2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ YES! $ $ $ $ $ $ $\endgroup$ – Did Feb 2 '15 at 10:47
  • 4
    $\begingroup$ $$\color{red}{bx}\sin(1/\color{red}{bx})-\color{blue}{ax}\sin(1/\color{blue}{ax})=f(\color{red}{bx})-f(\color{blue}{ax})$$ $\endgroup$ – Did Feb 2 '15 at 10:53
  • 1
    $\begingroup$ @Did You should really write down an answer with that. $\endgroup$ – Timbuc Feb 2 '15 at 11:08
1
$\begingroup$

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.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.