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Prove $$\int^\infty_0 \frac{\frac{1}{1+(bx)^2}-\frac{1}{1+(ax)^2}}{x}dx = \ln(\frac{a}{b})$$

I'm supposed to use Frulanni integrals and use the fact that $\int^\infty_0 \frac{f(bx)-f(ax)}{x}dx$ since this equals $[f(\infty)-f(0)] \ln(\frac{b}{a})$

Unfortunately, I can't figure out how to do such a transformation. Any help would be appreciated!

UPDATE: Here's what I tried.

Let $f(t)=\frac{1}{1+t^2}$ and $b^2=\frac{1}{d}$ and $a^2=\frac{1}{c}$

Then we have $$\int^\infty_0 \frac{f(td)-f(tc)}{x}=[f(\infty)-f(0)]\ln(\frac{d}{c})=(1-1)\ln(\frac{d}{c})=0.$$ but I don't think we're allowed to get 0 so I must have done something wrong.

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    $\begingroup$ Just use $f(t)=1/(1+t^2)$? It is already in the correct form. $\endgroup$ – mickep Feb 2 '15 at 8:53
  • $\begingroup$ I'll look into this. $\endgroup$ – chriskinda Feb 2 '15 at 8:54
  • $\begingroup$ After the update, the question becomes even more confused. What are $c$ and $d$ ? If you already known that $\int (f(ax)-f(bx))/x \;\mathrm dx=[f(\infty)-f(0)]\ln\frac ab$, it seems you have just nothing to do. $\endgroup$ – Tom-Tom Feb 2 '15 at 9:15
  • $\begingroup$ @Tom-Tom We have that function equals $ln(\frac{b}{a})$ not $ln(\frac{a}{b})$ as desired. $\endgroup$ – chriskinda Feb 2 '15 at 9:21
  • $\begingroup$ See Frullani's integral and Fubini's theorem. $\endgroup$ – Lucian Feb 2 '15 at 9:23
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So, let $f(t)=1/(1+t^2)$. Then, your integral can be written $$ \int_0^{+\infty}\frac{f(bx)-f(ax)}{x}\,dx $$ and, as you point out, it is a Frullani integral, with value $$ (f(+\infty)-f(0))\ln(b/a) $$ Now $$ \lim_{t\to+\infty}f(t)=0 $$ and $$ f(0)=1. $$ Hence, the value of your integral is $$ -\ln(b/a)=-(\ln b-\ln a)=\ln a-\ln b=\ln(a/b) $$ (I made that calculation in detail since I think that was your problem) as desired.

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HINT: $$\int^\infty_0 \frac{\frac{1}{1+(bx)^2}-\frac{1}{1+(ax)^2}}{x}dx =\int^\infty_0 {\frac{1}{x(1+(bx)^2)}dx-\int_0^{\infty}\frac{1}{x(1+(ax)^2)}}dx $$ and also $$\int\dfrac{1}{x(1+kx^2)}dx=\int\dfrac{1}{x}dx-\int\dfrac{kx}{1+kx^2}dx=\ln x-\dfrac{1}{2}\ln (1+kx^2)+c.$$

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  • $\begingroup$ Interesting, but we're trying to use the Frullani integrals $\endgroup$ – chriskinda Feb 2 '15 at 9:26

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