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I was struggling with the following integral: $$\displaystyle I=\int\limits_{0}^{+\infty}\dfrac{\arctan \pi x-\arctan x}{x}dx$$

And I found a question about it on this site where the answer said that we can rewrite the above integral as follows:

$$I=\int_0^\infty dx\int_1^\pi dy\ \frac{1}{1+x^2y^2}$$

But I have no clue why this is true, could anybody enlighten me?

3 Answers 3

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You can directly compute $$ \int_{1}^\pi\frac{1}{1+x^2y^2}\,dy=\frac{\arctan(\pi x)-\arctan x}{x} $$ Just substitute $u=xy$.

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$$\int_{0}^{+\infty}\frac{\arctan(ax)-\arctan(x)}{x}\,dx = \int_{0}^{+\infty}\frac{1}{x}\int_{x}^{ax}\frac{1}{1+t^2}\,dt\,dx $$ and by substituting $t=xu$, $dt=x\,du$ in the innermost integral we get: $$ \int_{0}^{+\infty}\int_{1}^{a}\frac{1}{1+x^2 u^2}\,du\,dx \stackrel{\text{Fubini}}{=}\int_{1}^{a}\frac{\pi}{2u}\,du=\color{red}{\frac{\pi}{2}\,\log a}.$$ The same can be deduced from the complex version of Frullani's theorem.

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    I love the complex generalization! By the way, have you ever seen my development elsewhere? – 2017-02-04
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    @Dr.MV: I think you should submit your proof of the complex version of Frullani's theorem to Mathworld (and maybe editing/making a Wikipedia page too). They have a little form for submissions hidden somewhere on their site, it would be a great improvement of the page dedicated to Frullani's theorem. Usually your lemma is just re-proved by differentiation under the integral sign or contour integration. At least, as far as I saw. – 2017-02-05
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    Jack, I did use contour integration. I'm not familiar with Mathworld. Whom do I contact? – 2017-02-05
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    @Dr.MV: this page should do the job: http://mathworld.wolfram.com/contact/ – 2017-02-05
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    Truth to be told, I once wrote an addendum for the page dedicated to Poncelet's porism and they never considered me, I wish you a better luck. – 2017-02-05
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    Thank Jack! Much appreciative. – 2017-02-05
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Another way you could do this is by differentiation under the integral $I(\alpha)=\int_{0}^{\infty}\frac{\tan^{-1} \alpha x}{x}\text{d}x\to I'(\alpha)=\int_{0}^{\infty}\frac{1}{1+(\alpha x)^2}\text{d}x=\frac{1}{\alpha}\tan^{-1} (\alpha x)\big\rvert_{0}^{\infty}=\frac{\pi}{2\alpha}.$ integrating it to get back to $I(\alpha)$ we get $\frac{\pi}{2}\ln \alpha+C$ because we have a difference of these 2 functions we know that the constant will cancel out. So using this information we have $\int\frac{\tan^{-1} ax-\tan^{-1} bx}{x}\text{d}x=\frac{\pi}{2}\ln \left (\frac{a}{b}\right )$