Improper integral

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January 4th 2012, 11:04 AM
Also sprach Zarathustra

Re: Improper integral

We know,

$\int^{\infty}_{0}\frac{\sin{yx}}{x}dx=\frac{\pi}{2 } (y>0)$

That is uniformly converges relatively to $y$ for $y\geqslant y_0>0$ , so:

$\int^{\infty}_{0}\frac{dx}{x}\int_a^b\sin{yx}\;{dy }=\int_0^{\infty} \frac{\cos{ax}-\cos{bx}}{x^2}dx=\frac{\pi}{2}(b-a)$

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