

[Tolaso J Kos]

Evaluate the integral: \( \displaystyle \int_0^\infty \frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, dx\).

[Grigorios Kostakos]

Frullani's theorem: Let \(D=\bigl\{{(x,y)\in{\mathbb{R}}^2\;|\; x\geqslant0,\; a\leqslant y \leqslant b}\bigr\}\,, 0<a<b\,,\) and \(f(xy): D\longrightarrow{\mathbb{R}}\) is a function which is continuously differentiable on \(D\). If \(f(xy)\) takes finite values at \(x=0\) and \(x=\infty\) for all \(y\in[a,b]\), denoted as \(f(0)\) and \(f(\infty)\) respectively, then \[\displaystyle \int_{0}^{\infty}{\frac{f(ax)-f(bx)}{x}\, dx}=\bigl({f(0)-f(\infty)}\bigr)\log\tfrac{b}{a}\,.\]

Proof: For the surface integral \[J=\iint_{D}{-f'(x,y)\,dx\,dy}\] we have that
\begin{align*}
J&=\displaystyle \int_{0}^{\infty}{\biggl({\int_{a}^{b}{-f'(x,y)\,dy}}\biggr)\, dx}\\
&=\int_{0}^{\infty}{\Bigl[{-f(xy)}\Bigr]_{y=a}^{y=b}\frac{1}{x}\, dx}\\
&=\int_{0}^{\infty}{\frac{f(ax)-f(bx)}{x}\, dx}
\end{align*} and \begin{align*}
J&=\displaystyle \int_{a}^{b}{\biggl({\int_{0}^{\infty}{-f'(x,y)\,dx}}\biggr)\, dy}\\
&=\int_{a}^{b}{\Bigl[{-f(xy)}\Bigr]_{x=0}^{x=\infty}\frac{1}{y}\, dy}\\
&=\bigl({f(0)-f(\infty)}\bigr)\int_{a}^{b}{\frac{1}{y}\, dy}\\
&=\bigl({f(0)-f(\infty)}\bigr)({\log{b}-\log{a}})\\
&=\bigl({f(0)-f(\infty)}\bigr)\log\tfrac{b}{a}\,.
\end{align*}

Applying the above theorem for the function \(\arctan\) and \(a=1,\,b=\pi\) we have that \begin{align*}
\displaystyle \int_0^\infty \frac{\arctan(\pi x)-\arctan x}{x}\, dx&=-\int_0^\infty \frac{\arctan x-\arctan(\pi x)}{x}\, dx\\
&=-\Bigl({0-\frac{\pi}{2}}\Bigr)\log\tfrac{\pi}{1}\\
&=\frac{\pi}{2}\log\pi\,.
\end{align*}

[Tolaso J Kos]

OK, Grigoris nice solution.
Let us see another two solutions , one without the Lemma and one with Liebniz.

2nd solution: We can transform the given integral into a double integral. We note that:

\begin{align*}
\int_{0}^{\infty }\frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, dx &=\int_{0}^{\infty } \left [ \frac{\tan^{-1}(xy)}{x} \right ]_1^\pi\, dx\\
&= \int_{0}^{\infty }\int_{1}^{\pi}\frac{dy \,dx}{1+x^2y^2}\\
&= \int_{1}^{\pi}\int_{0}^{\infty }\frac{dx\,dy}{1+x^2y^2}\\
&= \int_{1}^{\pi}\left [ \frac{\tan^{-1}(xy)}{y} \right ]_0^\infty\, dy\\
&= \int_{1}^{\pi}\frac{\pi}{2y}\, dy=\frac{\pi\ln\pi}{2}
\end{align*}

which gives the desired result.

3rd solution: We define \( \displaystyle I(a)=\int_{0}^{\infty }\frac{\tan^{-1}(ax)-\tan^{-1}x}{x}\, dx \) and we want to evaluate \( I(\pi) \). We differentiate with respect to \(a \) thus we get:

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

It is easy to notice that \(I(1)=0 \) . Thus: \( \displaystyle I(\pi)=I(1)+\int_{1}^{\pi}I'(a)\, da=\int_{1}^{\pi}\frac{\pi}{2a}\, da=\frac{\pi\ln\pi}{2} \).