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I want to calculate the integral which is dependent on the parameters $a$ and $b$ ($a,b \gt0)$.

$\int_{0}^{\infty} \frac {b \sin (ax) - a \sin (bx)}{x^2}dx$

This integral is solved in my textbook but I don't understand how and why like this:

$ab \int_{0}^{\infty} \frac {b \sin (ax- a \sin (bx)}{abx^2}dx=ab \int_{0}^{\infty} \frac {1}{x} (\frac {\sin (ax)}{ax} - \frac { \sin (bx)}{bx})dx = ab \ln (\frac {b}{a})$

$f(x) = \frac {1}{x}$, ${}$ $\lim_{\to \infty} \frac {sinx}{x}=0$, ${}$ $\lim_{\to 0^+} \frac {sinx}{x}=1$

I just can't see it how they calculated it. Thanks.

Leif
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    check this: https://math.stackexchange.com/questions/61828/proof-of-frullanis-theorem – farruhota Jan 19 '18 at 04:36
  • Well, I usually differentiate these types of integrals with the respect of the parameter, just I don't see it there. I have calculated many intgral like this, my problem is maybe more technical. It didn't help me to simplify it. – Leif Jan 19 '18 at 04:38

1 Answers1

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Using Frullani's integral formula, we see that \begin{align} \int^\infty_0 \frac{f(ax)-f(bx)}{x}\ dx = [f(\infty)-f(0)]\ln \frac{a}{b}. \end{align}

Jacky Chong
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