Evaluate the integral:
$$\int_0^\infty \frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, {\rm d}x$$
Solution
1st solution
We transform the integral in a double one, hence:
$$\begin{aligned} \int_{0}^{\infty }\frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, {\rm d}x &=\int_{0}^{\infty } \left [ \frac{\tan^{-1}(xy)}{x} \right ]_1^\pi\, {\rm }x\\ &= \int_{0}^{\infty }\int_{1}^{\pi}\frac{{\rm d}y \,{\rm d}x}{1+x^2y^2}\\ &= \int_{1}^{\pi}\int_{0}^{\infty }\frac{{\rm d}x\,{\rm d}y}{1+x^2y^2}\\ &= \int_{1}^{\pi}\left [ \frac{\tan^{-1}(xy)}{y} \right ]_0^\infty\, {\rm d}y\\ &= \int_{1}^{\pi}\frac{\pi}{2y}\, {\rm d}y=\frac{\pi\ln\pi}{2} \end{aligned}$$
2nd solution
We are invoking Frullani's Lemma:
Hence:
$$\begin{align*} \displaystyle \int_0^\infty \frac{\arctan(\pi x)-\arctan x}{x}\, dx&=-\int_0^\infty \frac{\arctan x-\arctan(\pi x)}{x}\, dx\\ &=-\Bigl({0-\frac{\pi}{2}}\Bigr)\log\tfrac{\pi}{1}\\ &=\frac{\pi}{2}\log\pi\ \end{align*}$$
$$\int_0^\infty \frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, {\rm d}x$$
Solution
1st solution
We transform the integral in a double one, hence:
$$\begin{aligned} \int_{0}^{\infty }\frac{\tan^{-1}(\pi x)-\tan^{-1} x}{x}\, {\rm d}x &=\int_{0}^{\infty } \left [ \frac{\tan^{-1}(xy)}{x} \right ]_1^\pi\, {\rm }x\\ &= \int_{0}^{\infty }\int_{1}^{\pi}\frac{{\rm d}y \,{\rm d}x}{1+x^2y^2}\\ &= \int_{1}^{\pi}\int_{0}^{\infty }\frac{{\rm d}x\,{\rm d}y}{1+x^2y^2}\\ &= \int_{1}^{\pi}\left [ \frac{\tan^{-1}(xy)}{y} \right ]_0^\infty\, {\rm d}y\\ &= \int_{1}^{\pi}\frac{\pi}{2y}\, {\rm d}y=\frac{\pi\ln\pi}{2} \end{aligned}$$
2nd solution
We are invoking Frullani's Lemma:
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 \[\displaystyle \int_{0}^{\infty}{\frac{f(ax)-f(bx)}{x}\, dx}=\bigl({f(0)-f(\infty)}\bigr)\log\tfrac{b}{a}\,.\]
Hence:
$$\begin{align*} \displaystyle \int_0^\infty \frac{\arctan(\pi x)-\arctan x}{x}\, dx&=-\int_0^\infty \frac{\arctan x-\arctan(\pi x)}{x}\, dx\\ &=-\Bigl({0-\frac{\pi}{2}}\Bigr)\log\tfrac{\pi}{1}\\ &=\frac{\pi}{2}\log\pi\ \end{align*}$$
Proof of Frullani's Theorem
ReplyDeleteFor 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*}