0

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?

YakSal Tafri
  • 763

3 Answers3

2

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

egreg
  • 238,574
1

$$\int_{0}^{+\infty}\frac{\arctan(ax)-\arctan(x)}{x}\,dx = \int_{0}^{+\infty}\frac{1}{x}\int_{x}^{ax}\frac{1}{1+t^2}\,dt\,dx $$ 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.

Community
  • 1
Jack D'Aurizio
  • 353,855
  • 1
    I love the complex generalization! By the way, have you ever seen my development elsewhere? – Mark Viola Feb 04 '17 at 23:59
  • @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. – Jack D'Aurizio Feb 05 '17 at 00:04
  • Jack, I did use contour integration. I'm not familiar with Mathworld. Whom do I contact? – Mark Viola Feb 05 '17 at 00:06
  • @Dr.MV: this page should do the job: http://mathworld.wolfram.com/contact/ – Jack D'Aurizio Feb 05 '17 at 00:10
  • 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. – Jack D'Aurizio Feb 05 '17 at 00:12
  • 1
    Thank Jack! Much appreciative. – Mark Viola Feb 05 '17 at 00:37
0

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 $\int\frac{\tan^{-1} ax-\tan^{-1} bx}{x}\text{d}x=\frac{\pi}{2}\ln \left (\frac{a}{b}\right )$

Teh Rod
  • 3,108