Prove $$\int^\infty_0 b\sin(\frac{1}{bx})-a\sin(\frac{1}{ax}) = -\ln(\frac{b}{a})$$

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

So I need to get the first equation into the form of the Frullani integral. I can't figure out how to make this transformation though because I'm no good at them.

  • $\begingroup$ Wait, you cannot find $f$ such that $b\sin(1/bx)-a\sin(1/ax)=(f(bx)-f(ax))/x$ for every $x$? $\endgroup$ – Did Feb 2 '15 at 10:45
  • $\begingroup$ @Did Exactly, am I missing something obvious? $\endgroup$ – chriskinda Feb 2 '15 at 10:47
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    $\begingroup$ YES! $ $ $ $ $ $ $\endgroup$ – Did Feb 2 '15 at 10:47
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    $\begingroup$ $$\color{red}{bx}\sin(1/\color{red}{bx})-\color{blue}{ax}\sin(1/\color{blue}{ax})=f(\color{red}{bx})-f(\color{blue}{ax})$$ $\endgroup$ – Did Feb 2 '15 at 10:53
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    $\begingroup$ @Did You should really write down an answer with that. $\endgroup$ – Timbuc Feb 2 '15 at 11:08

It suffices to take $ f(x) =\frac{\sin x}{x}$ and applies Frullani's theorem thereafter.

Setting, $u =\frac1x$ then $dx=-\frac{du}{u^2}$

$$\int^\infty_0 b\sin(\frac{1}{bx})-a\sin(\frac{a}{ax})dx =\int^\infty_0 \frac{b\sin(\frac{x}{b})-a\sin(\frac{x}{a})}{x^2}dx \\=\int^\infty_0 \frac{\frac bx\sin(\frac{x}{b})-\frac ax\sin(\frac{x}{a})}{x}dx = \frac{f(\frac{x}{b})- f(\frac{x}{a})}{x}dx =f(0)\ln\left(\frac{\frac1b}{\frac1a}\right).$$

Where, $$ f(x) =\frac{\sin x}{x}\to 1 ~~as ~~x\to 0$$ satisfies conditions of Frullani's Theorem.

$$\color{red}{\int^\infty_0 b\sin(\frac{1}{bx})-a\sin(\frac{a}{ax})dx =\ln\left(\frac{\frac1b}{\frac1a}\right) =\ln\left(\frac{a}{b}\right) }.$$


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