How would I show the result below using contour integration? $$\int_{-\infty}^{\infty} \frac{\cos bx - \cos ax}{x^2} dx = \pi (a-b)$$ where a>b>0 using contour integration. Any help would be greatly appreciated, thanks!
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Do you have any idea what contour to use? (There's an obvious one.) – Eric Towers May 28 '20 at 18:35
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Consider the region of a semi-circle in the first two quadrants of infinitely large radius which avoids the singularity at (0,0) by cutting out a semi-circle of infinitesimally small radius centered at (0,0). Once you draw the contour out and label the regions, use Cauchy's first theorem. If you still struggle, refer to @metamorphy 's post or say what you're struggling with. – Ty. May 28 '20 at 18:36
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I see, so it is a Cauchy principal value problem effectively? Thanks for your help with that! – May 29 '20 at 00:43
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See Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$? on performing contour integration to get $\int_0^\infty \frac{\sin x}x=\frac\pi2$. Then
\begin{align} & \int_{-\infty}^{\infty} \frac{\cos bx - \cos ax}{x^2} dx \overset{IBP} =2\int_0^\infty \frac{a\sin ax - b\sin bx}x=\pi (a-b) \end{align}
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