I would like to evaluate the integral $$\int_0^\infty\frac{\cos(ax)-\cos(bx)}{x}\mathrm dx$$ I think this could be done with complicated methods in singularity or other kinds of methods.
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4Please show how have you tried – zed111 Mar 17 '15 at 07:44
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1using Laplace transforms: $\log(|b/a|)$ – Math-fun Mar 17 '15 at 18:30
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A related question. – Lucian Mar 18 '15 at 00:51
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A variation of a Frullani’s integral
\begin{align}\int_0^\infty& \frac{\cos ax-\cos bx }{x}dx =\int_0^\infty\frac1x \ d\left(\frac{\sin ax}a- \frac{\sin bx}b \right)\\& \overset{ibp}=\int_0^\infty\frac{f(ax)-f(bx)}xdx= \ln\frac ba,\>\>\>\>\>\>\> f(x)=\frac{\sin x}x \end{align}
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Hint: Either use the formula for $\cos A\pm\cos B$, which will ultimately result in an expression
similar to that of Dirichlet's integral, or employ a trick similar to Frullani's integral, namely by
noticing that $~\dfrac{\cos ax-\cos bx}x~=~\displaystyle\int_a^b\sin(tx)~dt,~$ and then switch the order of integration
via Fubini's theorem.
Lucian
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1@RonGordon: Just because it doesn't converge, doesn't mean that it cannot be evaluated... ;-$)$ – Lucian Mar 17 '15 at 13:55
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5Yes, but it does mean you need to reconsider quoting Fubini's theorem. – Ron Gordon Mar 17 '15 at 14:06
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1@RonGordon: Or we could simply use it to justify the switch for a non-imaginary exponential integrand, and then mischievously and underhandedly generalize the real result for complex values of the exponent, via the Hand-Waving Theorem. – Lucian Mar 17 '15 at 14:15
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Using Laplace transforms,
$$\begin{align}\int_0^\infty\frac{\cos ax-\cos bx}x\mathrm dx&=\int_0^\infty\mathcal L(1)(x)\cdot(\cos ax-\cos bx)\mathrm dx\\&=\int_0^\infty\mathcal L(\cos ax-\cos bx)(x)\mathrm dx\\&=\int_0^\infty\frac{x}{x^2+a^2}-\frac{x}{x^2+b^2}\mathrm dx\\&=\frac12\ln\left(\frac{x^2+a^2}{x^2+b^2}\right)\Bigg|_0^\infty\\&=\ln\left|\frac{b}a\right|\end{align}$$
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